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Aleks04 [339]
3 years ago
6

How does 'g' vary from place to place?​

Physics
1 answer:
r-ruslan [8.4K]3 years ago
8 0

Explanation:

The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.

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A 2,000 kg car is parked at the top of a 30 m high hill. what is its potential energy?
Phoenix [80]
The PE for this question will be 588,000 because we take the mass (2,000 kg), multiply it by 9.8 which is Gravitational Acceleration and then multiply that by the height (30 meters)
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Determine the mechanical energy of this object a 1-kg ball rolls on the ground at <br> m/s
dedylja [7]
Mechanical energy = potential energy + kinetic energy
The ball is on the ground so it has no potential energy. that's all i know.
8 0
3 years ago
Please help quickly!
Naddik [55]

Answer:

○ D

Explanation:

A resistor is a passive electrical component that uses a circuit element with electric resistance. Resistors are used in electronics for reducing current flow, changing signal rates, dividing voltages, biasing active components, and terminating columns, among other applications.

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3 years ago
Uses of concave lens​
natita [175]

Answer:

Concave Lens Uses. Telescope and Binoculars Spectacles Lasers Cameras FlashlightsPeepholes. ...

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Concave lens used in peepholes.

3 0
2 years ago
A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th
Andre45 [30]

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

  • A uniform motion along the horizontal (x) direction
  • A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

s=u_y t + \frac{1}{2}at^2

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

u_y is the initial vertical velocity of the ball, which is given by

u_y = u sin \theta

where

u = 23.5 m/s is the initial velocity

\theta=33.5^{\circ} is the angle of projection

Substituting,

u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity, downward

Substituting everything into the equation we get:

-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s

Therefore, the horizontal distance covered in a time t is

d=v_x t

And substituting t = 3.59 s, we find

d=(19.6)(3.59)=70.4 m

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0
3 years ago
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