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In-s [12.5K]
3 years ago
13

30 points plz help ill do anything... literally anything.

Physics
1 answer:
ruslelena [56]3 years ago
3 0

Answer:

1. 2.5s

Explanation:

1. For time, divide Distance / speed

25m / 10

=2.5s

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A basketball player throws a ball at 60° to the horizontal at a net 4.0m away. The height of the net is 3.3m. The ball leaves hi
Hoochie [10]
Excellent work!
Your calculations are correct, but near the end, you have forgotten to cancel u.  If you cancel the u, the linear term becomes a constant, and the resultant equation becomes a simple quadratic which is much easier to solve.
I get u=7.468 m/s using g=9.81 (as you did)
You will rework and should get 7.468 m/s as well.

Congrats for the good work!

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A block of mass 10 kg moves from position A to position B shown in the figure above. The speed of the block is 10 m/s at A and 4
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We have that the block is moving horizontally. Hence, its potential energy due to gravity stays the same. The only change in its mechanical energy is the one due to the change of speed. This reduction of its kinetic energy, due to the conservation of energy, is equal to the work that friction does. We have that at A the kinetic energy is : K=1/2*m*u^2=10*10*10/2=500J. At B, we have that K=1/2*10*16=80J. Sine we have that the initial value is 500, the work from the friction force (opposite to the movement of the object) is 80-500=420J.
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In which stage of life will the sun undergo the most change?.
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Between the two asymptotic gigantic branches, the Sun changes the greatest in size, brightness, and temperature.
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A 12.0kg microwave oven is pushed 14.0m up the sloping surface of a loading ramp inclined at an angle 37 degrees above the horiz
Semenov [28]

Answer:

Answered

Explanation:

a) What is the work done on the oven by the force F?

W = F * x

W = 120 N * (14.0 cos(37))

<<<< (x component)

W = 1341.71

b) F_f=\mu_k N

F_f=0.25\times12\times9.8

= 29.4 N

W_f= F_f\times x

W_f= 29.0\times 14 cos(37)

W_f= 328.72 J = 329 J

c) increase in the internal energy

U_2 = mgh

= 12*9.81*14sin(37)

= 991 J

d) the increase in oven's kinetic energy

U_1 + K_1 + W_other = U_2 + K_2

0 + 0 + (W_F - W_f ) = U_2 + K_2

1341.71 J - 329 J - 991 J = K_2

K_2 = 21.71 J

e) F - F_f = ma

(120N - 29.4N ) / 12.0kg = a

a = 7.55m/s^2

vf^2 = v0^2 + 2ax

vf^2 = 2(7.55m/s)(14.0m)  

V_f = 14.5396m/s

K = 1/2(mv^2)

K = 1/2(12.0kg)(14.5396m/s)

K = 87.238J

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