C. Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis.
What is kinetic energy?
Kinetic energy is the energy possessed by a body due to its motion.
K.E = ¹/₂mv²
where;
m is mass of the objectThe kinetic energy of a body is directly proportional to the mass of the object.
Thus, the correct relationship between kinetic energy and mass is Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis.
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What is grandfather Paradox?
A grandfather paradox is a situation where individual travels to the past and then introduces a change which affects or contradicts the present.
What is a grandfather paradox?A paradox is a situation or statement which involves two contradictions.
A grandfather paradox is a situation which is defined by the ability of an individual to travel to a time in the past usually before the birth of their grandfather and then introduces a change which affects or contradicts the present. For example, killing the grandfather to prevent their birth.
In conclusion, a grandfather paradox is is an event which contradicts the present as a result of a change done to the past.
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At t=0s a small "upward" (positive y) pulse centered at x = 4.0 m is moving to the right on a string with fixed ends at x=0.0m and x = 13.0 m . The wave speed on the string is 3.5 m/s . At what time will the string next have the same appearance that it did at t=0s?
The next time the string will have the same appearance that it did at t=0s is 2.29 s.
Frequency of the wave
v = fλ
f = v/λ
where;
λ is wavelengthhalf of the upward pulse is a quarter of wavelength = ¹/₄ x 4 m = 1 m
f = 3.5/1
f = 3.5 Hz
Time of motion when the pulse is at 4 mt1 = 4/3.5 = 1.143 s
The next time the string will have the same appearance that it did at t=0s.
d = 4 m x 2 = 8 m
t2 = 8/3.5
t2 = 2.29 s
Thus, the next time the string will have the same appearance that it did at t=0s is 2.29 s.
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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet? (Planet-X doesn't have any atmosphere.)
2. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
3. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
Solutions for the three problems are is mathematically given as
v=7338.9349[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex][tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]What is the First Cosmic Speed i.e. the speed of a satellite on a low-lying circular orbit around this planet?(a) First cosmic speed (arbitral velocily)
[tex]v=\sqrt{\frac{G M}{r}}\\v=\sqrt{\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24}}{5870 \times 10^{3}}}[/tex]
v=7338.9349
(b) Second cosmic speed (escape velo.)
$$
\begin{gathered}
[tex]V=\sqrt{\frac{2 G M}{r}}\\\\V=\sqrt{2} \sqrt{\frac{G M}{r}}\\\\=V\sqrt{2} \times 7338.9349 \\[/tex]
[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex]
(c) In conclusion, in a circular orbit, the gravitational force is gets balanced by centripetal force
[tex]&m_{q \omega \omega^{2}}=\frac{G M M}{r^{2}} \\&r^{3}=\frac{G M}{\omega^{2}}=\frac{G M}{4 \pi^{2}} T^{2} \\&r^{3}=\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24} \times(16.6 \times 3600)^{7}}{4 \pi^{2}} \\[/tex]
[tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]
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A constant force F is applied on a body for a time interval of delta t.This force changes the velocity of the body from V1 to V2.then change in momentum in time interval will be
The change in momentum in time interval, given the data will be F × Δt
What is momentum?Momentum is defined as the product of mass and velocity. It is expressed as
Momentum = mass × velocity
What is impulse?This is defined as the change in momentum of an object.
Impulse = change in momentum
But
Impulse = force × time
Therefore
Force × time = change in momentum
How to determine the change in momentumInitial velocity = v₁ Finalvelocity = v₂Force = FChange in time = ΔtChange in momentum = ?Force × time = change in momentum
F × Δt = change in momentum
Change in momentum = F × Δt
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An object of height 5 cm is kept in front of a
lens. An inverted image of height 5 is formed. identity the lens and position of the
object and image.
Answer:
The type of lens involved is a convex lens and the object is positioned at the center of the curvature of the convex lens.
According to the question, a real, inverted, and same-sized image of the object is formed.
A concave lens is a diverging lens and always forms virtual images. But convex lenses are converging lenses and are the only type of lens that produce real and inverted images of the corresponding objects.
