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2 years ago

Answer the discussion question in complete sentences:

Physics

1 answer:

zhuklara [117]2 years ago

3 0

Earth is divided into two hemispheres/coast. Which is south, north, east, and west.
California is west coast.

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A solid weighs 16.5N on the surface of the moon. The force of gravity on the moon is 1.7N/Kg.

Fiesta28 [93]

Answer:

mass = 9.7 kg

Explanation:

Weight = Mass x Acceleration due to gravity (g)

16.5 = mass x 1.7

mass = $\frac{16.5}{1.7}$ = 9.7 kg

3 0

2 years ago

How many resistors are found in this circuit? A) 0 B) 1 C) 2 D) 3

Dmitriy789 [7]

Show us the pictures I don't see it

6 0

2 years ago

Two particles with the same charge will repel each other. True or false

FinnZ [79.3K]

True. Think of a magnet and how they only connect to the opposite charges.

7 0

3 years ago

A stone on ground is zero energy

NNADVOKAT [17]

Answer:

A stone on the ground does not have zero energy…there is an internal potential in every object. Aldo is not in action or in any mechanical motion it is being acted upon by gravity and also molecular forces and energy.

Hope this helps !

7 0

3 years ago

A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar

sesenic [268]

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

$v^2-u^2 = 2gd$

where

u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

h = 24 m is the distance covered

Solving for h,

$v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s$

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

$y_1 = h - \frac{1}{2}gt^2$

While the vertical position of the ball thrown upward is

$y_2 = ut - \frac{1}{2}gt^2$

The two balls meet when

$y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s$

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

$y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m$

So the distance below the top of the cliff is

$d=24.0 - 18.0 = 6.0 m$

4 0

3 years ago