Which of the following statements about a metal wire in the steady state are true? Select all that apply.

Physics

1 answer:

nata0808 [166]3 years ago

5 0

Answer:

Options: 1, 3, 5, 7, 8

Explanation:

In steady state at the point when the current at each point in the circuit is

consistent (doesn't change with time).

In numerous viable circuits, the steady state is accomplished in a brief period of time.

In the steady state, the charge (or current) streaming into any point in the circuit needs to equivalent the charge (or current) streaming out which is Kirchhoff's Node or Current law.

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A block is sent up a frictionless ramp along which an x axis extends upward. The figure below gives the kinetic energy of the bl

ss7ja [257]

Kinetic energy =1/2 mv^2

m=2ke/v^2 

m=2(34)/3.6^2 

m=5.24 

force normal = mg 
=5.24 x 9.8 
force normal = 51.4N

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5 0

3 years ago

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: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$