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3 years ago

In which condition the final velocity becomes zero?

Physics

1 answer:

alexgriva [62]3 years ago

7 0

Answer:

Terminal velocity is achieved, therefore, when the speed of a moving object is no longer increasing or decreasing; the object's acceleration (or deceleration) is zero.

Explanation:

hope it helps you

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A box with mass 25.14 kg is sliding at rest from the top of the slope with height 13.30 m

lutik1710 [3]

Answer:

Explanation:

The first thing we are asked to find is the Force experienced by the box. That is found in the formula:

F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:

$wsin\theta-f=ma$ where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:

First we need the weight of the box. Having the mass, we find the weight:

w = mg so

w = 25.14(10) so

w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:

251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so

F = 125.7 And in order to answer what a is equal to, we find f:

f = μ $F_n$ where Fn is the weight of the object.

f = .25(251.4) so

f = 62.85. Filling everything in now altogether to solve for a, the only missing value:

125.7 - 62.85 = 25.14a and

62.85 = 25.14a so

a = 2.5 m/s/s

Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that

a = -10 m/s/s

v₀ = 0 (it starts from rest)

Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:

Δx = v₀t + $\frac{1}{2}at^2$ and we are solving for time:

$-13.30=0t+\frac{1}{2}(-10)t^2$ and

$t=\sqrt{\frac{2(-13.30)}{-10} }$ so

t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.

3 0

3 years ago

A box slides downwards at a constant velocity on an inclined surface that has a coefficient of friction uK = .58 The angle of th

mojhsa [17]

Answer: The angle of inclination is nearly 30°

Explanation:

For a body on an inclined plane, the coefficient of friction between the body and the plane is equal to the ratio of the moving force applied to the body to the frictional force acting on the body.

If uK coefficient of friction;

Fm is the moving force

R is the normal reaction on the body

Mathematically uK = Fm/R

Fm = WSin(theta)

R = Wcos(theta)

uK = Wsin(theta)/Wcos(theta)

uK = tan(theta)

theta = arctan(uK)

If uK is 0.58

theta = arctan0.58

theta = 30°

The angle of the inclined will be 30°

7 0

3 years ago

A simple pendulum consists of a 2 kg bob attached to a 1.5 m long string. How much time (in s) is required for this pendulum to

Charra [1.4K]

Answer:

6.15 s

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

For the pendulum in this problem,

L = 1.5 m (length)

$g=9.8 m/s^2$ (acceleration due to gravity on Earth)

Therefore, its period is

$T=2\pi \sqrt{\frac{1.5}{9.8}}=2.46 s$

And therefore, the time taken for the pendulum to complete 2.5 oscillations is equal to 2.5 times the period:

$t=2.5T=(2.5)(2.46)=6.15 s$

3 0

3 years ago

What is the electric field due to a point charge of 20uC at a distance of 1 meter away from it?

Anettt [7]

The electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000 $\frac{N}{C}$ .

First, you have to know that the space surrounding a load suffers some kind of disturbance, since a load located in that space will suffer a force. The disturbance that this charge creates around it is called an electric field.

In other words, an electric field exists in a certain region of space if, when introducing a charge called witness charge or test charge, it undergoes the action of an electric force.

The electric field E created by the point charge q at any point P, located at a distance r, is defined as:

$E=K\frac{q}{r^{2} }$

where K is the constant of Coulomb's law.

In this case, you know:

K= 9×10⁹ $\frac{Nm^{2} }{C^{2} }$
q= 20 uC=20×10⁻⁶ C
r= 1 m

Replacing in the definition of electric field:

$E=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{20x10^{-6} C}{(1 m)^{2} }$

Solving:

E=180000 $\frac{N}{C}$

Finally, the electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000 $\frac{N}{C}$ .

Learn more:

5 0

2 years ago

PLEASE HELP!

WINSTONCH [101]

So you would first multiply 400 by 2 which equals 800, then add 30 which is 830.
Then you would subtract 1000-830=170.
The total force of the 6 other players would be 170N.
Hoped this helped ☺️

7 0

3 years ago

In which condition the final velocity becomes zero?​

In which condition the final velocity becomes zero?