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MAVERICK [17]
3 years ago
8

Which example best matches the term refraction? (17 Points)

Physics
1 answer:
Katyanochek1 [597]3 years ago
8 0

Answer:

B

Explanation:

Just answerd that question!!!

<u><em>Hope it helped!!!</em></u>

.

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The first Lunar Olympics is to be held on the moon inside a huge dome. Of the usual Olympic events track, field, swimming, gymna
mars1129 [50]
I think track would be most affected.
4 0
4 years ago
The velocity of a an object in linear motion changes from +25 meters per second to +15 meters per second in 2.0 seconds.
kodGreya [7K]

Answer:

-5.0 m/s^2

Explanation:

Missing question:

What is the object's acceleration?

Solution:

The acceleration of an object is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time taken for the change in velocity

For the object in this problem,

u = +25 m/s

v = +15 m/s

t = 2.0 s

Substituting,

a=\frac{15-25}{2}=-5.0 m/s^2

And the acceleration is negative because its direction is opposite to that of the velocity.

5 0
3 years ago
How many times can mechanical energy become chemical energy
maria [59]

Answer:

Chemical energy, Energy stored in the bonds of chemical compounds. Chemical energy may be released during a chemical reaction, often in the form of heat; such reactions are called exothermic.

3 0
3 years ago
What happens when electrical devices are added to the circuit ?
Ksivusya [100]

Answer:

they get power.

Explanation:

7 0
3 years ago
The 15 g head of a bobble-head doll oscillates in SHM at a frequency of 4.0 Hz.
Goshia [24]

Answer:

(a) 9.375 N/m

(b) 0.5024 m/s

(c) 0.01 kg/s

Explanation:

mass of head, m = 15 g = 0.015 kg

frequency, f = 4 Hz

Time period, T = 1 / f = 0.25 s

Let k is the spring constant.

(a)

The formula for the time period is

T=2\pi\sqrt{\frac{m}{K}}

0.25=2\times 3.14 \sqrt{\frac{0.015}{K}}

0.04=\sqrt{\frac{0.015}{K}}

K = 9.375 N/m

(b)

Amplitude, A = 2 cm

Let ω is the angular velocity.

Maximum velocity, v = A ω = A x 2πf

v = 0.02 x 2 x 3.14 x 4 = 0.5024 m/s

(c)

Let b is the damping constant.

A(t = 4s) = 0.5 cm

Ao = 2 cm

Using the formula of damping

\frac{A}{A_{0}}=e^{-\frac{bt}{2m}}

\frac{0.5}{2}}=e^{-\frac{b\times 4}{2\times 0.015}}

0.25=e^{-133.3 b}

Taking natural log on both the sides

ln (0.25) = - 133.3 b

- 133.3 b = - 1.386

b = 0.01 kg/s

3 0
3 years ago
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