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Sergio039 [100]
3 years ago
8

A large sphere has a mass of 175 kg and is suspended by a chain from the ceiling. The mass of the chain is 12.0 kg. What is the

tension in the chain?
1830, N1610, N1720, N1940 N.
Physics
1 answer:
Elina [12.6K]3 years ago
5 0

Each point in the chain supports the weight of all the mass below it.

At the bottom end of the chain, the weight is  (175 x 9.8) = 1,715 N .
At the top of the chain, the weight is  (175 + 12) x (9.8) = 1,833 N .

The tension in the chain varies linearly from  1,715N  at the bottom
to  1,833N  at the top.

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A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.
pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package.  Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

\sum F_y=m_1*a_m_i_n = T-m_1*g

If the package is barely lifted, that means that T=m_2*g; then:

\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g

Solving the equation for a_mín, we have:

a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: \sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a

For the package: \sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: \sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

m_1*a+m_1*g=m_2*g-m_2*a

Solving a, we have

(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2

We can then replace this value of a in one for the sums of force and find the tension T:

T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N

5 0
3 years ago
Which of the following is most useful to determine how much energy is being used by a circuit in a given amount of time?
user100 [1]

Answer:

The answer is A.

Explanation:

5 0
3 years ago
Winds that blow from the north and south poles
Gennadij [26K]
Winds that blow from the north and south poles would be called k<span>atabatic winds. I'm not sure if I spelled that right, but that's the answer I hope.</span>
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3 years ago
A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f
Zina [86]

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

A=\pi r^2

A =\pi (0.2)^2

A =0.1256 m^2

Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:

\mu=IA

\mu=2(0.1256)

\mu=0.25 A\cdot m^2

Therefore the magnetic dipole moment is 0.25A\cdot m^2

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You are an astronaut floating alongside your space ship. If you throw a baseball with a speed of 10 m/s, you will move at:
ZanzabumX [31]

Answer:

D

Explanation:

There is no friction to stop you from moving BC you are in space, however you have a larger mass than the ball, so it takes more force to get you up to the same speed as the baseball. You will move in the opposite direction of the ball because you exerted force on the ball in one direction and therefore yourself in the opposing direction.

4 0
3 years ago
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