Answer:
Nearest, the revolutions per minute will be 29.
Explanation:
Given that,
Radius of circle = 1 m
Acceleration a =g
We know that,
Angular frequency is defined as,
[tex]\omega=2\pi n[/tex]
Where, n = number of revolutions in one second
We need to calculate the revolutions in one second
Using formula of centripetal acceleration
[tex]a=\omega^2r[/tex]
Put the value of a and ω
[tex]g=(2\pi n)^2r[/tex]
[tex]n=\sqrt{\dfrac{g}{r}}\times\dfrac{1}{2\pi}[/tex]
Put the value into the formula
[tex]n=\sqrt{\dfrac{9.8}{1}}\times\dfrac{1}{2\pi}[/tex]
[tex]n=0.49[/tex]
We need to calculate the revolutions per minute
Using value for the revolutions per minute
[tex]n=0.49\times60[/tex]
[tex]n=29.4[/tex]
Hence, Nearest, the revolutions per minute will be 29.
BRAINLIEST. Agraph is probelow. The graph shows the speed of a car traveling east over a 12 second period. Based on the information in the graph, it can be
that in the first second
Answer:speeding up constantly
Explanation:
The graph between the time and the speed of the car shows that the speed is increasing constantly, so, option C is correct.
What is speed?A moving object's rate of change in distance traveled is measured as speed. Speed is a scalar, which implies it is a measurement with a magnitude but no direction.
A thing that moves quickly and with high speed, covering a lot of ground in a short time. On the other hand, a slow-moving object traveling at a low speed covers a comparatively small distance in the same amount of time. An object with zero speed does not move at all.
Given:
The graph shows the speed of a car traveling east over a 12-second period,
As you can see from the graph, at time t = 0 sec the speed is 10 m/s,
At t = 3 sec, the speed = 15.3 m/s
At t = 6 sec, the speed = 20.3 m/s
Thus, speed is increasing constantly.
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A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 555 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tension in the rope to the left of the mountain climber.
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The tension in the rope on the left of the mountain climber is [tex] T_a = 1106 \ N [/tex]
Explanation:
From the question we are told that
The weight of the mountain climber is m = 555 N
Generally from the diagram , the total amount of force acting on the rope along the vertical axis at equilibrium is mathematically represented as
[tex]T_a* cos 65 -555 + T_b * cos(85) = 0[/tex]
Here [tex]T_a, T_b[/tex] are the tension of the rope on the left and on the right hand side
So
[tex]0.423T_a + 0.0871T_b = 555[/tex]
=> [tex] 0.0871T_b = 555 - 0.423T_a[/tex]
=> [tex] T_b = \frac{555 - 0.423T_a}{0.0871}[/tex]
Generally from the diagram , the total amount of force acting on the rope along the horizontal axis at equilibrium is mathematically represented as
[tex]T_a* sin 65 - T_b * sin(85) = 0[/tex]
=> [tex] 0.9063T_a - 0.9962T_b = 0[/tex]
=> [tex] 0.9063T_a = 0.9962T_b [/tex]
=> [tex] 0.9063T_a = 0.9962[\frac{555 - 0.423T_a}{0.0871}] [/tex]
=> [tex] 0.9063T_a = [\frac{552.891 - 0.421T_a}{0.0871}] [/tex]
=> [tex] 0.0789T_a = [552.891 - 0.421T_a[/tex]
=> [tex] 0.4999T_a = 552.891 [/tex]
=> [tex] T_a = 1106 \ N [/tex]
Which of the organisms in the food web above is the top level carnivore
Answer:
apex consumers
Explanation:
they are top
Calculate the effective value of g, the acceleration of gravity, at 6700 m , above the Earth's surface. g
Answer:
The effective value of g at 6700 m above the Earth's surface is 9.79 m/s².
