The induced emf in the copper loop is -79.86 mV.
What is induced emf?The induced emf in the copper loop is given by the equation:
[tex]e = -(dΦ/dt) = - (d/dt)(BAcosθ)[/tex]
where Φ is the magnetic flux through the loop, B is the magnitude of the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the plane of the loop.
Given that the side of the square loop is 10.0 cm, the area of the loop is ([tex]10.0 cm)^2 = 100 cm^2.[/tex]
[tex]B(t) = 0.01T + (1.00×10^-3 T/s )t ,[/tex]
Given that the direction of the magnetic field makes an angle 37° with the plane of the loop, so the angle θ=37°
Substituting the values we have:
[tex]e = - (100 cm^2)( d/dt)((0.01T + (1.00×10^-3 T/s )t)cos37°)e = - (100 cm^2)(1.00×10^-3 T/s) cos37°e = - (100 cm^2)(1.00×10^-3 T/s) * 0.7986e = - (79.86×10^-3 T/s)[/tex]
The induced emf in the copper loop is -79.86 mV.
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Question 3 of 10
What is the primary means by which heat is transferred through fluids?
O A. Direct contact of particles
OB. Radiation
OC. Electromagnetic waves
OD. Convection currents
The primary means by which heat is transferred through fluids is convection currents (option D).
What is convection current?Convection is the transmission of heat in a fluid by the circulation of currents.
Heat can be transferred by different methods depending on the medium. Fluids like gases and liquids transfer heat through the process of convection.
Therefore, the primary means by which heat is transferred through fluids is convection currents.
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1.A virtual image is formed 20.5 cm from a concave mirror having a radius of curvature of 31.5 cm.
(a) Find the position of the object.
(b) What is the magnification of the mirror?
2.A contact lens is made of plastic with an index of refraction of 1.45. The lens has an outer radius of curvature of +2.04 cm and an inner radius of curvature of +2.48 cm. What is the focal length of the lens?
The position of the object is = -68cm
The magnification of the mirror= 0.3
Calculation of object distanceThe image distance = 20.5cm
The focal length= R/2 = 31.5/2= 15.75
The object distance= ?
Using the lens formula,1/f = 1/v-1/u
1/u = 1/v- 1/f
1/u = 1/20.5 - 1/15.75
1/u = 0.0489- 0.0635
1/u = -0.0146
u = -68cm
The magnification of the mirror is image size/object size
= 20.5cm/-68cm
= 0.3
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What are the balanced forces for someone in a parachute?
A. Gravity and air resistance
B. Centripetal and air resistance
C. Gravity and centripetal
D. Gravity and Earth
6.
A swimmer bounces straight up from a diving board and falls feet first into a pool. She
starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (3pt)
a) How long are her feet in the air?
b) What is her highest point above the board?
c) What is her velocity when her feet hit the water?
a) Her feet are in the air for 0.73+0.41 = 1.14 seconds
b) Her highest height above the board is 0.82 m
c) Her velocity when her feet hit the water is 7.16 m/s
Given,t = Time taken
u = Initial velocity = 4 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
a) Her feet are in the air for 0.73+0.41 = 1.14 seconds
s = ut + [tex]\frac{1}{2}[/tex]at²
2.62 = 0t + [tex]\frac{1}{2}[/tex] ₓ 9.8 ₓ [tex]t^{2}[/tex]
t = 0.73 s
b) Her highest height above the board is 0.82 m
The total height she would fall is 0.82+1.8 = 2.62 m
v = u + at
0 = 4 ₋ 9.8 ₓ t
t = 0.41 s
s = ut +[tex]\frac{1}{2}[/tex] at²
s = 4 ₓ 0.41 ₊ [tex]\frac{1}{2}[/tex] ₓ ₋9.8 ₓ 0.41 [tex]t^{2}[/tex]
c) Her velocity when her feet hit the water is 7.16 m/s
[tex]v = u + at \\v = 0 + 9.8[/tex] ₓ [tex]0.73[/tex]
v = 7.16 m/s
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The specific heat of copper is 0.385 J/g°C.
How much heat is needed to raise the temperature of 6.00 g of copper by 15.0°C?
35.0 J
90.0 J
234 J
34.7 J
The amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J (option A).
