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tia_tia [17]
2 years ago
11

A student attaches a block to a vertical spring so that the block-spring system will oscillate if the block-spring system is rel

eased from rest at a vertical position that is not the system’s equilibrium position. The system oscillates near Earth’s surface. The system is then taken to the Moon’s surface, where the gravitational field strength is nearly 16 that of the gravitational field strength near Earth's surface. Which of the following claims is correct about the period of oscillation for the system?
Physics
1 answer:
vodka [1.7K]2 years ago
4 0

Answer:

Time period of the motion will remain the same while the amplitude of the motion will change

Explanation:

As we know that time period of oscillation of spring block system is given as

T= 2\pi\sqrt{\frac{M}{k}}

now we know that

M = mass of the object

k = spring constant

So here we know that the time period is independent of the gravity

while the maximum displacement of the spring from its mean position will depends on the gravity as

mg = kx

x = \frac{mg}{k}

so we can say that

Time period of the motion will remain the same while the amplitude of the motion will change

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Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp
UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

6 0
2 years ago
The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,
KATRIN_1 [288]

Answer:

The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.

C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'

Explanation:

See attached picture.

3 0
2 years ago
You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af
djyliett [7]
I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
A: mgh_1+\frac{mv_1^2}{2}
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
W_f=F_f\cdot L
So, we know that the car is approaching the point B with the following amount of energy:
mgh_1+\frac{mv_1^2}{2}- F_fL
The law of conservation of energy tells us that this energy must the same as the energy at point B. 
The energy at point B is the sum of car's kinetic and potential energy:
B: mgh_2+\frac{mv_2}{2}
As said before this energy must be the same as the energy of a car approaching the loop:
mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL
Now we solve the equation for v_1:
v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\
v_1^2=39.23\\
v_1=\sqrt{39.23}=6.26\frac{m}{s}

4 0
2 years ago
Read 2 more answers
A sports car accelerates from 0 to 30 mph in 1.5 s. How long would it take to accelerate from 0 to 60 mph, assuming the power of
Crank

Answer:

6 s

Explanation:

given,

Sports car accelerate from 0 to 30 mph in 1.5 s

time taken to accelerate  0 to 60 mph = ?

The power of the engine is independent of velocity and neglecting friction

power =

P = constant  

the kinetic energy for 60 mph larger than this of 30 mph

 = \dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}

 = \dfrac{v_1^2}{v_2^2}

 = \dfrac{60^2}{30^2}

 = 4

gain in kinetic energy  = P x t

time = 4 x 1.5

       = 6 s

8 0
2 years ago
How much heat Q1 is transferred by 25.0 g of water onto the skin? To compare this to the result in the previous part, continue t
hodyreva [135]

Answer:

The heat transferred  from water to skin  is 6913.5 J.

Explanation:

Given that,

Weight of water = 25.0 g

Suppose that water and steam, initially at 100°C, are cooled down to skin temperature, 34°C, when they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a constant specific heat capacity c=4190 J/(kg°K) for both liquid water and steam.

We need to calculate the heat transferred  from water to skin

Using formula for stream

Q=mc\Delta T

Put the value into the formula

Q=25\times10^{-3}\times4190\times(373-307)

Q=6913.5\ J

Hence, The heat transferred  from water to skin  is 6913.5 J.

3 0
2 years ago
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