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dem82 [27]
3 years ago
8

The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone'

s velocity just before it strikes the ground A –1.75 meters/secondB –18 meters/secondC –79 meters/secondD –636 meters/second
Physics
1 answer:
Helen [10]3 years ago
5 0
We can use 3rd equation of motion to find the velocity of the stone just before it strikes the ground.

Height = S = 318 meters
Acceleration = a= 9.8 m/s²
Initial velocity = u = 0
Final velocity = v = ?

According to the 3rd equation of motion:

2aS = v² - u²

2(9.8)(318)=v²

v²=6232.8

⇒

v = 78.95 m/s

So, the velocity rounded of to nearest integer will be 79 meters per second.

Thus, C option is the correct answer
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Lorico [155]

Answer:

Ohms law states that current is directly proportional to the potential difference across the ends of a conductor provided that temperature and other factors kept constant

V=I(R)

6 0
3 years ago
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai
zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

8 0
3 years ago
A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to
irinina [24]

When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

<h3>Frictional force between the block and the horizontal surface</h3>

The frictional force between the block and the horizontal surface is determined by applying Newton's law;

∑F = ma

F - Ff = ma

Ff = F - ma

Ff = 4 - 2(1.2)

Ff = 4 - 2.4

Ff = 1.6 N

When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;

F - Ff = ma

5 - 1.6 = 2a

3.4 = 2a

a = 3.4/2

a = 1.7 m/s²

Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

Learn more about frictional force here: brainly.com/question/4618599

8 0
2 years ago
The ideal gas law includes all of the following properties, except the
Verizon [17]
The correct answer among the choices is option D. Density is not one of the properties included in the ideal gas law. The law is expressed as: PV=nRT. As we can see, the pressure, the volume and the temperature of the gas are included in the law.
3 0
3 years ago
Which law is associated with inertia
N76 [4]

Answer:

The focus of Lesson 1 is Newton's first law of motion - sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

4 0
3 years ago
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