How to calculate amount of heat?The amount of heat absorbed or released by a substance can be calculated using the following formula:
Q = mc∆T
Where;
Q = quantity of heat absorbed or releasedm = mass of the substancec = specific heat capacity∆T = change in temperatureQ = 6 × 0.385 × 15
Q = 90 × 0.385
Q = 34.65J
Therefore, the amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J.
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Hi I have a question it’s not about the subject but is at the same time what is Physics?
Answer:
the branch of science that is concerned with nature and properties of matter and energy.
Explanation:
a study of the basis of what does what in science.
A rocket takes off from Earth's surface, accelerating straight up at 63.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 84.4 kg, including her space suit. (Assume the rocket's initial motion parallel to the +y-direction. Indicate the direction with the sign of your answer.)
______N
From the calculation, the normal force is 6161.2 N.
What is the normal force?The normal force is given by the expression;
N - mg = ma
Then;
N = mg + ma
m = 84.4 kg
g = 9.8 m/s^2
a = 63.2 m/s2
Now we have;
N = m(g + a)
N = 84.4 (9.8 + 63.2)
N = 6161.2 N
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A 149-g baseball is dropped from a tree 15.0 m above the ground.
With what speed would it hit the ground if air resistance could be ignored?
Express your answer to three significant figures and include the appropriate units.
If it actually hits the ground with a speed of 9.00 m/s , what is the magnitude of the average force of air resistance exerted on it?
Express your answer to three significant figures and include the appropriate units.
1. The speed with which the ball hits the ground is 17.1 m/s
2. The magnitude of the average force of air resistance exerted on it is 0.77 N
1. How to determine the velocity with which the ball hits the groundInitial velocity (u) = 0 m/s Acceleration due to gravity (g) = 9.8 m/s²Height (h) = 15 m Final velocity (v) =?v² = u² + 2gh
v² = 2gh
Take the square root of both side
v = √(2 × 9.8 × 15)
v = 17.1 m/s
2. How to determine the forceWe'll begin by calculating the time to reach the ground. This is illustrated below:
Acceleration due to gravity (g) = 9.8 m/s²Height (h) = 15 m Time (t) =?h = ½gt²
15 = ½ × 9.8 × t²
15 = 4.9 × t²
Divide both side by 4.9
t² = 15 / 4.9
Take the square root of both side
t = √(15 / 4.9)
t = 1.75 s
Now we can determine the force. This can be obtained as illustrated below:
Mass (m) = 149 g = 149 / 1000 = 0.149 KgInitial velocity (u) = 0 m/sFinal velocity (v) = 9 m/sTime (t) = 5 ms = 1.75 sForce (F) = ?F = m(v –u) / t
F = 0.149(9 – 0) / 1.75
F = 0.77 N
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A stone is launched at an angle of 50 degree with initial velocity of 22m/s. Find out its initial and final velocity.
(a) The initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.
(b) The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.
Initial vertical velocity
The initial vertical velocity of the stone is calculated as follows;
Vi = Vsinθ
Vi = 22 x sin(50)
Vi = 16.85 m/s
Initial horizontal velocityVxi = V cosθ
Vxi = 22 x cos(50)
Vxi = 14.14 m/s
Final vertical velocity of the stoneVf = Vi - gt
where;
Vf is the final vertical velocity = 0 at maximum heightFinal horizontal velocity of the stoneVfx = Vxi = 14.14 m/s
Thus, the initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.
The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.
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A uniform meter stick of mass 0.20 kg is pivoted at the 40 cm mark. Where should one hang a mass of 0.50 kg to balance the stick?
Answer:
36 cm
Explanation:
Mass of stick; m1 = 0.20kg at midpoint.
Total length; L=1.0 m
Pivot at 0.40m
Atached mass m2 = 0.50kg
Applying rotational equilibrium we have;
Ʈnet = 0
(m1g) • r1 = (m2g) • r2
(0.2) (0.1m) = (0.5)(x)
x = 0.04m =4cm
measured away from 40cm mark gives a position on the stick of; 40cm - 4cm = 36 cm
A bullet of mass 50g moving with an initial speed of 500m/s penetrates a wall and comes to rest at in 0.2seconds. calculate the deceleration of the bullet over the 0.2second.