Explanation:
The value of g can be found using the following equation:
[tex] F = \frac{GmM}{r^{2}} [/tex]
[tex] ma = \frac{GmM}{r^{2}} [/tex]
[tex] a = \frac{GM}{r^{2}} [/tex]
Where:
a is the acceleration of gravity = g
G: is the gravitational constant = 6.67x10⁻¹¹ m³/(kg.s²)
M: is the Earth's mass = 5.97x10²⁴ kg
r: is the Earth's radius = 6371 km
Since we need to find g at 6700 m, the total distance is:
[tex] r_{T} = 6371000 m + 6700 m = 6377700 m [/tex]
Now, the value of g is:
[tex] a = \frac{GM}{r_{T}^{2}} = \frac{6.67\cdot 10^{-11} m^{3}/(kg*s^{2})*5.97 \cdot 10^{24} kg}{(6377700 m)^{2}} = 9.79 m/s^{2} [/tex]
Therefore, the effective value of g at 6700 m above the Earth's surface is 9.79 m/s².
I hope it helps you!
Two charged objects are separated by distance, d. The first charge has a larger magnitude (size) than the second charge. Which one exerts the most force?
Answer:
The two charged objects will exert equal and opposite forces on each other.
Explanation:
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of charges on the objects and inversely proportional to the square of the distance between the two objects.
This force of attraction or repulsion between the two charged objects is always equal and opposite.
Therefore, the two charged objects will exert equal and opposite forces on each other.
g A child bounces a 50 g super ball on the sidewalk. The velocity change of the super bowl is from 27 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
The average force exerted on the superball by the sidewalk is 0.00122 N.
Explanation:
Given;
mass of the super ball, m = 50 g = 0.05 kg
initial velocity of the super bowl, u = -27 m/s (assuming downward motion to be negative)
final velocity of the super bowl, u = 17 m/s (assuming upward motion to be positive)
time of motion, t = 1800 s
The average force exerted on the superball by the sidewalk is given by;
[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{0.05(17-(-27))}{1800}\\\\ F = \frac{0.05(44)}{1800}\\\\F = 0.00122 \ N[/tex]
Therefore, the average force exerted on the superball by the sidewalk is 0.00122 N.
Which of these statements best describes the impact of ocean thermal power and current power on the environment?
A. Current power may decrease the fish population.
B. Current power may decrease the gravitational pull of the moon.
C. Ocean thermal power may increase the fish population.
C. Ocean thermal power may increase the gravitational pull of the moon.
Answer:
A. Current power may decrease the fish population.
Explanation:
The statement that best describes the impact of ocean thermal power and current power on the environment is that current power may decrease the fish population.
The environment is made up of living and non-living components that co-exist and interact with one another.
Harnessing current power from ocean movement will seriously affect the fish population. Most fishes are not sedentary. They move and glide through the water. When current power causes a change in the environment of the fish. This will definitely affect the normal condition prevalent in the body of water.Answer:
A
Explanation:
I took the test good luck :D
Can anybody tell me what I'm suppose to do. I click start the numbers comes up to the right
am I right? be honest
Answer:
I chose c because it is the greater slope at point c
100 POINTS.
Please provide explanation.
Thank you
Answer:
(a) 0.829 m/s
(b) 3.27 m/s
(c) 0.000153 m²
55.8%
Explanation:
(a) Flow rate equals velocity times cross-sectional area. (1 L = 0.001 m³)
Q = vA
(0.001 m³ / 2.00 s) = v (48 × π (0.002 m)²)
v = 0.829 m/s
(b) Use Bernoulli equation. Choose point 1 to be the exit of the pump, and point 2 to be exit of the shower head. Choose 0 elevation to be at point 1.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
(1.50 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) v² + 0 = (1 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) (0.829 m/s)² + (1000 kg/m³) (10 m/s²) (5.50 m)
1.50×10⁵ Pa + (500 kg/m³) v² = 1×10⁵ Pa + 414.5 Pa + 55000 Pa
v = 3.27 m/s
(c) Flow rate is constant.
Q = vA
(0.001 m³ / 2.00 s) = (3.27 m/s) A
A = 0.000153 m²
Flow rate is proportional to the pressure difference and the radius raised to the fourth power.
Q ∝ ΔP r⁴
Q₂/Q₁ = (ΔP₂/ΔP₁) (r₂/r₁)⁴
Q₂/Q₁ = (1.120) (0.840)⁴
Q₂/Q₁ = 0.558
The flow decreases to 55.8% of the original value.
Answer:
Explanation:
Regarding the point of "Flow rate is proportional to the pressure difference and the radius raised to the fourth power", flow rate depends on pressure, cross-section area and speed. As speed also depends on cross-section area, flow rate becomes dependent on pressure and cross-section area squared.
