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dlinn [17]
3 years ago
14

When velocity is graphed with respect to time, what does the area under

Physics
2 answers:
denpristay [2]3 years ago
6 0
The Answer

Is Displacement
Juliette [100K]3 years ago
4 0

Answer:

the answer is D

Explanation:

Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object.

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George is applying a downward force of 50N to and object that has a mass of 50kg. What is the normal force (FN) of the object wh
attashe74 [19]

The normal force acting on the object is 500 N in the upward direction

<u>Explanation:</u>

As George is applying a downward force, the normal force will be in the upward direction. The normal force will be exerted due to the acceleration due to gravity exerted on the object.

So, as per Newton's second law, the normal force acting on the object can be measured by the product of mass of the object and the acceleration due to gravity acting on the object.

But as the acceleration due to gravity is a downward acting acceleration and the normal force is a upward acting force, so the acceleration will be having a negative sign in the formula.

Normal\ force = Mass \times Acceleration\ due\ to\ gravity

Here, acceleration due to gravity g = -10 m/s² and mass is given as 50 kg, then

Normal force = 50 × (-10) = -500 N

So, the normal force acting on the object is 500 N in the upward direction.

3 0
3 years ago
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th
Ket [755]

Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

5 0
4 years ago
A car initially at rest, accelerates at a constant rate of 4.0 m/s for 6s. How fast will the car be traveling at 6s
Katen [24]
It will be traveling exactly 24 miles per hour <span />
7 0
3 years ago
A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.
zzz [600]

Answer:

1000 Hz

Explanation:

<em>The frequency would be 1000 Hz.</em>

The frequency, wavelength, and speed of a wave are related by the equation:

<em>v = fλ ..................(1)</em>

where v = speed of the wave, f = frequency of the wave, and λ = wavelength of the wave.

Making f the subject of the formula:

<em>f = v/λ.........................(2)</em>

Also, speed (v) = distance/time.

From the question, distance = 900 m, time = 3.0 s

Hence, v = 900/3.0 = 300 m/s

Substitute v = 300 and λ = 0.3  into equation (2):

f = 300/0.3 = 1000 Hz

6 0
3 years ago
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