Which best describes a force?

Which one of the following statements concerning kinetic energy is true? a The kinetic energy of an object always has a positive

hodyreva [135]

a The kinetic energy of an object always has a positive value.

Explanation:

The kinetic energy of an object is the energy possessed by an object due to its motion, and it can be calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

Let's now analyze each statement:

a The kinetic energy of an object always has a positive value. --> TRUE. In fact, the mass of the object is always positive, and the term $v^2$ is always positive as well, so the kinetic energy is always positive.

b The kinetic energy of an object is directly proportional to its speed. --> FALSE. Looking at the formula, we see that the kinetic energy is proportional to the square of the speed, $K\propto v^2$ .

c The kinetic energy of an object is expressed in watts. --> FALSE. Watts is the units for measuring power, while the kinetic energy is measured in Joules, the units for the energy.

d The kinetic energy of an object is a quantitative measure of its inertia. --> FALSE. The inertia of an object depends only on its mass, not on its speed.

e The kinetic energy of an object is always equal to the object’s potential energy. --> FALSE. The potential energy depends on the altitude from the ground, not from the speed, so the two energies can be different.

Learn more about kinetic energy:

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8 0

3 years ago

Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1200 km/s , measured relative to the

Radda [10]

Answer:

The maximum electrical force is $2.512\times10^{-2}\ N$ .

Explanation:

Given that,

Speed of cyclotron = 1200 km/s

Initially the two protons are having kinetic energy given by

$\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2$

When they come to the closest distance the total kinetic energy is converts into potential energy given by

Using conservation of energy

$mv^2=\dfrac{kq^2}{r}$

$r=\dfrac{kq^2}{mv^2}$

Put the value into the formula

$r=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{1.67\times10^{-27}\times(1200\times10^{3})^2}$

$r=9.57\times10^{-14}\ m$

We need to calculate the maximum electrical force

Using formula of force

$F=\dfrac{kq^2}{r^2}$

$F=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{(9.57\times10^{-14})^2}$

$F=2.512\times10^{-2}\ N$

Hence, The maximum electrical force is $2.512\times10^{-2}\ N$ .

8 0

3 years ago

A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it t

navik [9.2K]

Answer:

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

Explanation:

We have given initial velocity of the bus u = 24 m/sec

And final velocity v = 16 m/sec

Distance traveled in this process s = 50 m

From third equation of motion we know that $v^2=u^2+2as$

$16^2=24^2+2\times a\times 50$

$a=-3.2m/sec^2$

(a) Now as the bus finally stops so final velocity v = 0 m/sec

So $v^2=u^2+2as$

$0^2=24^2-2\times 3.2\times s$

s= 90 m

So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration $a=-3.2m/sec^2$

So time $t=\frac{v-u}{a}=\frac{0-24}{-3.2}=7.5sec$

7 0

3 years ago

write a motivational speech to someone who wants to become a scientist please write this as it is for kids :) PLEASEE HELP!!!!

Gekata [30.6K]

Answer:

A motivational speech for a to be scientist. Crazy experiments and more you have the right to be whatever you adore. Your young yet strong everything will come along. I believe in you and so should you since your becoming a scientist you’ll acquire it soon. You’ll discover new things and sight to see when you become a scientist make sure to call me. :)

Explanation:

Hope it’s good I made up a couple words that rhyme good luck!

6 0

3 years ago

Read 2 more answers

To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp

motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

$\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0$

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that $f_{max}=\mu N$

Now, we find the acceleration:

$\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}$

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

$x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s$

8 0

3 years ago