In a round pipe like blood vessel, the cross-section area is equal to pi*radius squared. So flow rate is proportional to the pressure difference and (radius squared) squared; i.e. the radius raised to the fourth power.
The new flow rate = (1.12)*(0.84)^4
=0.5576 or 55.76% of the original flow rate
How long is a day in Neptune
Answer: the long day in neptune would be .18383562 years!
Explanation:also for every day is 16 hours
Calculate the ratio of the mechanical energy at B and mechanical energy at a (eb,ea) and (ec,ea). What do these ratios tell you about the conservation of energy?
A) is the mechanical conserved between a and b? explain
B) is the mechanical energy conserved between b and c ?explain
Answer:
Yes at A the mechanical energy is conserved.
Yes at B the part of mechanical energy is conserved potential energy and kinetic energy and some is lost as frictional force.
Explanation:
Ratio = Eb/ Ea= 1058.3 J/2940 J= 0.3599
Ratio = Ec/ Eb= 0J/ 1058.3 J= 0
At point A the skater is at rest or it is the starting point and the whole energy is due to the position of the skater i.e= mgh = 50 *9.8*6= 2940 J
Since there's no movement there is no Kinetic energy = 0 J
Yes at A the mechanical energy is conserved.
At point B the skater has traveled for some of the distance . It has potential energy and kinetic energy.
Yes at B the part of mechanical energy is conserved as potential energy and kinetic energy.
The total Mechanical energy = 1058.3 J
At point B Total Mechanical energy = PE+ KE
1058.3J = 980 J + 78.3 J
1058.3 J = mgh + 1/2mv²
= 50*2*9.8 + 1/2 *50*(8.85)²
= 980 J + 78.3 J
As the total energy of the system must remain the same some of the mechanical energy is lost as frictional force at point B .
2940 J-1058.3 J= 1881.7
At Point C the skater has arrived at the end point and the height , speed, PE, KE and ME all are zero.
(a) The ratio of the mechanical energy at B and mechanical energy at A is 0.36.
(b) The ratio of the mechanical energy at C and mechanical energy at A is 0.
(c) mechanical energy is conserved between a and b.
(d) mechanical energy is not conserved between b and c.
The given parameters;
mechanical energy at A, [tex]E_a = 2,940 \ J[/tex]mechanical energy at B, [tex]E_b =1,058.3 \ J[/tex]mechanical energy at C, [tex]E_c = 0[/tex]The ratio of the mechanical energy at B and mechanical energy at A;
[tex]ratio = \frac{E_b}{E_a} = \frac{1058.3}{2940} = 0.36[/tex]
The ratio of the mechanical energy at C and mechanical energy at A;
[tex]ratio = \frac{E_c}{E_a} = \frac{0}{2940} = 0[/tex]
The change mechanical energy between A and B from the given position;
[tex]\Delta E = mg(h_b - h_a) - \frac{1}{2}m(v_b^2 - v_a^2)\\\\ \Delta E = 50\times 9.8(2-6) \ - \ \frac{1}{2} \times 50(8.85^2 - 0)\\\\\Delta E =- 1960 + 1960\\\\\Delta E = 0 \ J[/tex]
Thus, we can conclude that mechanical energy is conserved between a and b.
The change mechanical energy between A and B from the given position;
[tex]\Delta E = mg(h_c - h_b) - \frac{1}{2}m(v_c^2 - v_b^2)\\\\ \Delta E = 50\times 9.8(0-2) \ - \ \frac{1}{2} \times 50(0^2 - 8.85^2)\\\\\Delta E = -980 + 1960 \\\\\Delta E = 980 \ J[/tex]
Thus, we can conclude that mechanical energy is not conserved between b and c.
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1. What quantity of heat is required to raise?
the temperature of 450 grams of water
from 35°C to 85°C?
capacity of water is 4.18 J/g °C.
Answer: Calculate the energy required in joules to raise the temperature of 450 grams of water from 15°C to 85°C? (The specific heat capacity of water is 4.18 J/g/°C)
Explanation:
The quantity of heat is required to raise the temperature of of water is 94050 joule.
What is law of conservation of energy?Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.
The law of energy conservation is adhered to by all energy forms. The law of conservation of energy essentially says that the total energy of the system is conserved in a closed system, also known as an isolated system.