When an object is placed at the center of the curvature of a convex lens, its corresponding image is formed on the opposite side of the convex lens. The image formed is real and inverted.
The distance of the image from the lens is equal to the distance of the object from the lens.
For a convex lens, the distance of the center of curvature from the lens is double the focal length of the lens.
That's why the convex lens and center of curvature are the correct answer to this question.
Explanation:
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.17 times a second. A tack is stuck in the tire at a distance of 0.351 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed
The tangential speed of the wheel is determined as 4.786 m/s.
Tangential speed of the wheel
The tangential speed of the wheel is calculated as follows;
v = ωr
where;
ω is angular speed in rad/sr is radius of the circular pathv = (2.17 x 2π rad)/s x 0.351 m
v = 4.786 m/s
Thus, the tangential speed of the wheel is determined as 4.786 m/s.
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A certain force gives mass m1, an acceleration of 12 m/s² and mass m2 an acceleration of 3.3. m/s². What acceleration will this force give to the combined mass?
Acceleration is the rate of change of velocity. Usually, acceleration means the speed is transforming, but not always. When an object moves in a circular path at a steady speed, it is still accelerating, because the focus of its velocity is changing.
a = 4,552 m / s², b) a = 2,588 m / s²
Newton's second law is
F = ma
a = F / m
in this case the force remains constant
indicate us
* for a mass m₁
a₁ = F/m₁
a₁ = 12, m/ s²
* for a mass m₂
a₂= 3.3 m / s²
acceleration m = m₂-m₁
substitute
[tex]$\begin{aligned}&a=\frac{F}{m_{2}-m_{1}} \\&1 / a=\frac{m_{2}}{F}-\frac{m_{1}}{F}\end{aligned}$[/tex]
let's calculate
1/a=1/3.3 - 1/12 = 0.21969
a = 4,552 m / s²
a= 4,552m/s²
What is speed and acceleration?Speed estimates the rate of movement of an object, that is, the distance traveled per unit of time. Acceleration calculates the rate of change of velocity, that is, the change in velocity between two different moments
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a body has a mass of 2kg.it accelerats from 20m/s to 40m/s in 4 seconds.the resultant force is
The resultant force is 8N
Given that mass is 2kg , v= 40m/s, u =20m/s and we need to calculate resultant force
F=ma
m is given
so for a
v-u/t=a { first equation of motion }
40-20/4= 4
so a=4
F = ma =2*4 = 8N
The difference between the forces that are acting on an object as part of a system is known as the resultant force.
v = u + at is the first equation of motion. Here, v denotes the end speed, u the starting speed, an acceleration, and t the passage of time. The first equation of motion is provided by the velocity-time relation, which may be used to calculate acceleration.
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The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.
The initial speed of the rock as it left the astronaut's hand is 168 m/s.
The speed of the satellite is 60.2 m/s.
Acceleration due to gravity of the satellite
g = GM/R²
where;
M is mass of the satelliteR is radius of the satelliteg = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(6.32 x 10⁴)²
g = 0.132 m/s²
initial speed of the rock when it reaches maximum heightv² = u² - 2gh
0 = u² - 2gh
u² = 2gh
u = √2gh
u = √(2 x 9.8 x 1440)
u = 168 m/s
Speed of the satellitev = √GM/r
v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(1.45 x 10⁵)]
v = 60.2 m/s
Thus, the initial speed of the rock as it left the astronaut's hand is 168 m/s.
The speed of the satellite is 60.2 m/s.
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The three displacement vectors have magnitude of A=5.00m,B=5.00m and C=4.00m.Find the resultant (magnitude and directional angle) of the three vectors?
3.00 m is the magnitude of the resultant vector
137.1° is the directional angle of the resultant vector.
1) To find the magnitude of the resultant
Discover each vector's individual components first.
We must consider the signs of the components because the angles in the figure are measured in various ways.
In this case, both the y component of vector C and the x component of vector A are negative.
