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adoni [48]
3 years ago
15

Please answer the question number 2 in this pic in detail

Physics
1 answer:
soldier1979 [14.2K]3 years ago
3 0
It would be C ............
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Descibe the real-world examples of Newton's third lawthat were idenified in "Applications of Newton's Laws."
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Answer:

A horse pulls a cart, a person walks on the ground

Explanation:

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Screws and wedges are modified blank
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This is what I know 
<span>Splitting securing A wedge is a triangular shaped tool, a compound and portable inclined plane, and one of the six classical simple machines. </span>
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You are in a planet where the acceleration due to gravity is known to be 3.28 m/s^2. You drop a ball and record that the ball ta
notsponge [240]
B) 7.87 m/s

The gravitational pull is the rate of change of velocity which is the acceleration. Formula for acceleration is;
a =  \frac{final \: velocity - initial \: velocity}{time \: taken}

Given:

• Initial velocity = 0m/s; I dropped the ball, and didn't throw it, so it was at rest firstly
• Time taken = 2.40s
• Acceleration = 3.28m/s^2

We're require to find the final velocity, at which the ball hit the ground with. Ignoring air resistance, keep in mind that the velocity of an object increases as it comes closer to the ground.

3.28  =  \frac{final \: velocity - 0}{2.40}
3.28 \times 2.40 = final \: velocity
final \: velocity = 7.872 \frac{m}{ {s} }

5 0
3 years ago
Which of the following has the largest m/mentum?
ASHA 777 [7]
A: The total building of Campbell high school, including the trailers and the construction area
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3 years ago
the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force
-Dominant- [34]

Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

F_{net} = 51 N

now we also know that friction force due to ground is given as

F_f = 24 N

now we have

F_{net} = F_{ext} - F_f

51 = F_{ext} - 24

F_{ext} = 75 N

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

a = \frac{F_f}{m}

a = \frac{24}{24} = 1 m/s^2

now we can find the distance by kinematics

v_f^2 - v_i^2 = 2 a d

0 - 1.5^2 = 2(-1)d

d = 1.125 m

so the distance moved will be 1.125 m

6 0
3 years ago
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