All categories

2 years ago

1.55 J of work was done on the object that moved 5 meters. How much force was applied to the

Physics

1 answer:

Firdavs [7]2 years ago

3 0

Answer:

Work=force x displacement

Force =work/ displacement

F=1.55/5

F=0.31N

Explanation:

You might be interested in

Which best describes the current atomic model?

Lorico [155]

<span>A: It is not an exact representation of the atom, but is close enough to be very useful.

Hope this helps!</span>

3 0

3 years ago

Read 2 more answers

Describe the relationship between kinetic energy and speed and give an example of how changing and object speed would affect its

Andreas93 [3]

Answer:

I'm not sure

Explanation:

I have had that question to Uchida c r go crew in to go be

6 0

2 years ago

A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.

NemiM [27]

Answer:

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, $\omega=2.8\ rad/s$

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

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

$K_t=\dfrac{1}{2}m(r\omega)^2$

$K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2$

The rotational kinetic energy of the ball is :

$K_r=\dfrac{1}{2}I \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2$

Ratio of translational to the rotational kinetic energy as :

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0

3 years ago

You want to make a ride so you do not want to exceed 1.1g’s, if the radius of the turns are 10m, then what is the maximum speed

Citrus2011 [14]

The maximum speed is 10.4 m/s

Explanation:

For a body in uniform circular motion, the centripetal acceleration is given by:

$a=\frac{v^2}{r}$

where

v is the linear speed

r is the radius of the circular path

In this problem, we have the following data:

- The maximum centripetal acceleration must be

$a=1.1 g$

where $g=9.8 m/s^2$ is the acceleration of gravity. Substituting,

$a=(1.1)(9.8)=10.8 m/s^2$

- The radius of the turn is

r = 10 m

Therefore, we can re-arrange the equation to solve for v, to find the maximum speed the ride can go at:

$v=\sqrt{ar}=\sqrt{(10.8)(10)}=10.4 m/s$

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

8 0

3 years ago

What is the maximum number of electrons an oxygen atom can hold in its outer energy level?

gulaghasi [49]

The answer is 8.
Hope I helped.

6 0

3 years ago