The vectors' components are as follows, using a little trigonometry:
Magnitude of A = 5 m
[tex]A_{x}[/tex] = - (5.00m) cos20° = -5 ×0.408 = -4.698 m
[tex]A_{y}[/tex] = + (5.00m)sin20° = +5 × 0.342 = +1.710 m
Magnitude of B = 5m
[tex]B_{x}[/tex] = +(5.00m)cos60° = 5 × 0.5 = +2.5m
[tex]B_{y}[/tex] = +(5.00m)sin60° = 5 × √3/2 = +4.33 m
Cx = 0
Cy = -4.00m
The sum of all three vectors, which we refer to as R, produces components.
Rₓ = Aₓ + Bₓ + Cₓ
= -4.698 + 2.5 + 0
= -2.198 m
[tex]R_{y}[/tex] = [tex]A_{y} + B_{y} + C_{y}[/tex]
= +1.71 +4.33 - 4.00
= 2.040 m
R = [tex]\sqrt{R_{x^{2} }+R_{y^{2} }[/tex]
= [tex]\sqrt{(-2.198)^{2 } + (2.040)^{2} }[/tex]
= 3.00 m
2) To find the directional angle of resultant
tan θ = 2.040/-2.198 = -0.928
θ = -42.9°
Such a vector would be in the so-called "fourth quadrant," as it is well known. However, we discovered that R has a negative x component and a positive y component, indicating that such a vector must reside in the "second quadrant.
The calculator accidentally returned an angle that is 180 degrees off, thus we must add 180 degrees to the naïve angle in order to get the right angle.
Therefore, R's actual direction is determined by-
θ = -42.9° + 180° = 137.1°
Hence, the magnitude of resultant vector is 3.00 m and directional angle is 137.1°
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Figure 21.62 shows a box on whose surfaces the electric field is measured to be horizontal and to the right. On the left face (3 cm by 2 cm) the magnitude of the electric field is 400 V/m, and on the right face the magnitude of the electric field is 1000 V/m. On the other faces only the direction is known (horizontal). Calculate the electric flux on every face of the box, the total flux, and the total amount of charge that is inside the box.
(a) The electric flux on left face is 0.24 Vm and on the right face is 1 Vm.
(b) The total flux is 1.24 Vm.
(c) The total amount of charge that is inside the box is 1.1 x 10⁻¹¹ C.
Area of the left face
The area of the left face is calculated as follows;
A1 = 0.03 m x 0.02 m = 0.0006 m²
Electric flux on the left faceФ1 = EA1
Ф1 = (400 V/m)( 0.0006 m²) = 0.24 Vm
Let the dimension of the right face = 5 cm by 2 cm
Area of the right faceA2 = 0.05 m x 0.02 m = 0.001 m²
Electric flux on the right faceФ2 = EA2
Ф2 = (1000 V/m)( 0.001 m²) = 1 Vm
Total fluxФ = Ф1 + Ф2
Ф = 0.24 Vm + 1 Vm = 1.24 Vm
Total charge inside the boxФ = Q/ε
Q = εФ
Q = (8.85 x 10⁻¹²)(1.24)
Q = 1.1 x 10⁻¹¹ C
Thus, the electric flux on left face is 0.24 Vm and on the right face is 1 Vm.
The total flux is 1.24 Vm and the total amount of charge that is inside the box is 1.1 x 10⁻¹¹ C.
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two particles P and Q are shot vertically up. Particle P is first up with a velocity of 40m/s. After 4 seconds, particle Q is also shot up. Find a)where the two particles meet,if at the point of meeting,P has a velocity 15m/s. b ) the velocity with which Q is shot. take g=10m/s^2
The velocity with which the particle Q is fired is 15m/s upwards.
What is velocity?The vector quantity velocity (v), represented by the equation v = s/t, quantifies displacement (or change in position, s) over the change in time (t). Speed (or rate, r) is a scalar number, denoted by the equation r = d/t, that quantifies the distance traveled (d) over the change in time (t).
Assume the upward direction is positive and the downward direction is negative.
