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kiruha [24]
3 years ago
7

A 1.0-kg mass (mA) and a 6.0-kg mass (mB) are attached to a lightweight cord that passes over a frictionless pulley. The hanging

masses are free to move. Find the acceleration of the larger mass.
Physics
2 answers:
RideAnS [48]3 years ago
8 0

Answer:

Explanation:

Given

m_1=1\ kg

m_2=6\ kg

Pulley mass String

For m_1=1\ kg

T-m_1g=m_1a

T=m_1(g+a)

for other body m_2

m_2g-T=m_2a

T=m_2(g-a)

Equating value of Tension

m_2=m_1\times \frac{g+a}{g-a}

6=\frac{g+a}{g-a}

6(10-a)=10+a

50=7a

a=\frac{50}{7}

a=7.142\ m/s^2

IrinaVladis [17]3 years ago
8 0

Answer:

Answer:

7 m/s^2

Explanation:

mA = 1 kg

mB = 6 kg

Let a be the acceleration and T be the tension in the string

By use of Newton's second law

T - mA g = mA x a  .... (1)

mBg - T = mB x a ..... (2)

Adding both the equations

(mB - mA) g = (mA + mB) x a

(6 - 1) x 9.8 = (6 + 1) x a

7 a = 9.8 x 5

a = 7 m/s^2

Thus, the acceleration in the system is 7 m/s^2.

Explanation:

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Answer:

e. 400 Hz

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I have attached the image of the rod.

We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.

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Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.

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