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Deffense [45]
4 years ago
6

A 0.060 kg ball hits the ground with a speed of -32m/s. The ball is in contact with the ground for 45 milliseconds and the groun

d exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?
A. 9.3m/s
B.12m/s
C.41m/s
D.73m/s
Physics
2 answers:
Dominik [7]4 years ago
9 0
Using the kinematic equation V=Vo + at, the initial velocity (Vo) is -32, the acceleration (a) is found by dividing force by mass 55N/.06kg=916.66 m/s/s, and the time is 0.045 sec.

V=Vo+at
V=-32+916.66*.045
V=9.25

The answer is A.
aleksklad [387]4 years ago
7 0

it's A. i took the test right now

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Consider the possible scenarios and determine which of them are true for positive feedbacks or negative feedbacks.A. Increased p
avanturin [10]

Answer:

Scenario A, B and E is True.

Explanation:

Scenario A) True. Removing carbon dioxide from atmosphere decreases greenhouse effect of atmosphere. Thus, temperature rise decreases.

Scenario B) True. The more evaporation creates the more greenhouse effect. Therefore, temperature rise increases.

Scenario C) False. Removing carbon dioxide from atmosphere decreases greenhouse effect of atmosphere. Thus, temperature rise decreases.

Scenario D) False. The more evaporation creates the more greenhouse effect. Therefore, temperature rise increases.

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5 0
3 years ago
A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet
astraxan [27]

Answer:

Acceleration = 2.35 m/s^{2}

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = \frac{mg}{2} = 0.5 mg

Friction along incline = umg cos30° = mg 0.3\times \frac{\sqrt{3}}{2}

Friction along incline  = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = \frac{net force}{mass} = 0.24 g = 2.35 m/s^{2}

The height of incline = 8 m

Length of the inclined edge = 16 m

v^{2}=u^{2}+2as

v^{2}= 2\times 0.24 \times 9.8\times 16

v= 8.67 m/s

5 0
4 years ago
Calculate the work done by an applied force of 76.0 N on a crate for the following. (Include the sign of the value in your answe
telo118 [61]

Answer:

a) 400.4Joules

b) 262.69Joules

Explanation:

Work is said to be done if the force applied to an object cause the object to move through a distance

Workdone = Force × Distance

Given

Force = 76N

Distance= 5.2m

Work done = 77 × 5.2

Work done = 400.4Joules

b) If the force is exerted at an angle of 41°

Work done = Fdsin theta

Work done = 77(5.2)sin41

Work done = 400.4sin41

Work done = 262.69Joules

6 0
3 years ago
A family communicating using a cell phone at dinner. Using a cell phone is more addicted to technology things than missed opport
tatyana61 [14]

Answer:

The family may be addicted to technology by using their cellphones to communicate is still communicating in a way, but it would be better to use social face- to - face interactions.

Explanation:

4 0
3 years ago
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A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s,
Andreyy89

Answer:

-450 m/s

Explanation:

Momentum is conserved.

p₀ = p

0 = (1.5 kg) (1.5 m/s) + (0.005 kg) v

v = -450 m/s

6 0
3 years ago
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