Velocity P, = 40m/s
Distance traveled by P,
Using the first equation of motion for particle P,
v = u + at
⇒ 0 = 40 + (-10)t
⇒ t = 4s
This is the time it takes for P to rise
Now, the maximum height(s) reached by the particle P is,
Using the second equation of motion,
s = ut + 1/2at²
⇒ s = 40×4 + 1/2 × (-10) × 4²
⇒ s = 80m
a) A particle P falls when Q is shot after 4 seconds from the initial time:
V² = U² + 2aH₁
⇒ H₁ = 15² - 0/ 2(-10)
⇒ H₁ = 11.25m
Particles P and Q meet at a distance from the ground (H₂).
Height, H₂ = s - H
⇒ H₂ = 80 - 11.25
= 68.75m
Particles P and Q meet at a distance of 68.75m from the ground.
v = u + at₁
⇒ 15 = 0 + (-10) t₁
⇒ t₁ = 1.5s
It is equal to P's fall time and Q's rise time.
b) For particle Q
H₂ = u₂t₁ + 1/2at₁
⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5
⇒ u₂ = 15 m/s
Therefore,
The velocity with which the particle Q is fired is 15m/s upwards.
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Question 2
A photon of green light has a wavelength of 520 nm. Find the green photon's frequency in Hz?
Hints: C=fa ; this will give you the frequency in Hz; 1 nm = 1x10-⁹ nm
5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .
The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.
Given:Wavelength of green light = 520 nm
f = c / λ
where, f = Frequency
c = Speed of light = 3 × [tex]10^8[/tex] m/s
λ = Wavelength of light
∴ f = c / λ
f = [tex]\frac{3*10^8}{520 * 10^-^9}[/tex]
= 5.77 ×[tex]10^1^4[/tex] Hz
Therefore, 5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .
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A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable. (You'll need to get the various positions from the graph. Many are exactly on one of the tic marks.)
323.5 N is the tension in the cable.
Given
Mass of crate(M) = 175.5 kg
Mass of boom(m) = 94.7 kg
The tension, T in the cable can be calculated by taking moments of force about the central point of marked X.
The Angle of the boom with the horizontal can be calculated by
tanθ = 5/10
θ = tan⁻¹(5/10) = 26.56°
Angle of the boom with horizontal is 26.56°
The angle of cable with horizontal can be calculated by
tan B = 4/10
B = tan⁻¹(4/10) = 21.80°
Angle of cable with horizontal is 21.80°
Taking moments of force about the point X
(Mcosθ + mcosθ) 0.5 = T(sin(θ +B)1
(175.5 × cos 26.56 + 94.7 × cos 26.56 )× 0.5 = T (sin(26.56 + 21.80) X 1
By calculating, we get
Tension(T) = 241.68/0.747
Tension(T) = 323.5 N
Hence, 323.5 N is the tension in the cable.
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The Earth’s diameter is about 8,000 miles; our Moon’s diameter is about 2,000 miles; how
many Moon’s would fit inside of the volume of the Sun?
Three moons can fit inside the volume of the sun.
What is the moon?The moon is a non luminous body found in the space. It could cause a solar eclipse when it comes between the sun and the earth.
Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;
8,000 miles/ 2,000 miles = 3
Hence, three moons can fit inside the volume of the sun.
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a)Calculate the effective value of g, the acceleration of gravity, at 7900 m above the Earth's surface.
b)Calculate the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.
The solution for the acceleration of gravity is given as
[tex]g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex][tex]g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]This is further explained below.
What is the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.?Generally,
Mass of earth [tex]$M=5.97 \times 10^{24} \mathrm{~kg}$[/tex]
Radius of earth [tex]$R=6371 \mathrm{~km}$[/tex]
Gravitational const. [tex]$G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$[/tex]
height [tex]$h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$[/tex]
[tex]R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}[/tex]
In conclusion, acceleration due to gravity at this point will be
[tex]g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex]
for [tex]$h_{2}=7900 \mathrm{~km}$[/tex]
[tex]R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]
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A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.