The deceleration of the bullet over 0.2 second, given the data from the question is –2500 m/s²
What is acceleration?This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
a is the acceleration
v is the final velocity
u is the initial velocity
t is the time
NOTE: Deceleration is the opposite of acceleration
With the above equation for acceleration, we can obtain the deceleration of the bullet. Details below:
How to determine the deceleration of the bulletInitial velocity (u) = 500 m/sFinal velocity (v) = 0 m/sTime (t) = 0.2 sDecelration (a) =?a = (v – u) / t
a = (0 – 500) / 0.2
a = –500 / 0.2
a = –2500 m/s²
Thus, the deceleration of the bullet is –2500 m/s²
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light travel
3, 00,000 km/s. Is it velocity or speed?
If you speed through a construction zone while workers are present, your fines could be:.
If you speed through a construction zone while workers are present, your fines could be as much as one thousand dollars.
What is a Fine?
This is referred to as the amount which is instructed by a court or authority to be paid as a result of it being a penalty for various types of offences. each crime has its specific fine which helps to serve as deterrent for unlawful behavior in the community.
it is always best not to speed when within a construction zone which has workers present in the area. This is ideal as it helps to prevent incidences of accidents or death.
it is therefore the reason why a fine of 1000 dollars is to be paid so that people can think of such high amount before performing certain types of activities when driving and makes it the most appropriate choice.
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Which of the following changes would increase the force between two
charged particles to 9 times the original force?
A. Decreasing the distance between the particles by a factor of 3
B. Decreasing the amount of charge on one of the particles by a
factor of 9
C. Increasing the distance between the particles by a factor of 3
D. Increasing the amount of charge on each particle by a factor of 9
The answer is A. Decreasing the distance between the particles by a factor of 3.
The Universal Law of Gravitation is :
F = Gm₁m₂ / r² (where 'r' is the distance between them)
Since force is inversely proportional to the square of the distance between them, distance has to be decreased by a factor of 3 to increase the force to 9 times the original force.
For an air bag to work, it has to inflate full of nitrogen incredibly fast-within to
milliseconds of the collision. For a 60-liter cylindrical air bag to work property, the
nitrogen gas has to reach a pressure of 2.37 atm. At 25°G, how many moles of
nitrogen gas are needed to pressurize the air bag? Given, 0.0821 L-atm/mol-Kl
5.8 moles of nitrogen gas are needed to pressurize the air bag.
What's the expression of Ideal gas equation?Ideal gas equation is PV=nRTP= pressure, V = volume, n= no. of moles of gas, R= universal gas constant, T = temperature of the gasWhat's the no. of moles of nitrogen present in a 60L air bag at 2.37 atm pressure and 25°C temperature?P= 2.37 atm, V = 60L, R= 0.0821 L-atm/mol-K, T = 25°C = 298Kn= PV/RT= (2.37×60)/(0.0821×298)
= 5.8 moles
Thus, we can conclude that 5.8 moles of nitrogen gas are needed to pressurize the air bag.
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A 500 N force accelerates an object at 20 m s-2. What is its mass?
Answer: The mass of the object is 25kg.
The given question deals with Newton's second law of motion and its applications.
Explanation: Given force, F=500N
acceleration, a=20 m/[tex]s^{2}[/tex]
From Newton's 2nd law of motion , we have
F=ma where m=mass of the object
⇒500=m×20
⇒m=500/20=25
∴ Mass of the object is 25 kg .
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The escape velocity of a bullet from the surface of planet Y is 1695.0 m/s. Calculate the escape velocity from the surface of the planet X if the mass of planet X is 1.59 times that of Y, and its radius is 0.903 times the radius of Y.
The escape velocity from the surface of the planet X is 2,249.2 m/s.
Escape velocity of planet X[tex]v = \sqrt{\frac{2GM}{r} } \\\\v^2 = \frac{2GM}{r}\\\\v^2r = 2GM\\\\G = \frac{v^2r}{2M}[/tex]
where;
M is mass of the planetr is radius of the planetG is universal gravitation constant[tex]\frac{v_x^2 \ r_x}{2M_x} = \frac{v_y^2 \ r_y}{2M_y} \\\\\frac{v_x^2 \ r_x}{M_x} = \frac{v_y^2 \ r_y}{M_y} \\\\v_x^2 = \frac{v_y^2 \ r_yM_x}{M_yr_x}\\\\v_x^2 = \frac{(1695)^2 (r_y)(1.59M_y)}{M_y(0.903r_y)} \\\\v_x^2 = 5,058,814.78\\\\v_x = \sqrt{5,058,814.78} \ \ = 2,249.2 \ m/s[/tex]
Thus, the escape velocity from the surface of the planet X is 2,249.2 m/s.
