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Strike441 [17]
3 years ago
7

@alexrobin13 A 62 kg box is lifted 12 meters off the ground. How much work was done?

Physics
2 answers:
34kurt3 years ago
6 0
7291.2! I'm for sure this is the right answer.
nevsk [136]3 years ago
3 0
Given: 
weight=62 kgs
 height= 12 m
 gravity= 9.8 
 To find:
 work done= mass*gravity*height 
 solution: 
 w=62*12*9.8
 workdone=7291.2 J
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A car of mass 1500 kg is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s.
Lilit [14]

Answer:

Explanation:

a. The source of centripetal force on the car is  (3) the static friction force.

b. ac = v²/R = (20²)/50 = 8 m/s²

c.  Fc = m(ac) = 1500(8) = 12 kN

d. μ = Fc/N = Fc/mg = 12000 / 1500(9.8) = 0.8163... ≈ 0.82

6 0
3 years ago
Can someone explain to how to calculate this
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answer

option d is the correct answer

explanation

as we know frequency is equal to 1 /t

f= 457 Hz

t=1

SO, 1/457

=0.0022sev

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3 years ago
Is the voltage of two identical lamps the same?​
Dahasolnce [82]

Answer:

It depends if they have the same lightbulb in them.

Explanation:

8 0
2 years ago
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Fatima is watching her pet cat, Winter, napping in the sun. Fatima is curious about the heart rate of Winter when she is napping
svetoff [14.1K]

Answer:

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There are two hypotheses she could test:

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Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil
laila [671]

Answer:

W_n_e_t=7.648512 \approx 7.6J

K.E=0.8J

v=0.7844645406 \approx 0.78m/s

Explanation:

From the question we are told that

Mass of pitcher   M= 2.6kg

Force on pitcher f=8.8N

Distance traveled 48cm=>0.48m

Coefficient of friction \mu=0.28

a)Generally frictional force is mathematically given by

F=\mu N

F=0.28*2.6*9.8

F=7.1344N

Generally work done on the pitcher is mathematically given as

W_n_e_t=W_f+W_F

W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N

W_n_e_t=4.224-3.424512

W_n_e_t=0.799488\approx 0.8J

b)Generally K.E can be given mathematically as

K.E= W_n_e_t

Therefore

K.E=0.8J

c)Generally the equation for kinetic energy is mathematically represented by

K.E=1/2mv^2

0.8=1/2mv^2

Velocity as subject

v=\sqrt{\frac{K.E*2}{m} }

v=\sqrt{\frac{0.8*2}{2.6} }

v=0.7844645406 \approx 0.78m/s

6 0
3 years ago
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