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kati45 [8]
3 years ago
5

A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is

the period if the amplitude of the motion is increased to 2A?(a) 2T(b) T(c) T/2(d) 2√T(e) 4T
Physics
1 answer:
frutty [35]3 years ago
6 0
<h2>Option B is the correct answer.</h2>

Explanation:

Period of a spring mass arrangement is given by

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

       where m is mass and k is spring constant.

So period of spring mass arrangement is independent of amplitude of motion.

Here amplitude changes from A to 2A.

Period for amplitude A is given by T.

Since period remains same for amplitude 2A also, the period is T.

Option B is the correct answer.

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Perform the calculation and report your answer using sig figs. 657.70 - 26.543
Anton [14]

Answer:

The answer is 631.157

Explanation:

The question requested that the answer to the subtraction of 26.543 from 657.70 must be written using significant figures.

Here are a few tips about how to Identify significant figures.

1) It should be noted that <u>the number "0" is what is usually (but not always) affected</u> while trying to identify significant figures. Hence, <u>all other numbers/digits are always significant</u>. For example, 26.543 has five significant figures.

2) The zeros found between these "other numbers/digits" are also significant. For example, 2202 has four significant figures.

3) In the case of a decimal, the tailing zeros or the final zero is also significant. 657.70 and 657.07 have five significant figures.

Now, back to the question

657.70  - 26.543  = 631.157.

Our final answer does not have a zero, hence all the digits (six) are significant.

8 0
3 years ago
An ice skater is spinning at 6.00 rev/s with his moment of inertia being 0.400 kg/m2. Calculate his new moment of inertia if he
labwork [276]

Answer:

New moment of inertia will be I=1.92kgm^2

Explanation:

It is given initially angular velocity \omega =6rev/sec=6\times 2\pi =37.68rad/sec

Moment of inertia I=0.4kgm^2

Angular momentum is equal to L=I\omega =37.68\times 0.4=15.072kgm^2/sec

Now angular velocity is decreases to \omega =1.25rev/sec=1.25\times 2\times 3.14=7.85rad/sec

As we know that angular momentum is conserved

So 15.072=I\times 7.85

I=1.92kgm^2

So new moment of inertia will be I=1.92kgm^2

4 0
3 years ago
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal
AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

v=\sqrt{v_x^2+v_y^2}           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

v_y^2=v_{oy}^2+2gh   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}

vx is calculated by using the information about the horizontal range of the ball:

R=v_o\sqrt{\frac{2h}{g}}    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0
4 years ago
An electric heater rated 600 w operates 6 hours per day find the cast to operate it for 30 days , at rs. 4.00 per unit​
djverab [1.8K]

Answer:

Rs. 432*10^3 (In kilowatts per hour)

I hope it will be useful.

3 0
3 years ago
In a cyclic process, a gas performs 123 J of work on its surroundings per cycle. What amount of heat, if any, transfers into or
Margaret [11]

Answer:

123 J transfer into the gas

Explanation:

Here we know that 123 J work is done by the gas on its surrounding

So here gas is doing work against external forces

Now for cyclic process we know that

\Delta U = 0

so from 1st law of thermodynamics we have

dQ = W + \Delta U

dQ = W

so work done is same as the heat supplied to the system

So correct answer is

123 J transfer into the gas

8 0
3 years ago
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