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An 87-kg football player traveling 5.2 m/s is stopped in 1.0 s by a tackler.
What is the original kinetic energy of the player?
Express your answer to two significant figures and include the appropriate units.
What average power is required to stop him?
Express your answer to two significant figures and include the appropriate units.
The original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.
What is Kinetic Energy ?The energy possessed by a body in motion is known as Kinetic Energy. The S. I unit is Joule.
Given that an 87-kg football player traveling 5.2 m/s is stopped in 1.0 s by a tackler. The given parameters are;
Mass m = 87 KgVelocity v = 5.2 m/sTime t = 1 sThe original kinetic energy of the player can be calculated by using the formula K.E = 1/2m[tex]v^{2}[/tex]
Substitute all the parameters into the formula
K.E = 1/2 x 87 x [tex]5.2^{2}[/tex]
K.E = 1176.24
K.E = 1200 J
Power is the rate at which work is done.
Work done = energy
The average power is required to stop him can be calculated by using the formula P = E/t
Substitute all the parameters into the formula
P = 1200/1
P = 1200 W
Therefore, the original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.
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100m with a constant speed 200km/hr the pilot drops abomb from the plane. determine (neplect air resistance of friction) X Q
The horizontal distance XQ traveled by the bomb is 250 m.
Distance X Q
Let the XQ be the horizontal distance traveled by the bomb.
Time for the bomb to drop from 100 mh = vt + ¹/₂gt
Let the vertical velocity = 0
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
t = √(2h/g)
t = √(2 x 100 / 9.8)
t = 4.5 s
Horizontal distance traveled by the bombXQ = vx(t)
where;
vx is horizontal speed, = 200 km/hr = 55.56 m/s
XQ = 55.56 x 4.5
XQ = 250 m
Thus, the horizontal distance XQ traveled by the bomb is 250 m.
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Two hockey pucks are moving towards each other. (Assume no friction.) The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path. Find the final speed and angle of the first puck.
The final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.
What happened in an Elastic Collision ?In an elastic collision, both momentum and energy are conserved. But only momentum is conserved in inelastic collision.
Given that two hockey pucks are moving towards each other. The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path.
The given parameters are;
M1 = 0.13 kgM2 = 0.16 kgU1 = 1.11 KgU2 = 1.21 KgV1 = ?V2 = 1.16 kgФ1 = ?Ф2 = 42°The mathematical representation of the above question will be in two components.
Horizontal component
M1U1 - M2U2 = M1V1cosФ - M2V2cosФ
Substitute all the parameters
0.13 x 1.11 - 0.16 x 1.21 = 0.13 x V1 cosФ - 0.16 x 1.16cos42
0.1443 - 0.1936 = 0.13V1cosФ - 0.1379
0.13V1cosФ = 0.0886
V1cosФ = 0.0886/0.13
V1cosФ = 0.6815 ........ (1)
Vertical component
0 = M1V1sinФ - M2V2sinФ
M1V1sinФ = M2V2sinФ
Substitute all the parameters
0.13 x V1 sinФ = 0.16 x 1.16sin42
V1 sinФ = 0.1242/0.13
V1 sinФ = 0.9553 ......... (2)
Divide equation 2 by 1
V1 sinФ / V1 cosФ = 0.9553/ 0.6815
Tan Ф = 1.40
Ф = [tex]Tan^{-1}[/tex](1.4)
Ф = 54.5°
Substitute Ф into equation 2
V1 sin54.5 = 0.9553
V1 = 0.9553 / 0.8141
V1 = 1.17 m/s
Therefore, the final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.
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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.
The tension in the cable is equal to 323.5 N.
What is the tension in the cable?The tension, T in the cable is determined by taking moments about the pivot marked X.
The angles of the boom and the cable with the horizontal are first calculated.
Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°
The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80
Taking moments about the pivot:
175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1
Tension = 241.68/0.747
Tension = 323.5 N
In conclusion, the tension in the cable helps to suspend the crate.
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A projectile leaves the ground at an angle of 60° the horizontal.Its kinetic energy is E.Neglecting air resistance, find in terms of E its kinetic energy at the highest point of the motion
The kinetic energy of the projectile at the highest point of its motion will be E/4.
What is Projectile Motion?When a projectile will be thrown obliquely near the surface of the earth, it travels a curved path under uniform acceleration directed toward the center of the Earth. The path of a particle is called a projectile while the motion of a projectile is projectile motion.
