The response of an object to the gravity is called ___

On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed

madreJ [45]

Answer:

The length of the rough patch is 4.345 meters.

Explanation:

According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:

$K_{1} = K_{2} + W_{loss}$

Where:

$K_{1}$ , $K_{2}$ - Initial and final kinetic energy, measured in joules.

$W_{loss}$ - Work losses due to friction.

By applying definitions of kinetic energy and work, the expression described above is expanded:

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + f \cdot \Delta s$

Where:

$v_{1}$ , $v_{2}$ - Initial and final speed of the skater, measured in meters per second.

$m$ - Mass of the skater, measured in kilograms.

$\Delta s$ - Length of the rough patch, measured in meters.

$f$ - Friction force, measured in newtons.

According to the statement, friction force is represented by the following expression:

$f = r \cdot m \cdot g$

Where:

$r$ - Ratio of friction force to weight, dimensionless.

$g$ - Gravitational constant, measured in meters per square second.

Then,

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + r \cdot m \cdot g \cdot \Delta s$

The equation is simplified algebraically and patch length is cleared afterwards:

$\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = r \cdot g \cdot \Delta s$

$\Delta s = \frac{v_{1}^{2}-v_{2}^{2}}{2 \cdot r \cdot g }$

Given that $v_{1} = 5\,\frac{m}{s}$ , $v_{2} = 2.5\,\frac{m}{s}$ , $r = 0.22$ and $g = 9.807 \,\frac{m}{s^{2}}$ , the length of the rough patch is:

$\Delta s = \frac{\left(5\,\frac{m}{s} \right)^{2}-\left(2.5\,\frac{m}{s} \right)^{2}}{2\cdot (0.22)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta s = 4.345\,m$

The length of the rough patch is 4.345 meters.

6 0

3 years ago

How fast must a train accelerates from rest to cover 518 m in the first 7.48 s?

Rus_ich [418]

Heya!

For this problem, use the formula:

s = Vo * t + (at^2) / 2

Since the initial velocity is zero, the formula simplifies like this:

s = (at^2) / 2

Clear a:

2s = at^2

(2s) / t^2 = a

a = (2s) / t^2

Data:

s = Distance = 518 m

t = Time = 7,48 s

a = Aceleration = ¿?

Replace according formula:

a = (2*518 m) / (7,48 s)^2

Resolving:

a = 1036 m / 55,95 s^2

a = 23,34 m/s^2

The aceleration must be <u>23,34 meters per second squared</u>

8 0

3 years ago

What are the two factors that determine the strength of the gravitational force between two objects?

Lena [83]

Answer:

mass and distance

Explanation:

took the test

4 0

2 years ago

A sample of an ideal gas is heated, and its kelvin temperature doubles. What happens to the average speed of the molecules in th

BabaBlast [244]

A sample of an ideal gas is heated, and its kelvin temperature doubles. The average speed of the molecules in the sample will increases by a factor of $\sqrt{2}$

The root-mean square (RMS) velocity is the value of the square root of the sum of the squares of the stacking velocity values divided by the number of values. The RMS velocity is that of a wave through sub-surface layers of different interval velocities along a specific ray path.

Root mean square speed is a statistical measurement of speed.

The root mean square speed can be calculated as : V1 : $\sqrt{3 R T / Mo}$

if temperature becomes double

let T1 is initial temperature

So , T2 = 2 * T1

now ,

Root mean square speed will be (V2) = $\sqrt{(3 R (2T)) / Mo}$

= $\sqrt{2}$ * $\sqrt{3 R T / Mo}$

= $\sqrt{2}$ V1

Thus when temperature becomes double, the root mean square speed increases by a factor of $\sqrt{2}$

To learn more about root mean square velocity here

brainly.com/question/13751940

#SPJ4

3 0

1 year ago

If a force of 5 n to left and 9 n to right act on a object what is the net force

emmainna [20.7K]

4 n to the right is the net force

7 0

3 years ago

The response of an object to the gravity is called ____