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3 years ago

In terms of energy transfer, how is a rise in

Physics

1 answer:

KATRIN_1 [288]3 years ago

5 0

Answer:

Let's look deeper into it.

Explanation:

Heat energy corresponds to the transfer of thermal energy between two given objects. Temperature itself gives off the measurement of how much thermal energy any given object has.

The higher the measurement, the more heat there is. The lower the measurement, the less heat there is.

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10. A 25 kg apple cart is being pushed with an applied force of 115 N. The coefficient of friction between the ground and the ca

ziro4ka [17]

Answer:

1.1 m/s²

Explanation:

From the question,

F -mgμ = ma.................... Equation 1

Where F = applied force, m = mass of the apple cart, g = acceleration due to gravity, μ = coefficient of friction., a = acceleration of the apple cart.

Given: F = 115 N, m = 25 kg, μ = 0.35

Constant: g = 10 m/s²

Substitute these values into equation 2

115-(25×10×0.35) = 25×a

115-87.5 = 25a

25a = 27.5

a = 27.5/25

a = 1.1 m/s²

8 0

3 years ago

Which of the following biomes might be found at a latitude of 33° South?

cupoosta [38]

The answer is going be desert.

5 0

3 years ago

A layer of oil (n = 1.38) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid.

Vinil7 [7]

Answer:

Refractive index of unknown liquid = 1.56

Explanation:

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 65.0° )

${\theta_r}$ is the angle of refraction ( 53.0° )

${n_r}$ is the refractive index of the refraction medium (unknown liquid, n=?)

${n_i}$ is the refractive index of the incidence medium (oil, n=1.38)

Hence,

$1.38\times {sin65.0^0}={n_r}\times{sin53.0^0}$

Solving for ${n_r}$ ,

Refractive index of unknown liquid = 1.56

4 0

3 years ago

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

The horizontal component of the pulling force, $F cos \theta$ , where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the forward direction
The force of friction, $F_f$ , acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$