Given, the angle of projection with horizontal, [tex]\theta = 60 ^o[/tex]
Consider that 'E' is the initial value of the kinetic energy of the projectile.
The equation for the initial kinetic energy is : [tex]E =\frac{1}{2}mu^2[/tex]
where m is the mass of the given projectile.
The component of the velocity of the projectile in the horizontal direction:
uₓ = u cosθ
uₓ = u cos 60°
uₓ = u/2
From the equation of motion: v = u +at
v = (u/2) + (0) t
v = u/2
The final kinetic energy of the projectile:
[tex]E_f = \frac{1}{2}mv^2[/tex]
[tex]E_f = \frac{1}{2}m(\frac{u}{2} )^2[/tex]
[tex]E_f = \frac{1}{4} (\frac{1}{2}mu^2 )[/tex]
[tex]E_f = \frac{E}{4}[/tex]
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The thermal emission of the human body has maximum intensity at a wavelength of approximately 9.5 μm.What photon energy corresponds to this wavelength?
Answer:
Explanation:
2.1 x 10^2 - 20J
Part B
A roller coaster ride starts with the roller coaster car being pulled to the top of the first hill with pulley system. The car is
released from the top with an initial velocity close to zero, then accelerates downward. From that first hill, the roller coaster just
coasts; there is no driving force, other than gravity, to keep It going. Assuming no friction, what can you say about the height of
the other hills in the roller coaster ride?
The highest point of a roller coaster is almost always the first hill. In the majority of roller coasters, the hills get smaller as the train travels down the track.
To find the answer, we have to know more about the mechanical energy of a system.
How to find the answer?Since it influences the mechanical energy of the system, the first hill must be the highest.One of the fundamental tenets of physics is that, in the absence of friction, mechanical energy must be conserved. Mechanical energy is the product of kinetic energy and potential energy.When the vehicles ascend the first hill on the roller coaster, mechanical energy is provided to the system because the speed is zero at this point.Mechanical energy = U = mgh
Where m represents the car mass, g represents gravity, and h represents height
If the system is to continue moving, the other hills on the mountain must be lower than the first hill. When the vehicles are released, this energy is converted into kinetic and potential energy when it lowers and ascends, but the sum of these two cannot be larger than the starting energy.Finally, by applying the principle of energy conservation, we may determine that, the initial hill must be the highest.
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A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car moves 30.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car.
(a) the speed v
______m/s
(b) the horizontal force exerted on the car (Enter the magnitude.)
_______N
The horizontal force applied is 160 N while the velocity is 2.03 m/s.
What is the speed of the car?The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
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A solid cylinder of uniform density of 0.85 g/cm3 floats in a glass of water tinted light blue by food coloring.
Its circular surfaces are horizontal. What effect will the following changes, each made to the initial system, have on X, the height of the upper surface above the water? The liquids added do not mix with the water, and the cylinder never hits the bottom.
1. The cylinder is replaced with one that has the same density and diameter, but with half the height.
2. Some of the water is removed from the glass.
3. A liquid with a density of 1.06 g/cm3 is poured into the glass.
4. The cylinder is replaced with one that has the same height and diameter, but with density of 0.83 g/cm3.
5. A liquid with a density of 0.76 g/cm3 is poured into the glass.
6. The cylinder is replaced with one that has the same density and height, but 1.5× the diameter.
Options are: Increase, Decrease, No change
The buoyant force acting on the cylinder is, [tex]Fb = \rho Ahg[/tex]. Here A is the cross-sectional area of the cylinder, h is the height of the cylinder, ρ is the density of the cylinder, and g is the acceleration due to gravity.
This buoyant force is also equal to the volume of the fluid displaced. [tex]Fb = \sigma h(A-x)g[/tex]. Here σ is the density of the fluid.
Equate the above two equations and solve for x.
[tex]\rho Ahg = \sigma A(h-x)g\\\rho h = \sigma h - \sigma x\\x = \frac{(\sigma - \rho)h}{\sigma}[/tex]
So, the distance x depends on the density of the fluid, density of the cylinder and the height of the cylinder.
1. The density of the cylinder is same and distance x is independent of the diameter of the cylinder. Therefor, there will be no change in the distance x. Hence, the correct answer is No change.
2. Now the height is changing keeping the density same. As the distance x is directly proportional to the height, the distance x will increase.
3. The density of the added liquid is greater than of the water and it does not mix with the water. So, the liquid will settle down and there will be no change in the distance x.
