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almond37 [142]
3 years ago
7

A small block of mass m on a horizontal frictionless surface is attached to a horizontal spring that has force constant k. The b

lock is pushed against the spring, compressing the spring a distance d. The block is released, and it moves back and forth on the end of the spring. During its motion, what is the maximum speed of the block?
Physics
1 answer:
Rashid [163]3 years ago
5 0

Answer:

v_{max} = |d|\cdot \sqrt{\frac{k}{m} }

Explanation:

The maximum speed of the block occurs when spring has no deformation, that is, there is no elastic potential energy, which can be remarked from appropriate application of the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot d^{2} = \frac{1}{2}\cdot m \cdot v^{2}

k \cdot d^{2} = m\cdot v^{2}

v_{max} = |d|\cdot \sqrt{\frac{k}{m} }

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Three joules of work is needed to shift 10 C of charge from one place to another. The potential difference between the places is
dimaraw [331]

Answer:

The potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

Explanation:

Given

Work done W = 3J

Amount of Charge q = 10C

To determine

We need to determine the potential difference V between the places.

The potential difference between the two points can be determined using the formula

Potential Difference (V) = Work Done (W) / Amount of Charge (q)

or

\:V\:=\:\frac{W}{q}

substituting W = 3 and q = 10 in the formula

V=\frac{3}{10}

V=0.3 V

Therefore, the potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

4 0
2 years ago
Please help me 5-11 help help
Illusion [34]
Your answer is -6 did that answer your question
5 0
3 years ago
Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and
tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1)  </u>

dT/dx= (-20-50)/0.35==> -280 K/m

 q_x  =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (3)

q_x  =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x  =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x  =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0
3 years ago
How to understand that an inductor behaves like short circuit and capacitor like an open circuit in steady state in a network?
garik1379 [7]
It depends on the steady-state frequency. At zero frequency an inductor behaves like an open circuit. As the frequency increases, the inductor acts more like an open circuit and a capacitator acts more like a short circuit
8 0
3 years ago
Two machines do thae same may have different power why
marissa [1.9K]

The quantity work has to do with a force causing a displacement. Work has nothing to do with the amount of time that this force acts to cause the displacement. Sometimes, the work is done very quickly and other times the work is done rather slowly. For example, a rock climber takes an abnormally long time to elevate her body up a few meters along the side of a cliff. On the other hand, a trail hiker (who selects the easier path up the mountain) might elevate her body a few meters in a short amount of time. The two people might do the same amount of work, yet the hiker does the work in considerably less time than the rock climber. The quantity that has to do with the rate at which a certain amount of work is done is known as the power. The hiker has a greater power rating than the rock climber.

Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation.


Power = Work / time


or


P = W / t

5 0
3 years ago
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