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Tanzania [10]
3 years ago
15

You hold a barbell in a horizontal position. The barbell’s center of mass is exactly mid-way between your two hands. A friend wa

lks up and attaches a weight on the end near your left hand. How does the torque on the bar exerted by your left hand compare to the torque on the bar from your right hand if the bar remains horizontal and at rest?

Physics
1 answer:
Harrizon [31]3 years ago
6 0

Explanation:

First consider that each hand works as a fulcrum: a pivot point where the barbell can rotate.

Now consider only the left hand. If the center of mass of the barbell is between hands (in the middle) it is displaced respect the fulcrum, therefore the weight which is pushing the bar downwards becomes a rotational force. The same thing happens to the other hand. Now, if more weight is added to the left hand the center of mass is displaced towards the left hand and depending how much weight is added, the center of mass will change its position and therefore the torque each hand experiences changes.

If the center of mass is still between hands: The torque remains almost the same changing only the magnitudes but not the direction.

If the center of mass is on the hand: there is no torque for the left hand because there is no leaver.

If the center of mass is to the left: now the torque changes direction and both hands need to stop it in the same direction.

(see diagram below)

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uysha [10]

Answer:

Time=8.23880597 seconds

Explanation:

Quantity of charge(q)=2.76c

Current(I)=0.335A

Time(t)=?

t=q/I

t=2.76/0.335

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7 0
3 years ago
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the
Novay_Z [31]

Answer:

It will be cut in half

Explanation:

The diffraction of a slit is given by the formula

a sin θ = m where

a = width of the slit,

λ = wavelength and

m = integer that determines the order of diffraction.

Next we divide both sides by a, we have

sin θ = m λ / a

Also, recall that

a’ = 2 a

Then we substitute in the previous equation

2asin θ' = m λ, if divide by 2a, we have

sin θ' = (m λ / 2a).

Now again, from the first equation, we said that sin θ = m λ / a, so we substitute

sin θ ’= sin θ / 2

Then we use trigonometry to find the width, we say

tan θ = y / L

Since the angle is small, we then have

tan θ = sin θ / cos θ

tan θ = sin θ, this then means that

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we will then substitute

y’ / L = y/L 1/2

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4 0
3 years ago
A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact w
aliya0001 [1]

Answer:

   f = 878,080 N

Explanation:

mass of pile driver (m) = 2100 kg

distance of pile driver to steel beam (s) = 5 m

depth of steel driven (d) = 12 cm = 0.12 m

acceleration due to gravity (g0 = 9.8 m/s^{2}

calculate the average force exerted on the pile driver by the beam.

  • from work done = force x distance
  • work done = change in potential energy of the pile driver
  • equating the two equations above we have

               force x distance = m x g x (s - d)

              f x 0.12 = 2100 x 9.8 x (5- (-0.12))

              d = - 0.12 because the steel beam went down at we are taking its  

              initial position to be an origin point which is 0

              f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12

                   f = 878,080 N

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