PHYSICS COLLEGE

A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the bottom. It then slides on the horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.20. How far does the block slide on the horizontal surface before it comes to rest?

Answers

Answer 1

Answer:

The block slide on the horizontal surface is "24.99 m" far.

According to the question,

Vertical height = 5.0 m
Coefficient of friction = 0.20

Let,

The time taken be "t".

Now,

→ $s = ut+ (1)/(2) at^2$

By substituting the values, we get

$5 = (1)/(2)* 9.8* t^2$

$t = 1.01 \ sec$

The final velocity will be:

→ $v_1 = gt$

$= 9.8* 1.01$

$= 9.899 \ m/s$

Now,

→ $t = (u)/(a)$

$= (9.899)/(0.2* 9.8)$

$= 5.05 \ seconds$

hence,

The distance will be:

→ $s = ut+0.5* at^2$

$= 9.899(5.05)-0.5* (0.2* 9.8* 5.05^2)$

$= 24.99 \ m$

Thus the above approach is right.

Learn more about friction here:

brainly.com/question/18851133

Answer 2

Answer:

Answer:

The block slides on the horizontal surface 25 m before coming to rest.

Explanation:

Hi there!

For this problem, we have to use the energy-conservation theorem. Initially, the block has only gravitational potential energy (PE) that can be calculated as follows:

PE = m · g · h

Where:

m = mass of the block.

g = acceleration due to gravity.

h = height at which the block is located.

As the block starts to slide down the track, its height diminishes as well as its potential energy. Due to the conservation of energy, energy can´t disappear, so the loss of potential energy is compensated by an increase of kinetic energy (KE). In other words, as the block slides, the potential energy is converted into kinetic energy. The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

m = mass of the block.

v = speed of the block.

Then, at the bottom of the ramp, the kinetic energy of the block will be equal to the potential energy that the block had at the top of the ramp.

Initial PE = KE at the bottom

When the block starts sliding horizontally, friction force does work to stop the block. According to the energy-work theorem, the change in the kinetic energy of an object is equal to the net work done on that object. In other words, the amount of work needed to stop the block is equal to its kinetic energy. Then, the work done by friction will be equal to the kinetic energy of the block at the bottom, that is equal to the potential energy of the block at the top of the track:

initial PE = KE at the bottom = work done by friction

The work done by friction is calculated as follows:

W = Fr · Δx

Where:

W = work

Fr = friction force.

Δx = traveled distance.

And the friction force is calculated as follows:

Fr = μ · N

Where:

μ = coefficient of friction.

N = normal force.

Since the block is not accelerated in the vertical direction, in this case, the normal force is equal to the weight (w) of the block:

Sum of vertical forces = ∑Fy = N - w = 0 ⇒N = w

And the weight is calculated as follows:

w = m · g

Where m is the mass of the block and g the acceleration due to gravity.

Then, the work done by friction can be expressed as follows:

W = μ · m · g · Δx

Using the equation:

intial PE = work done by friction

m · g · h = μ · m · g · Δx

Solving for Δx

h/μ = Δx

5.0 m / 0.20 = Δx

Δx = 25 m

The block slides on the horizontal surface 25 m before coming to rest.

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The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 0.078 V exists across the membrane. The thickness of the membrane is 7.1 x 10-9 m. What is the magnitude of the electric field in the membrane?

Answers

Answer:

$10.99* 10^6\ V/m$

Explanation:

Given:

Potential difference across the membrane (ΔV) = 0.078 V

Thickness of the membrane (Δx) = 7.1 × 10⁻⁹ m

Magnitude of electric field (|E|) = ?

We know that, the electric field due to a potential difference (ΔV) across a distance of Δx is given as:

$E=-(\Delta V)/(\Delta x)$

So, the magnitude of the electric field is calculated by ignoring the negative sign and thus is given as:

$|E|=(\Delta V)/(\Delta x)$

Plug in the given values and solve for '|E|'. This gives,

$|E|=(0.078\ V)/(7.1* 10^(-9)\ m)\n\n|E|=10.99* 10^6\ V/m$

Therefore, the magnitude of the electric field in the membrane is $10.99* 10^6\ V/m$ .

