How does a bicycles kinetic energy change when it slows down?

An ice block measuring 5cm by 5cm by 5cm has a desity of 5g/cm3. What is the mass of the ice block?

Svetradugi [14.3K]

Density = (mass) / (volume)

Volume of your block = (5cm x 5cm x 5cm) = 125 cm³

5 gm/cm³ = (mass) / (125 cm³)

Multiply each side by (125 cm³): Mass = 625 gm

Note:
I don't know what kind of frozen substance you're working with,
but maybe you ought to be careful handling it ... I know that's
not water ice you have there. The density of water ice is not
5 gm/cm³. In fact, it's a little less than 1 gm/cm³. That's why
the ice floats in your soda.

4 0

3 years ago

Read 2 more answers

A parachutist with a camera, both descending at a speed of 10.8 m/s, releases that camera at an altitude of 50 m. In this proble

Vadim26 [7]

Answer:

The velocity of the camera is 33.11 m/s.

Explanation:

Given that,

Speed = 10.8 m/s

Altitude = 50 m

Suppose determine the velocity of the camera just before it hits the ground?

We need to calculate the velocity of the camera

Using equation of motion

$v^2=u^2+2gs$

Where, v = final velocity of camera

u = initial speed of camera

s = distance

Put the value into the formula

$v^2=(-10.8)^2+2\times(-9.8)\times(-50)$

$v=\sqrt{1096.64}$

$v=33.11\ m/s$

The direction will be downward so it is the negative velocity.

Hence, The velocity of the camera is 33.11 m/s.

8 0

3 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

3 years ago

Which explains upwelling in the oceans?

Leokris [45]

Winds blowing across the ocean surface push water away. Water then rises up from beneath the surface to replace the water that was pushed away. This process is known as “upwelling.”

Upwelling occurs in the open ocean and along coastlines. The reverse process, called “downwelling,” also occurs when wind causes surface water to build up along a coastline and the surface water eventually sinks toward the bottom.

Water that rises to the surface as a result of upwelling is typically colder and is rich in nutrients. These nutrients “fertilize” surface waters, meaning that these surface waters often have high biological productivity. Therefore, good fishing grounds typically are found where upwelling is common.

7 0

3 years ago

a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after

STatiana [176]

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

7 0

3 years ago