The magnitude of the force on the left-hand pole is mathematically given as
f'=0.167N
Q=45 degrees
What is the magnitude of the force on the left-hand pole.?Generally, the equation for Force is mathematically given as
F=mg/sinФ
Therefore
F(17.1*10^{-3})*9.8/sin 45
F=0.237N
Considering horizontal axis or plane
f'-fcos=0
Therefore
f'=0.237*cos45
f'=0.167N
In conclusion, the slope
tanФ=30/30
tanФ=1
Ф=45
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A space craft is moving relative to the earth , an observer on the earth finds that, between 1pm and 2pm according to her clock, 3601 seconds elapse on the space craft clock . What is the space craft speed relative to the earth?c=2.998×10^8ms
The speed of the space craft relative to the earth is given as: 0.024c. This is solved using the the equation for time dilation.
What is time dilation?
Time dilation is the "slowing down" of a clock as determined by an observer in relative motion with regard to that clock under the theory of special relativity.
The formula is given as :
Δt = [Δr]/ √ 1 - (v²/c²)
Thus,
v = c√1 - (Δr/Δt)²
= c √(1 - (3600/3601)²
v = 0.024c
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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×103 km, mass of Earth = 5.98×1024 kg, G = 6.67×10-11 Nm2/kg2.)
The speed of a satellite in a circular orbit around the Earth is 4,188 m/s.
Speed of the satelliteThe speed of the satellite is calculated as follows;
v = √GM/r
where;
M is mass of Earthr is radius of satellitev = √[(6.67 x 10⁻¹¹ x 5.98 x 10²⁴) / (3.57 x 6.37 x 10⁶)]
v = 4,188 m/s
Thus, the speed of a satellite in a circular orbit around the Earth is 4,188 m/s.
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A man pushing a crate of mass
m = 92.0 kg
at a speed of
v = 0.845 m/s
encounters a rough horizontal surface of length
ℓ = 0.65 m
as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.351 and he exerts a constant horizontal force of 280 N on the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude_____N
What is the direction?
1. Opposite as the motion of the crate
2. Same as the motion of the crate
(b) Find the net work done on the crate while it is on the rough surface.
______J
(c) Find the speed of the crate when it reaches the end of the rough surface.
_______m/s
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is 23.7 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
Magnitude of net force on the crateF(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
Net work done on the crateW = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
Acceleration of the cratea = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
Speed of the cratev² = u² + 2as
v² = 0.845² + 2(-0.396)(0.65)
v² = 0.199
v = √0.199
v = 0.45 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is 23.7 J.
The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×10^3 km, mass of Earth = 5.98×10^24 kg, G = 6.67×10^-11 Nm^2/kg^2.)
Answer: The speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth is 4188.11 m/s.
Explanation: To find the answer, we need to know about the equation of motion of a satellite around earth.
What is the equation of motion of a satellite around earth?We have gravitational force of attraction between the satellite of mass m and earth of mass M as,[tex]F_g=\frac{GMm}{r^2}[/tex]
The expression for centripetal force of,[tex]F_c=\frac{mv^2}{r} \\[/tex]
These two forces are equal for a satellite around earth.[tex]\frac{GMm}{r^2} =\frac{mv^2}{r} \\thus,\\v=\sqrt{\frac{GM}{r} }[/tex]
How to solve the problem?Given that,[tex]r=3.57 R_E=3.57*6.37*10^3=22.74*10^3 km\\M=5.98*10^24kg\\G=6.67*10^{-11}Nm^2/kg^2[/tex]
Thus, the speed of the satellite will be,[tex]v=\sqrt{\frac{6.67*10^{-11}*5.98*10^{24}}{22.74*10^6m} } =4188.11 m/s[/tex]
Thus, we can conclude that the speed of satellite will be 4188.11 m/s.
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In a circular orbit with a radius that is 3.57 times the mean radius of the Earth, a satellite moves at a speed of 132.43km/s.
In order to get the solution, we must understand the satellite's planetary motion equation.