Which statement about thin lenses is correct? In each case, we are considering only a single lens. A. A diverging lens always produces a virtual inverted image. B. A converging lens always produces a real inverted image. C. A converging lens sometimes produces a real erect image. D. A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E. A diverging lens always produces a virtual erect image.

Answers

A diverging lens always produces a virtual erect image.

The general lens formula is given as;

$(1)/(F) = (1)/(U) + (1)/(V)$

Where;

U = object distance
V = image distance
F = focal length of the lens

A lens can be converging or diverging.

A converging lens produces a virtual image when the object is placed in front of the focal point. The image can also be real when the object is placed beyond focal point.

The image produced by a diverging lens is always virtual and upright.

Thus, we can conclude that a diverging lens always produces a virtual erect image.

Learn more here:brainly.com/question/11788630

Answer:

E) true. The image is always virtual and erect

Explanation:

In this exercise we are asked to find the correct statements,

for this we can use the constructor equation

1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

In diverging lenses, the focal length is negative and the image is virtual and erect

In convergent lenses, the positive focal length, if the object is farther than the focal length, the image is real and inverted, and if the object is at a shorter distance than the focal length, the image is virtual and straight.

With this analysis let's review each statement

A) False. The image is right

B) False. The type of image depends on where the object is with respect to the focal length

C) False. The real image is always inverted

D) False. The image is always virtual

E) true. The image is always virtual and erect

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The snowball falls straight back down in to a 6 inches of snow. The snowball feels a deceleration of 100 m/s2 upon impact with the snow bank before coming to rest. (a) When does the snowball hit the top of the snow bank? (b) How far from the ground does the snowball come a rest?

Answers

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\n\Rightarrow 0=4.5-32.1* t\n\Rightarrow (-4.5)/(-32.1)=t\n\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+(1)/(2)at^2\n\Rightarrow s=4.5* 0.14+(1)/(2)* -32.1* 0.14^2\n\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+(1)/(2)at^2\n\Rightarrow 4.315=0t+(1)/(2)* 32.1* t^2\n\Rightarrow t=\sqrt{(4.315* 2)/(32.1)}\n\Rightarrow t=0.518\ s$

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

$v=u+at\n\Rightarrow v=0+32.1* 0.518\n\Rightarrow v=16.62\ ft/s$

$v^2-u^2=2as\n\Rightarrow s=(v^2-u^2)/(2a)\n\Rightarrow s=(0^2-16.62^2)/(2* -100* 3.28)\n\Rightarrow s=0.42\ ft$

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

Use the worked example above to help you solve this problem. A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 15.0 m/s. The point of release is h = 46.0 m above the ground.(a) How long does it take for the ball to hit the ground?
(b) Find the ball's speed at impact.
(c) Find the horizontal range of the ball.

Answers

Answer:

Explanation:

Please helpa car is driven 200 km west and then 80 km southwest. what is the displacement of the car from the point of orgin (magnitude and direction)? draw a diagram.

Answers

Let's take east and west to be positive and negative, respectively, and north and south to be positive and negative, respectively. Then in terms of vectors (using ijk notation), the car first moves 200 km west,

r = (-200 km) j

then 80 km southwest,

s = (-80/√2 km) i + (-80/√2 km) j

so that its total displacement is

r + s = (-80/√2 km) i + ((-200 - 80/√2) km) j

r + s ≈ (-56.6 km) i + (-256.6 km) j

This vector has magnitude

√((-56.6 km)² + (-256.6 km)²) ≈ 262.7 km

and direction θ such that

tan(θ) = (-256.6 km) / (-56.6 km) ==> θ ≈ -102.4º

relative to east, or about 12.4º west of south.

If a coil stays at rest in a very large static magnetic field, no emf is induced in this coil. Group of answer choices True False

Answers

Answer:

True

Explanation:

Faraday's Law says that there is a emf induced in a conductor when the vector flux of the magnetic field across it changes in time.
This can be true due to one of two facts, either the magnitude of the magnetic field changes in time, or the area through which the flux occurs changes due to the movement of the object.
In this case, due to the magnetic field is constant, and the coil stays at rest, there is no possible change in flux, so emf induced is zero.

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