Mass of water: m = 450 grams
Initial temperature of water = 35° C
Final temperature of water = 85° C
Capacity of water is 4.18 J/g °C.
Hence, the quantity of heat is required to raise = mass × Capacity × raise in temperature
= 450 × 4.18 × (85 - 35) joule
= 94050 joule.
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The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm
Answer:
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
Explanation:
The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:
[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]
[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.
[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.
And we expand the equation above by definitions of elastic potential energy and kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.
[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.
If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:
[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]
[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be Imax
The complete question is;
A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 16-kV power supply. What is the power dissipated in his body?
A) If the internal resistance of the power supply is 1600 Ω , what is the current through the person's body?
B) What is the power dissipated in his body?
C) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be I_max = 1.00mA or less?
Answer:
A) I = 1.379 A
B) P = 19016.41 W
C) r = 15990000 Ω
Explanation:
A) We are given;
Internal resistance of the power supply; r = 1600 Ω
Body resistance between hands; R = 10kΩ = 10000 Ω
Power supply voltage; E =16 kV = 16000 V
Formula for the current through the person's body with internal resistance is given by;
I = E/(R + r)
Thus;
I = 16000/(10000 + 1600)
I = 1.379 A
B) Formula for power dissipated is;
P = I²R
P = 1.379² × 10000
P = 19016.41 W
C) Now, we are told that the maximum current should be I_max = 1.00mA or less. So, I_max = 0.001 A
Thus, from I = E/(R + r) and making r the subject, we have;
r = (E/I) - R
r = (16000/0.001) - 10000
r = 15990000 Ω
A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance rl (r
(A) E/8
(B) E 78.
(C) E/2
(D) 2E
(E) 8E
Answer:
A ) E/8
Explanation:
If the sphere of radius R carries charge Q, then the volumetric charge density is:
ρ₁ = [Q/ (4/3)*π*R³]
Therefore the net charge inside r ( r < R ) is:
q₁ = ρ * (4/3)*π*r³
And E = K * q₁/r K = 9,98 *10⁹ [N*m²/C²]
E = K * ρ * (4/3)*π*r³/r
E = K * ρ * (4/3)*π*r²
If now the charge is distributed over a sphere of radius 2R
ρ₂ = [Q/ (4/3)*π*(2R)³]
ρ₂ = [Q/ (4/3)*π*8*R³]
Then ρ₂ < ρ₁ in fact ρ₂ = (1/8)*ρ₁
The electric field depends on the net charge enclosed by a gaussian surface, and the distance between the net charge and the considered point, ( considering the net charge as being at the center of the gaussian surface) In this case, there was no distance change then
E₂ = E₁/8
The right answer is lyrics A ) E/8
An airplane, starting at rest, takes off on a 600. m long runway accelerating at a rate of 12 m/s/s. How many seconds does it take to reach the end of the runway?
x = 1/2 at²
where x = length of runway, a = acceleration, and t = time.
600 m = 1/2 (12 m/s²) t²
t² = (1200 m) / (12 m/s²)
t² = 100 s²
t = 10 s
On what part of the eye are rods and cones found?
Answer:
retina.
Explanation:
Light is described as having a dual wave-particle nature. Which piece of evidence provides support for the model of light as a particle?
. Young’s double slit experiment showed that light waves show interference.
. Light reflects when it hits a surface.
. Light refracts when it moves from one medium to another.
. Light does not need a medium to travel.
Answer:
light reflects when it hits the surface
Explanation:
Youngs double slit is a evidence for wave nature,
The properties refraction are attributed as properties of waves. The phenomena of interference and diffraction also fall in this category.
So,
the answer must be B
Answer: Light does not need a medium to travel.
Explanation: I took the test and got it right :]
Two protons are a distance 3 10-9 m apart. What is the electric potential energy of the system consisting of the two protons
Answer:
The electric potential energy of the system is 7.87x10⁻²⁰ J.
