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dimulka [17.4K]
3 years ago
11

A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest posi

tion and released, it oscillates with a period of 2 seconds.Four English majors are discussing what would happen to the period of oscillation if the cart was displaced 12 cm from its rest position instead of 6 cm and again released.With which, if any, of these students do you agree?
Physics
1 answer:
Marianna [84]3 years ago
8 0

Answer:

Time period of horizontal Oscillation  = T = 2\pi\sqrt{\frac{m}{k} }

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Explanation:

Solution:

As we know that:

F = Kx

Here,

K = Spring constant

x = displacement.

First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.

But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.

First of all, let us see the equation of the time period of the oscillation.

We need to check, if time period does depend on the displacement or not.

As we know,

Time period of horizontal Oscillation  = T = 2\pi\sqrt{\frac{m}{k} }

As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.

Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.

Hence the Time period will remain same.

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A ball is dropped from a rooftop 60m high. <br> How long is the ball in the air?
alina1380 [7]

Answer: 3.49 s

Explanation:

We can solve this problem with the following equation of motion:

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)

Where:

y=0 m is the final height of the ball

y_{o}=60 m is the initial height of the ball

V_{o}=0 m/s is the initial velocity (the ball was dropped)

g=9.8 m/s^{2} is the acceleratio due gravity

t is the time

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (2)

t=\sqrt{\frac{2 (60 m)}{9.8 m/s^{2}}} (3)

Finally we find the time the ball is in the air:

t=3.49 s (4)

7 0
3 years ago
One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i
kati45 [8]

Answer:

The force exerted on an electron is 7.2\times10^{-18}\ N

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance  = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}

E_{1}=1.0183\times10^{3}\ N/C

Using formula of electric field again

E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}

E_{2}=-1.064\times10^{3}\ N/C

We need to calculate the resultant electric field

Using formula of electric field

E=E_{1}+E_{2}

Put the value into the formula

E=1.0183\times10^{3}-1.064\times10^{3}

E=-0.045\times10^{3}\ N/C

We need to calculate the force exerted on an electron

Using formula of electric field

E = \dfrac{F}{q}

F=E\times q

Put the value into the formula

F=-0.045\times10^{3}\times(-1.6\times10^{-19})

F=7.2\times10^{-18}\ N

Hence, The force exerted on an electron is 7.2\times10^{-18}\ N

8 0
3 years ago
What is the work done by the friction when the body slides against a rough horizontal surfaces?
katovenus [111]

Negative work

Hope this helps :)

4 0
2 years ago
PLEASE HELP! I'LL GIVE BRAINLEST​
melisa1 [442]

Answer:

Weight = 8.162 Newton.

Explanation:

Given the following data;

Mass = 2.2 kg

Acceleration due to gravity = 3.71 N/kg

To find the weight of the textbook;

Weight = mass * acceleration due to gravity

Weight = 2.2 * 3.71

Weight = 8.162 N

Therefore, the weight of the science textbook in mars is 8.162 Newton.

3 0
3 years ago
Help pls, see picture. Will mark Brainliest
Pani-rosa [81]

Answer:

a ) option 2 is correct

b) -ve acceleration for upward motion ,0 acceleration at top point ,+ve acceleration on downward motion ...

Explanation:

mark me as brainliest ❤️

5 0
3 years ago
Read 2 more answers
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