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Elena L [17]
3 years ago
10

Assuming that the acceleration of an object due to gravity is ag = 10 m/s2, it is possible to find the distance an object falls

when dropped by using the equation ay = \frac{1}{2} ag t2. Using this value for gravity, how far would an object drop after 5.0 s?
105 m
125 m
135 m
115 m
Physics
1 answer:
qwelly [4]3 years ago
7 0

Answer:

125 m

Explanation:

i just did the edpuzzle

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Riders on a ferris wheel move in a circle with a speed of 4.0 m/s. As they go around, they
Vanyuwa [196]
Answer: D = 16m

Explanation: given values: a = 2 m/s2, v = 4 m/s
In this case we have to determine the diameter of the Ferris wheel.
Diameter of circle is given as: D = 2.r.
First we have to find radius of wheel. The best way to find that is using the centripetal acceleration equation: a = v2/r
Plug in values in above equation to find radius: 2 m/s2 = (4 m/s)2/r 2 m/s2 = (16 m2/s2)/r r = (16 m2/s2)/2 m/s2
r = 8.0m
Diameter of Ferris wheel is:
D = 2.r.
D = 2.8m
D = 16m
6 0
3 years ago
How is amplitude changed in an instrument or tuning fork
azamat

Amplitude is affected by the energy wave in the instrument. High energy wave means high amplitude and low energy wave means low amplitude.

<u>Explanation:</u>

The amplitude of a periodic variable is a measure of its change over a single period. There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.

The amount of energy carried by a wave is related to the amplitude of the wave. Amplitude of an instrument is directly affected by the wave of the energy in the instruments. High energy wave means high amplitude and low energy wave means low amplitude in the instrument.

7 0
3 years ago
Calculate the accleration of a car if its velocity increases from 15m/s to 75m/s in 5 second​
andrezito [222]

Answer:

I think the acceleration is 12m/s

3 0
3 years ago
Knowledge and skills learned through socialization are an example of
ziro4ka [17]

Answer:

I think no.2 the answer

Because socialization and social resources are both for me

3 0
3 years ago
A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
3 years ago
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