Explanation:
The electric potential energy is given by:
[tex]E = \int{Fdr} = \frac{Kq_{1}q_{2}}{r}[/tex]
Where:
q₁ = q₂ is the charge of the protons = 1.62x10⁻¹⁹ C
r is the distance = 3x10⁻⁹ m
K: is the electrostatic constant = 9x10⁹ Nm²/C²
[tex] E = \frac{Kq_{1}q_{2}}{r} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*(1.62 \cdot 10^{-19} C)^{2}}{3\cdot 10^{-9} m} = 7.87 \cdot 10^{-20} J [/tex]
Therefore, the electric potential energy of the system is 7.87x10⁻²⁰ J.
I hope it helps you!
The electric potential energy of the system should be 7.87x10⁻²⁰ J.
Calculation of the electric potential energy:SInce We know that
fdr = kq1q2/r
Here
q₁ = q₂ i.e. is the charge of the protons = 1.62x10⁻¹⁹ C
r should be the distance = 3x10⁻⁹ m
K should be the electrostatic constant = 9x10⁹ Nm²/C²
Now electric potential energy should be
= (9x10⁹ Nm²/C² * 1.62x10⁻¹⁹ C) / 3x10⁻⁹ m
= 7.87x10⁻²⁰ J.
hence, The electric potential energy of the system should be 7.87x10⁻²⁰ J.
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My parrot has a mass of 1.33kg, what is it's weight here on earth
Answer:
Your parrot, from earth, that weighs 1.33kg is 1.33kg on earth. as far as i'm aware there is only one earth and everything always weighs the same on one planet as it did on that same planet.
Explanation:
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east. what is the distance and displacement
Explanation:
Let east = E, and, west = opposite to east = - E.
Here, displacement:
=> 2m east + 4m west + 1m east
=> 2E + 4(-E) + 1E
=> 2E - 4E + 1E
=> - 1E
=> 1(-E)
=> 1m west
And, distance,
=> 2m + 4m + 1m = 7m
The distance of a person is 7 m and the displacement of the person is 1m west.
To find the distance and displacement, the given values are,
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east.
What is the distance and the displacement?Displacement:
The displacement is shortest distance between initial and final position or we can say it is the straight line distance between initial and final position.If object moves in a straight line path without any turn then the path length and the displacement is always same.Distance:
The distance is the total path length of the object while it will move from initial to final position.If the object move on curved path then displacement is smaller than the distance moved by the object.Let us consider East = E and west = opposite to east = - E.
Calculating the displacement:
= 2m east + 4m west + 1m east
= 2E + 4(-E) + 1E
= 2E - 4E + 1E
= - 1E
= 1(-E)
= 1m west.
The displacement is 1m west.
Now calculating the distance,
= 2m + 4m + 1m
= 7m
The distance is 7m.
Thus, the displacement and the distance is found as 1 m west and 7m.
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Luck walked to a store that is 250m away and it took him 50secs while Layne walked to the mall that is 1000m away and took her 200s to do. What do they have in common?
A. Average speed
B. Acceleration
C. Displacement
D.mass
Answer:
Average speed
Explanation:
250/50=5
1000/20=5
12, The error in measurement may occur due to
S
a, inexperience of a person
b, The faulty apparatus
c, Inappropriate method
d, Due to all reasons in a, b and c
Answer:
d, Due to all reasons in a, b and c
Explanation:
All of the reasons from the choices can lead to errors in measurement. An error is an uncertainty introduced to a scientific process.
Errors can be due to the following reasons;
It can be to a poor technician; an inexperienced scientist can introduce serious into their set up. Even when recording their observation, a dearth of experience can lead to error. Faulty apparatus can also lead to errors in measurement. An instrument that is poorly calibrated can also result in measurement errors. An inappropriate scientific method or measuring guidelines can also lead to errors in measurements.block of mass m sits at rest on a rough inclined ramp that makes an angle with the horizontal. What must be true about normal force F on the block due to the ramp
Answer:
Explanation:
For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.
The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction
Taking the sum of forces along the y component
Sum Fy = -W+R = ma
Since acceleration is zero
-W+R = m(0)
-W+R = 0
-W = -R
W = R
Hence the Normal reaction force acting on the on the body is equal to normal force
How do I proton and and electron compared
A 849-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 35.0 m/s. What is the average force exerted on the car during this time
Answer:
The average force exerted on the car during this time is 5,943 N
Explanation:
Given;
mass of the car, m = 849 kg
initial velocity of the car, u = 0
time of motion of the car, t = 5.00 s
final velocity of the car, v = 35 m/s
The average force exerted on the car during this time is given by;
[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]
Therefore, the average force exerted on the car during this time is 5,943 N
Answer:
5943N
Let's say (+x) = eastward
Average horizontal acceleration
ax = vx -v0x/5.00s
= 35.0m/s-0/5.00s
= +7.09m/s
From here we apply the second law of newton
During this period average horizontal force acting on car
Summation x = max = (849kg)(+7.09m/s²)
= 5943N
+5.943x10³N
= 5.94kN east ward.