What is the satellite's orbital motion equation?The earth's mass M and the satellite's mass M are attracted to one another by gravity.[tex]F_g=\frac{GMm}{r^2}[/tex]
The term used to describe centripetal force of,[tex]F_c=\frac{MV^2}{r}[/tex]
When a satellite orbits the earth, these two forces are equivalent. Thus, the velocity will be,[tex]\frac{GMm}{r^2}=\frac{mV^2}{r}\\V=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10{-11}*5.98*10^{24}}{22.74*10^3} } \\V=132.43*10^3m/s[/tex]
As a result, we may estimate that the satellite will move at a speed of 132.43 km/s.
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A 45-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
_______m
The altitude or height of the pole vaulter as she crosses the bar is 4.04 m.
What is the height of the pole vaulter?The height of the pole vaulter is determined from the change in kinetic energy which is equal to the potential energy at that height.
Potential energy = Change in kinetic energymgh = m(v - u)²/2h = (v - u)²/2g
h = (10 - 1.1)²/2 * 9.8
h = 4.04 m.
In conclusion, the height is determined from the potential energy at that height.
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Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.
a) For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.
b) What magnetic field does the line produce at ground level as a percent of the earth's magnetic field, which is 0.50 G .
c) Is this value of magnetic field cause for worry?
Yes. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
No. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth's field.
No. Since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.
(a) The magnetic field the line produced at ground level is 3.75 x 10⁻⁶ T.
(b) The lines magnetic field as a percent of the earth's magnetic field is 7.5%.
(c) No, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.
Magnetic field the line produced at ground levelB = (μ x I) / (2πr)
B = (4π x 10⁻⁷ x 150) / (2π x 8)
B = 3.75 x 10⁻⁶ T
Percent of the earth's magnetic fieldx = 3.75 x 10⁻⁶ /0.5G
x = (3.75 x 10⁻⁶ ) / (0.5 x 10⁻⁴)
x = 0.075
x = 0.075 x 100% = 7.5%
Thus, we can conclude that, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.
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Consider a message signal m(t) with the spectrum shown in the following Figure. The message signal bandwidth W=1KHz, This signal is applied to a product modulator, together with a carrier Accos(2πfc t)wave producing the DSB-SC modulated wave S(t). This modulated wave is next applied to a coherent detector. Assuming perfect synchronism between the carrier waves in the modulator and detector, determine the spectrum of the detector output when: (a) The carrier frequency fc=1.25 KHz. And (b) The carrier frequency fc=0.75 KHz. What is the lowest carrier frequency for which each component of the modulated wave S(t) is uniquely determined by m(t)?
The lowest carrier frequency for which each component of the modulated wave S(t) is uniquely determined by m(t) is
[tex]-Fc,-Fc+W,-Fc-wW = > -1.5k,-0.5k,-2.5kHz[/tex]"-w,w -1kHz,1kHz" are the frequency components of the detector output in this case.
[tex]-Fc,-Fc+w,-Fc-W = > -0.75k,0.25k,-1.75kHz[/tex]This is further explained below.
What is the lowest carrier frequency for which each component of the modulated wave S(t) is uniquely determined by m(t)?Generally, the equation for the modulated wave containing frequency components is mathematically given as
[tex]Fc,Fc+w,Fc-W = > 1.5k,2.5k,0.5kHz[/tex]
[tex]-Fc,-Fc+W,-Fc-wW = > -1.5k,-0.5k,-2.5kHz[/tex]
"-w,w -1kHz,1kHz" are the frequency components of the detector output in this case.
b)
In conclusion,, then the modulated wave contains frequency components as
[tex]Fc,Fc+W,Fc-W = > 0.75k,1.75k,-0.25kHz[/tex]
[tex]-Fc,-Fc+w,-Fc-W = > -0.75k,0.25k,-1.75kHz[/tex]
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What force pulls a falling apple down to the ground?
A. Spring force
B. Tension
C. Normal force
D. Gravity
Answer:
d
Explanation:
Gravitational force is a type of force that pulls objects to the Earth's surface.
The following table lists the speed of sound in various materials. Use this table to answer the question.
