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lina2011 [118]
4 years ago
12

Can you please help me???

Physics
1 answer:
zhenek [66]4 years ago
8 0
Question 12 would be: C

question 13 would be: B
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During a drive by golfer Tiger Woods, the angular velocity of his club is zero at the top of his backswing and 30 rad/s at the b
ddd [48]

Explanation:

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4 years ago
Which has more volume feathers or rocks
Nataly_w [17]
Volume doesn't depend on what the substance is, only on how much of it there is.
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3 years ago
The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo
Lina20 [59]
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
At point A, all this energy has converted into kinetic energy, which is:
K= \frac{1}{2}mv^2
And since K=7.35 J, we can find the velocity, v:
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3 0
3 years ago
a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an
Mademuasel [1]
We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
T_1 - W_p - W_s + T_2 =0
where
W_p = (90kg)(9.81 m/s^2)=883 N is the weight of the person
W_s = (200kg)(9.81 m/s^2)=1962 N is the weight of the scaffold
Re-arranging, we can write the equation as
T_1 = 2845 N-T_2 (1)

2) Equilibrium of torques:
T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using W_p = 883 N and replacing T1 with (1), we find
2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0
from which we find
T_2 = 1128 N

And then, substituting T2 into (1), we find
T_1 = 1717 N
8 0
3 years ago
A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c
mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

7 0
4 years ago
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