Marisa’s car accelerates at an average rate of 2.6m/s^2. Calculate how long it takes her car to accelerate from 24.6m/s to 26.8m/s? Show your work.
given info is... Acceleration(a)=2.6m/s^2
final velocity(v)=26.8m/s
initial velocity(u)=24.6m/s
need to find.... time(t)=?
[tex]a=\frac{v-u}{t} \\2.6=\frac{26.8-24.6}{t} \\\\[/tex]
[tex]t=\frac{v-u}{a}[/tex]
[tex]t=\frac{26.8-24.6}{2.6}[/tex]
[tex]t=0.846s[/tex]
Explanation:
It takes 0.84 second her car to accelerate from 24.6m/s to 26.8m/s.
What is acceleration?Acceleration is the rate at which speed and direction of velocity vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting.
Given parameters:
Initial speed of the car: u = 24.6 m/s
Final speed of the car: v = 26.8 m/s.
Acceleration of the car: a = 2.6 m/s²
Time interval: t = ?
change is speed = final speed - initial speed
= 26.8 m/s - 24.6 m/s
= 2.2 m/s
From the definition of acceleration,
acceleration = change is speed / time interval
So, time interval = change is speed / acceleration
= 2.2 m/s/2.6 m/s²
= 0.84 second.
Hence, it takes 0.84 second her car to accelerate from 24.6m/s to 26.8m/s.
Learn more about acceleration here:
brainly.com/question/12550364
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It takes 3.8 x 10^-5 for a pulse of the radio waves from a radar to reach a plane and bounce back. How far is the plane from the radar?
Answer: 11400 m
Explanation:
Given:
t = 3.8 x 10^-5 s
v = 3 x 10^8 m/s
d = ?
Formula:
d = vt
= (3.8 x 10^-5 s)(3 x 10^8 m/s)
= 11400 m
hope this helps :)
The uniform movement allows to find the results for the distance from the radar to the plane is: 5.7 10³ m or 5.7 km
Kinematics studies the motion of objects looking for relationships between position, velocity and acceleration, in the special case that the acceleration is zero is called uniform motion and is described by the expression
[tex]v = \frac{d}{t}[/tex]
d = v t
Where v is the velocity, d the displacement and t the time.
Radar waves are electromagnetic waves with constant velocity
v = 3 10⁸ m/s
They indicate that the time of the waves to go to the plane and return is 3.8 10⁻⁵ s, therefore if the speed is constant, the time to reach the plane is half of the total time.
t = [tex]\frac{t_{total} }{ 2}[/tex]
t = [tex]\frac{3.8 \ 10^{-5}}{2}[/tex]
t = 1.9 10⁻⁵ s
Let's calculate
d = 3 10⁸ 1.9 10⁻⁵
d = 5.7 10³
In conclusion with the uniform movement we can find the results for the distance from the radar to the plane is: 5.7 10³ m or 5.7 km
Learn more here: brainly.com/question/20369552
g An angry rhino with a mass of 2700 kg charges directly toward you with a speed of 3.70 m/s. Before you start running, as a distraction, you throw a 0.180 kg rubber ball directly at the rhino with a speed of 9.05 m/s. Determine the speed of the ball (in m/s) after it bounces back elastically toward you.
Answer:
9.05m/s
Explanation:
given data
m1= 2700kg
v1=3.7m/s
m2=0.18kg
v2=9.05m/s
v3=?
We know that the velocity of the rhino will remains unchanged after impact as the mass of the rubber ball is negligible
m1v1+m2v2=m1v1+m2v3
2700*3.7+0.18*9.05=2700*3.7+0.18*v3
9990+1.629=9990+0.18v3
9991.629-9990=0.18v3
1.629=0.18v3
v3=1.629/0.18
v3=9.05m/s