Substance Speed (m/s)
Glass 5,200
Aluminum 5,100
Iron 4,500
Copper 3,500
Salt water 1,530
Fresh water 1,500
Mercury 1,400
Hydrogen at 0°C 1,284
Ethyl Alcohol 1,125
Helium at 0°C 965
Air at 100°C 387
Air at 0°C 331
Oxygen at 0°C 316
Sound will travel fastest in air at _____.
-5°C
0°C
10°C
15°C
Sound will travel fastest in air at 15°C.
Speed of sound in air
The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;
Air at 100°C 387 m/s
Air at 0°C 331 m/s
From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.
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Piston 1 in the figure has a diameter of 1.87 cm. Piston 2 has a diameter of 9.46 cm.
In the absence of friction, determine the force F, necessary to support an object with a mass of 991 kg placed on piston 2. (Neglect the height difference between the bottom of the two pistons, and assume that the pistons are massless).
Force necessary to support the object on piston 2 is 24× 10⁴ N.
To find the answer, we need to know about the force and pressure on piston 1 and piston 2.
What's the pressure on piston 2?The force on piston 2= mass × acceleration due to gravity= 991 Kg × 9.8 = 9414.5N
Mathematically, force= pressure/areaPressure= force × area of piston= 9414.5N × π(9.46² cm² /4)
= 9414.5N × π(9.46²× 10^(-4)m²/4)
= 66.2 N/m²
What's the force needed to held the mass on piston 2?Pressure on piston 2 = pressure on piston 1Force on piston 1= pressure on piston 1/area of piston 1= 66.2/ π(1.87² cm² /4)
= 66.2/ π(1.87²×10^(-4)m² /4)
= 24× 10⁴ N
Thus, we can conclude that force necessary to support the object on piston 2 is 24× 10⁴ N.
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On planet Zog, Mr. Spock measures that it takes 1.38 s for a mass of 0.5 kg to hit the ground when released from rest from a height of 2.85 m.
1. Calculate the size of the acceleration of gravity on that planet.
2. He decides to repeat the experiment. Calculate the work he must do to move the mass from the ground back up to its initial height.
The acceleration due to gravity is 3 m/s^2 and the work done is -4.3 J.
What is the acceleration due to gravity?Now we must use the formula;
h = ut + 1/2gt^2
Since it was dropped from a height u = 0 m/s
h = height
u = initial velocity
g = acceleration due to gravity
t = time
h = 1/2gt^2
g = 2h/t^2
g = 2 * 2.85 /(1.38)^2
g = 5.7/1.9
g = 3 m/s^2
The work that must be done is against gravity hence;
W = -(mgh)
W = - (0.5 kg * 3 m/s^2 * 2.85 m)
W = - 4.3 J
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A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at the other end as shown. The cable is attached at a height H=1.70 m above the pivot and the plank's CM is located a distance d=0.700 m from the pivot.
Calculate the tension in the cable.
Hello!
Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.
[tex]\Sigma \tau = 0[/tex]
We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)
- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)
Both of these act in opposite directions. Let's use the equation for torque:
[tex]\tau = r \times F[/tex]
Doing the summation using their respective lever arms:
[tex]0 = L Tsin\theta - dF_g[/tex]
[tex]dF_g = LTsin\theta[/tex]
Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:
[tex]tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o[/tex]
Now, let's solve for 'T'.
[tex]T = \frac{dMg}{Lsin\theta}[/tex]
Plugging in the values:
[tex]T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}[/tex]
An object 10mm in height is located 20cm from a crown glass spherical surface whose power is +10.00DS. Locate the image
The image is present at 20cm from the crown glass spherical surface.
To find the answer, we need to know about the lens formula.
What's the lens formula?It's (1/V)-(1/U)= (1/f)V= image distance from the lens, U= object distance, f= focal length of the lensWhat's the image distance, if object is present at 20cm from crown glass of power 10DS?Focal length (f)= 1/ power = 1/10 = 0.1 m U= -20cm = -0.2m (-ve sign due to sign convection)(1/V)-(-1/0.2)= (1/0.1)=> (1/V)+5=10
=> 1/V= 5
=> V=0.2m = 20cm
Thus, we can conclude that the image is present at 20cm.
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