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2 years ago

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Please Help! A ball is thrown straight up from the ground. What way does its acceleration point at the top?

1 answer:

QveST [7]2 years ago

3 0

Answer: A. Vertical

Explanation:

If the ball is thrown straight up from the ground (asuming the ground as height 0), this means its initial velocity is greater than zero.

While the ball rises and gains height, its velocity decreases until it reaches its maximum height where it stops (velocity equal to zero) and then it begins to fall to the ground.

Now, <u>during all the movement, the ball has an acceleration</u>, which is the acceleration due gravity (9.8 m/s^{2} on Earth), even at the top or maximum height (when the ball stops just for fraction of time), and this acceleration points vertical and downward to the Earth's center.

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Help please, what does gravitational force do according to these answer choices?

laiz [17]

Answer:

the answer is C

Explanation:

gravity forces down not up or sideways.

5 0

2 years ago

How many types of electrical charge are there in all materials and what are the charges

JulsSmile [24]

There are two types of electric charges; positive and negative

- If you need more info than this let me know

- hope this helps

4 0

3 years ago

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

suter [353]

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

$F=G\frac{m_{E}m}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

$m$ are the mass of each communications satellite

$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

$F=mg$ (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

Isolating $g$ :

$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

Finally:

$g=0.223 m/s^{2}$

5 0

2 years ago

Q6) A speed skater moving to the left across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She slows steadil

77julia77 [94]

Answer:

-2.8 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity The S. I unit of acceleration is m/s²

Using the equation of motion,

v² = u² + 2as................... Equation 1

Where v = Final velocity, u = initial velocity, a = acceleration, s = distance,

Given: v = 6.0 m/s, u = 8.0 m/s, s = 5.0 m.

Substituting into equation 1

6² = 8²+2(a)5

36 = 64 + 10a

10a = 36-64

10a = -28

10a/10 = -28/10

a = -2.8 m/s²

Note: a is negative because because the skater decelerate on the rough ice

Hence the magnitude of her acceleration is = -2.8 m/s²

6 0

2 years ago

A5 kg frisbee is thrown from rest to a final speed of 12 m/s. What is the impulse of the frisbee?

lesya692 [45]

Answer:

60kgm/s

Explanation:

Given parameters:

Mass of frisbee = 5kg

Final speed = 12m/s

Unknown:

Impulse of the frisbee = ?

Solution:

The impulse of the frisbee is the same as the change in momentum.

It is given as:

Impulse = mass (Final velocity - Initial velocity)

Impulse = 5(12 - 0) = 60kgm/s

3 0

2 years ago