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3 years ago

Energy that flows from an objective with a higher temperature to an object with a lower temperature

Physics

1 answer:

dalvyx [7]3 years ago

7 0

That's THERMAL energy, often referred to as "heat".

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The speed of an object in a set direction is called its what?

alisha [4.7K]

The speed of an object in a set direction is its velocity.

3 0

3 years ago

Read 2 more answers

A mother pats a child every time he throws a chocolate wrapper in the dustbin. This is an example of (A) Observational Learning

TEA [102]

D. I think, not sure

8 0

4 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

An object is pushed from rest across a sheet of ice, accelerating at 8.0 m/s^2 [E] over a displacement of 1.05 m [E]. The object

tatiyna

Answer:

$D_T=18.567m$

Explanation:

From the question we are told that:

Acceleration $a=8.0 m/s^2$

Displacement $d=1.05 m$

Initial time $t_1=6.0s$

Final Time $t_2=2.5s$

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

$V^2=2as$

$V=\sqrt{2*6*1.05}$

$V=4.1m/s$

Generally the equation for Distance traveled before stop is mathematically given by

$d_2=v*t_1$

$d_2=3.098*4$

$d_2=12.392$

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity $v_3=0 m/s$

Initial velocity $u_3=4.1 m/s$

Therefore

Using Newton's Law of Motion

$-a_3=(4.1)/(2.5)$

$-a_3=1.64m/s^2$

Giving

$v_3^2=u^2-2ad_3$

Therefore

$d_3=\frac{u_3^2}{2ad_3}$

$d_3=\frac{4.1^2}{2*1.64}$

$d_3=5.125m$

Generally the Total Distance Traveled is mathematically given by

$D_T=d_1+d_2+d_3$

$D_T=5.125m+12.392+1.05 m$

$D_T=18.567m$

6 0

3 years ago

A ball is dropped from 8.5 meters above the ground. If it begins at rest, how long does it take to hit the ground?

Anna35 [415]

Answer:

Explanation:

Givens

d = 8.5 meters

vi = 0

a = 9.81

t = ?

Formula

d = vi * t + 1/2 a t^2

Solution

8.5 = 0 + 1/2 9.81 * t^2 multiply both sides by 2

8.5 = 4.095 t^2 Divide both sides by 4.095

8.5/4.095 = t^2

1.7329 = t^2 Take the square root of both sides

t = 1.316

It takes 1.316 seconds to hit the ground.

6 0

3 years ago