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Luda [366]
3 years ago
6

A uniform meter stick (with a length of 1.00 meter) has a mass of 108 g. It is supported at its midpoint by a vertical rigid blu

e rod (the two are connected by a frictionless hinge). A weight (mass 1) has a mass of 21g and it hangs from a (massless) string at a distance of 24 cm from the left end of the stick. Another weight (mass 2) has a mass of 27 g and it hangs from a string at a distance of 91 cm from the left end of the stick. The system was held motionless by the (vertical) blue rod and by a vertical red string at the right end of the stick (one meter from the left end of the stick). Red string Then somebody cut the red string. The blue rod still holds the center of the stick in place but the system can now rotate about the midpoint of the stick Immediately after the red string is cut.
Find the vertical tension in the blue rod.
1. 508 N
2. 143 N
3.153 N
4. 0.4704N
5. 0.7644N
6. 1.539 N

Physics
1 answer:
alexandr1967 [171]3 years ago
4 0

Answer:4

Explanation:

Given

m_1=21\ gm

m_2=27\ gm

Mass of stick is m=108\ gm

Let T be the tension in the red string

Now if the red string is cut , suppose T is the tension in the blue rod immediately after cut

Therefore

T=m_1g+m_2g

T=(0.021+0.027)\times 10

T=0.48\ N

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B)
Vlad1618 [11]

Answer:

a) 16m/s b) 192m

Explanation:

v1=32m/s a=-2m/s^2 t=8s v2=? d=??

a) I will use this equation v2= v1 + a*t

v2= 32m/s + -2m/s^2 * 8s

v2= 32m/s + -16m/s

v2= 16m/s

b) v2^2=v1^2 + 2ad

rearranging

v2^2-v1^2=2ad

v2^2-v1^2/2= a d

v2^2-v1^2/2a=d

16m/s^2 - 32m/s^2/ 2 x-2m/s^2 =d

d=192m

5 0
2 years ago
Preston tossed a red ball upward and it reaches a maximum height of 3.0. What is the final velocity when it returns to prestons
Leokris [45]
That will depend on the units of the 3.0. We need to know if it's 3 feet, 3 yards, 3 meters, or 3 miles. Each one will have a different answer.
5 0
3 years ago
CAN SOMEBODY PLEASE HELP ME! i need help and i wanna pass
umka21 [38]

Answer:it would be C

Explanation:

8 0
3 years ago
If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
The near point of an eye is 56.0 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan
irakobra [83]

Answer:

Explanation:

Near point = 56 cm .

near point of healthy person = 25 cm

person suffers from long sightedness

convex lens will be required .

object distance u  = 25 cm

image distance   v = 56 cm

both will be negative as both are in front of the lens.

lens formula

I/v - 1 / u = 1/f

- 1/56 +1/25 = 1/f

- .01785 + .04 = 1/f

1/f  = .02215

f = 45.15 cm .

4 0
3 years ago
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