4. The density of the added liquid is less than that of the water and it does not mix with the water. So, the liquid will not settle down and the distance x will change. The change in distance x can be determined as follow:
[tex]\rho Agh = \sigma' Axg + \sigma A(h-x)g\\\rho h=\sigma' x + \sigma h - \sigma x\\x=(\frac{\sigma - \rho}{\sigma - \sigma'})h[/tex]
Here, σ' is the density of the added liquid.
From the above relation it is clear, that on adding the liquid of the density less than that of water, the denominator term become small ando so the value of x will increase.
5. On removing some of the water inside the glass, the height of the water column will decrease, but the value of x does not depend on the height of the water column. So, there will be no change in the distance x.
6. The density of the new cylinder is smaller than that of the earlier one. So, the numerator term will increase. Therefore, the value of x will increase.
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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).
The force F, necessary to support an object with a mass of 991 kg placed on piston 2. is 20.61J.
How to calculate the value?It should be noted that by Pascal Law, the pressure on piston 1 will have the same value as the pressure on piston 2.
This will be:
(991 × 10) /(π × 0.0946/2)²
= 9910/0.022
= 450454.6 Pa
F1 = A1 × 450454.6 = 3.14 × (0.0187/2)² × 450454.6
= 123.64
F = 123.64/6
F = 20.61
Therefore, the force F, necessary to support an object with a mass of 991 kg placed on piston 2. is 20.61J.
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emily is standing 150 feet from a circular target with radius 3 inches. will she hit the target if her aim is off by 0.2 degrees?
Answer:
no
Explanation:
The angle subtended by the radius of the target at Emily's distance can be found using the tangent relation.
tan(α) = opposite/adjacent = (1/4 ft)/(150 ft) = 1/600
The angle is found using the inverse relation -
α = arctan(1/600) ≈ 0.095°
If Emily's aim is off by 0.2°, she will miss the target by several inches.
Emily's projectile will miss her aiming point by ... (150 ft)tan(0.2°) ≈ 0.524 ft ≈ 6.28 in
A copper transmission cable 50.0 km long and 10.0 cm in diameter carries a current of 105 A. What is the potential drop across the cable? Let ρcopper = 1.72 × 10—8 Ω • m.
A) 5.75 V
B) 5.48 V
C) 11.5 V
D) 16.9 V
5.48 V is the potential drop across the cable for a copper transmission cable of 50.0 km long and 10.0 cm in diameter carries a current of 105 A
Ohm's Law states that the potential drop is determined by the equation: V = IR, where I is the current and R is the wire resistance.
R=PL/A
Under the assumption that all physical parameters and temperatures remain constant, Ohm's law asserts that the voltage across a conductor is directly proportional to the current flowing through it.
Only when the given temperature and the other physical variables remain constant does Ohm's law apply. Increasing the current causes the temperature to rise in some components. The filament of a light bulb serves as an illustration of this, where the temperature increases as the current increases. Ohm's law cannot be applied in this situation. The filament of the lightbulb defies Ohm's Law.
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
The vertical component of force exerted by the hi.nge on the beam will be,142.10N.
To find the answer, we need to know more about the tension.
How to find the vertical component of the force exerted by the hi.nge on the beam?Let's draw the free body diagram of the system.To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.[tex]F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\[/tex]
To find the answer, we have to find the tension,[tex]Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N[/tex]
Thus, the vertical component of the force exerted by the hi.nge on the beam will be,[tex]F_V=(29*9.8)-(169.43*sin57)=142.10N[/tex]
Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.
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The hi.nge will apply a force of 142.10N on the beam in the vertical direction.
We must learn more about the tension in order to find the solution.
How can I determine the vertical component of the force the hi.nge has on the beam?Let's create the system's free body diagram.We must balance the total vertical force to zero in order to get the vertical component of the force applied to the beam by the height.[tex]F_V=mg-Tsin\alpha[/tex]
We must identify the tension in order to find the solution.[tex]Tlsin\alpha =mg\frac{l}{2}sin\beta \\T=\frac{mgsin90}{2sin57} =169.43N[/tex]
Consequently, the force that the height exerts on the beam will have a vertical component that is,[tex]F_v=(29*9.8)-(169.43*sin57)=142.10N[/tex]
This leads us to the conclusion that the vertical component of the force the hi.nge exerts on the beam will be 142.10N.
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