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larisa86 [58]
3 years ago
8

An object’s motion remains constant when acted upon by what?

Physics
1 answer:
igomit [66]3 years ago
6 0

Answer:

An outside force

Explanation:

Newton's law an object in motion stays in motion an object at rest stays at rest unless acted on by an outside force.

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Light rays in a material with index of refrection 1.29 1.29 can undergo total internal reflection when they strike the interface
deff fn [24]

Answer:

The second material's index of refraction is 1.17.

Explanation:

Given that,

Refractive index of the material, n = 1.29

Critical angle is 65.9 degrees.

We need to find the second material's index of refraction. We know that at critical angle of incidence, angle of refraction is equal to 90 degrees. Using Snell's law as:

n_1\sin \theta_c=n_2\sin (90)\\\\n_2=n_1\sin \theta_c\\\\n_2=1.29\times \sin (65.9)\\\\n_2=1.17

So, the second material's index of refraction is 1.17.

7 0
3 years ago
Match the correct term with the statement that is true about it:
icang [17]
ANSWER:
1. Cardiovascular Fitness
2. Muscular Fitness
3. Flexibility
4. Motivation
5. Money
6.
7. Family Behaviors
8. Understanding

is there another option for number 6? i could only find 7 word options
6 0
4 years ago
Read 2 more answers
Sound waves cannot carry energy through. A water B air C a mirror D a vacuum
Ber [7]
I looked up the question and got D- a vacuum
3 0
3 years ago
If a force of 12 N is applied to an object
Varvara68 [4.7K]

Answer:

<h3>The answer is 3 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{12}{4}  \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

6 0
4 years ago
A cable passes over a pulley. Because the cable grips the pulley and the pulley has nonzero mass, the tension in the cable is no
topjm [15]

Answer:

\alpha=214.8 rad/s^2

Explanation:

We are given that

F_1=137 N

F_2=43 N

Net force=F=F_1-F_2=137-43=94 N

Mass,m=1.21 kg

Radius,r=0.723 m

We have to find the magnitude of its angular acceleration.

Moment of inertia ,I=\frac{1}{2}mr^2

Substitute the values

Torque ,\tau=I\alpha

F_{net}\times r=\frac{1}{2}mr^2\alpha

\alpha=\frac{2F_{net}}{mr}

\alpha=\frac{2\times 94}{1.21\times 0.723}

\alpha=214.8 rad/s^2

4 0
3 years ago
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