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VikaD [51]
2 years ago
9

Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushe

s with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible.
Physics
1 answer:
Goryan [66]2 years ago
8 0

Answer:

The second child must exert a force of magnitude 23.3N to keep the door from moving.

Explanation:

We have to find the moment that the first child exerts and then match it to that exercised by the second child.

F1= 17.5N

d1= 0.6m

F2= ?

d2= 0.45m

M= F * d

M1= 17.5N * 0.6m

M1= 10.5 N.m

M1=M2

M2= F2 * 0.45m

10.5 N.m= F2 * 0.45m

10.5 N.m/0.45m = F2

F2=23.3 N

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A 70kg man runs at a pace of 4 m/s and a 50g meteor travels at 2 km/s. Which has the most kinetic energy?.
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4 0
2 years ago
While traveling from Boston to Hartford, Person A drives at a constant speed of 55 mph for the entire trip. Person B drives at 6
Ne4ueva [31]

Answer:

B will take 1.034 times the time of A from Boston to Hartford.

Explanation:

Let the distance from Boston to Hartford be S.

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Time taken by person A

             t_A=\frac{S}{55}

Person B drives at 65 mph for half the distance and then drives 45 mph for the second half of the distance.

Time taken by person B

            t_B=\frac{\frac{S}{2}}{65}+\frac{\frac{S}{2}}{45}=\frac{S}{130}+\frac{S}{90}=\frac{220S}{130\times 90}=\frac{11S}{585}

Ratio of time of arrival of B to A

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B will take 1.034 times the time of A from Boston to Hartford.

8 0
2 years ago
A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha
Alecsey [184]

Answer:

<h2>9.375Nm</h2>

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given  F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ  = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm

6 0
2 years ago
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