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3 years ago

14

Random kinetic energy possessed by objects in a material at finite temperature. An object that feels hot has a lot of this.

2 answers:

gregori [183]3 years ago

3 0

Internal energy or thermal energy.

KATRIN_1 [288]3 years ago

3 0

Answer:

Thermal Energy

Explanation:

This is an example of kinetic energy. It results in an object having a temperature that can be measured. Also, thermal energy can be transferred from one object to another in the form of heat.

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The captain told the passengers that the plane is flying at 450 miles per hour. This information describes the plane's

marta [7]

The speed of the plane

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3 years ago

Use the information to answer the following question.

Nostrana [21]

Answer:

The answer is B. :)

Explanation:

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2 years ago

What quantity is represented by a unit called a newton (N)?

zavuch27 [327]

$\textsf {C. Force}$

$\textsf {The quantity which is represented by a unit called Newton (N)}\\\textsf {is Force.}$

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2 years ago

In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e

Zanzabum

Answer:

The velocity is $v_t = 0.02175 \ m/s$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 0.024 \ kg$

The initial speed of the bullet is $u_b = 1200 \ m/s$

The mass of the target is $m_t = 320 \ kg$

The initial velocity of target is $u_t = 0 \ m/s$

The final velocity of the bullet is is $v_b = 910 \ m/s$

Generally according to the law of momentum conservation we have that

$m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t$

=> $0.024 * 1200 + 320 * 0 = 0.024 * 910 + 320 * v_t$

=> $v_t = 0.02175 \ m/s$

3 0

3 years ago

A tank holds a 1.44-m thick layer of oil that floats on a 0.98-m thick layer of brine. Both liquids are clear and do not intermi

denpristay [2]

Answer:

Angle of ray makes with the vertical is 62.1 degree

Explanation:

As per the ray diagram we know that the angle of incidence on oil brine interface will be given as

$tan\theta_i = \frac{0.7}{0.98}$

$\theta_i = 35.5^0$

now by Snell'a law at that interface we have

$\mu_1 sin\theta_i = \mu_2 sin \theta_r$

now we will have

$1.52 sin35.5 = 1.40 sin\theta_r$

$\theta_r = 39.12^0$

now this is the angle of incidence for oil air interface

so now again by Snell's law we will have

$\mu_2 sin\theta_i' = \mu_{air} sin\theta$

$1.40 sin39.12 = 1 sin\theta$

$\theta = 62.1^0$

3 0

2 years ago