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Alex_Xolod [135]
2 years ago
13

A uniform meter stick balances on a fulcrum placed at the 40 cm mark when a weight W is placed at the 30 cm mark. What is the we

ight of the meter stick?

Physics
2 answers:
lora16 [44]2 years ago
6 0

Answer:

The answer is W

Explanation:

Since the metre stick is a metre rule, the length of the stick is 1m which is equivalent to 100cm.

Check the attachment for diagram.

Note that the weight of the metre rule will be positioned at the centre of the metre rule i.e at 50cm from one end of the metre rule as shown in the diagram.

Using the principle of moments to solve for W. It States that the sum of clockwise moments is equal to the sum of anticlockwise moments.

Since Moment = Force × perpendicular distance

Let the weight of the meter rule be M

Clockwise moment = M × 10

Anticlockwise moment = W × 10

Equating both to get M;

10M = 10W

M = W

This shows that the weight M of the metre rule is equal to the weight placed at the 30cm mark

statuscvo [17]2 years ago
4 0

Answer:

The weight of the meter rule is equal to the weight W placed at the 30cm mark

Explanation:

Detailed explanation and calculation is shown in the image

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The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s2. Neptune has mass 1.02×1026 kg and radiu
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Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N

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Explanation: Please see the attachments below

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a child hits a ball with a force of 350 N. (a) If the ball and bat are in contact for 0.12 is, what impulse does the ball receiv
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Explanation:

Given that,

Force with which a child hits a ball is 350 N

Time of contact is 0.12 s

We need to find the impulse received by the ball. The impulse delivered is given by :

J=F\times t\\\\J=350\times 0.12\\\\J=42\ N-m

So, the impulse is 42 N-m..

We know that he change in momentum is also equal to the impulse delivered.

So, impulse = 42 N-m and change in momentum =42 N-m.

7 0
2 years ago
Cassy shoots a large marble (Marble A, mass: 0.06 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marbl
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Conservation of linear momentum:

m*v inital = m*v final

0.06*0.7 + 0.03*0 = 0.06*(-0.2) + 0.03*v

(my algebra, or use ur calculator: 0.06*.07=0.042, etc ... or ur teacher may think you got some help)

0.06*(0.7+0.2)=0.03*v, v = 0.06*0.9/0.03=1.8 m/s

Answer 1.8 m/s (positive, to the right).

 

4 0
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A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas
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Answer:

v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

F=qE

F_p=F_e

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

\frac{m_ev_e^2}{2d}=qE

we can equals these expressions for both proton and electron, because the forces qE are the same:

\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

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2 years ago
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