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3 years ago

Find fq, the vertical force that pier q exerts on the right end of the bridge.

1 answer:

5 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
<span>View the entire bridge as the system, and find the torque about a pivot chosen to eliminate the torque from one of the unknowns
</span>Find the sum of the torques ∑ τ Q about the right pier Q of the bridge. Remember that counterclockwise torque is positive
∑τQ - 2MgL− 3FpL

<span>The sum of the torques is zero
</span>
FP = 2/3 Mg

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What is force of attraction ?

Otrada [13]

Answer

a force between two bodies, molecules, particles, or the like, that attracts each to the other. Gravity is the force of attraction between two bodies that have mass.

Explanation:

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3 years ago

Nerve impulses in a human body travel at a speed of about 100 m/s. Suppose a woman accidentally steps barefoot on a thumbtack. A

sasho [114]

T = 0.017 s

From the foot to the brain is almost the same as the height. We are not given the height of the woman, but to find "about" how much time, we need a height to work with.

She *could* be 1.7 m <- height = distance

Formula for speed, where k = speed, d = distance, t = time
k = d/t
Rearrange to solve for time:
t = d/k
Substitute known values:
t = (1.7 m) / (100 m/s)
Solve:
t = 0.017 s

Therefore, it takes about 0.017s for the impulse to travel from the foot to the brain.

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3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

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3 years ago

A comet orbits a star in a strongly elliptical orbit. The comet and star are far from other massive objects. As the comet travel

Kruka [31]

Answer:

the kinetic energy decreases and the potential energy Increases.

Explanation:

as the comet travels away from the star it gains an energy which it posses because of its position or state.once the comet moves away form the star the kinetic decreases until its lost all together to where the potential energy starts increasing.

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3 years ago

Read 2 more answers

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Answer:

$1.8\cdot 10^{-11} N$