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MAXImum [283]
3 years ago
11

A ball is rolled uphill a distance of 5 meters before it slows, stops, and begins to roll back. The ball rolls downhill 9 meters

before coming to rest against a tree. What is the amount of displacement of the ball?
Physics
1 answer:
Zepler [3.9K]3 years ago
6 0
The magnitude of the ball's displacement when the ball is rolled uphill with a 5 meters before it slows, stops and begins to roll back and the ball rolls downhill 9 meters before coming to rest is 4 meter. Just subtract 9 meters with 5 meters. Therefore the answer is 4 meters. 
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What two factors affect the rate of acceleration of an object?
Masja [62]

For help with this answer, we look to Newton's second law of motion:

     Force = (mass) x (acceleration)

Since the question seems to focus on acceleration, let's get
'acceleration' all alone on one side of the equation, so we can
really see what's going on.

Here's the equation again:

                                                 Force = (mass) x (acceleration)

Divide each side by 'mass',
and we have:                            Acceleration = (force) / (mass) .

Now the answer jumps out at us:  The rate of acceleration of an object
is determined by the object's mass and by the strength of the net force
acting on the object.


5 0
3 years ago
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg
IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

M = \frac{q}{p}

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m

Now using thin lens formula:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\

<u>f = 1 m</u>

6 0
3 years ago
Arunning back practices for his upcoming football game by running drills. He runs forward 62.4 yards, and runs backward for
Mama L [17]

Answer: He has only move 0.2 yards

Explanation: When you subtract 18.3 from 18.5 you get 0.2 and that is how much he's moved

4 0
3 years ago
A communications channel has a bandwidth of 4,000 hz and a signal-to-noise ratio (snr of 30 db. what is the maximum possible dat
levacccp [35]
Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as 

C = B × log₂(1 + S/N)

where B is the bandwidth of the channel and S/N is its signal-to-noise ratio.

Since the given SN ratio is in decibels, we must first express it as a ratio with no units as

SN (in decibels) = 10 × log (S/N)
30 = 10log(S/N)
log(S/N) = 3
S/N = 10³ = 1000

Now that we have S/N, we can solve for its capacity (in bits per second) as 

C = 4000 × log₂(1 + 1000) 
C = 39868.91 bps

Thus, the maximum capacity of the channel is 39868 bps or 40 kbps.

Answer: 40 kbps

 
4 0
3 years ago
A sound wave is an example of an electromagnetic wave in nature. true or false?
Natasha2012 [34]
False. A sound wave is an example of a mechanical wave, not an EM wave.

Hope this helps! :)
5 0
3 years ago
Read 2 more answers
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