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Dmitriy789 [7]
4 years ago
15

You attach a 1.90 kg mass to a horizontal spring that is fixed at one end. You pull the mass until the spring is stretched by 0.

400 m and release it from rest. Assume the mass slides on a horizontal surface with negligible friction. The mass reaches a speed of zero again 0.600 s after release (for the first time after release). What is the maximum speed of the mass (in m/s)
Physics
1 answer:
Gre4nikov [31]4 years ago
6 0

Answer:

2.09 m/s

Explanation:

As the  spring is stretched initially , and the mass released from rest i.e v=0. Also, The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.600 s. This illustrates half oscillation of the system.

Therefore, for the period of a full oscillation of the system

T= 2t => 2(0.6)=> 1.2 s

As the frequency is the reciprocal of the period, we have

f= 1/T => 1/1.2

f= 0.833 Hz

The angular frequency'ω' is given by,

ω= 2πf => 2π x 0.833=>5.23 rad/s

For the maximum velocity of the object  in a spring-mass system:

V(_{max} )= Aω

where A is the amplitude of the oscillation. As here, the amplitude of the motion corresponds to the initial displacement of the object (A=0.400 m)

V(_{max} )= 0.4 x 5.23 =>2.09 m/s

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A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J              

Explanation:

We have given the player turntable initially rotating at speed of 33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec

Now speed is reduced by 75 %

So final speed \frac{3.49\times 75}{100}=2.6175rad/sec

Time t = 5.5 sec

From first equation of motion we know that '

\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2

(a) Now final velocity \omega =0rad/sec

So time t to come in rest  t=\frac{0-3.49}{-0.158}=22sec

(b) The work done in coming rest is given by

\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J

4 0
3 years ago
An ideal parallel-plate capacitor consists of a set of two parallel plates of area Separated by a very small distance 푑. This ca
dolphi86 [110]

Answer:

doubled the initial value

Explanation:

Let the area of plates be A and the separation between them is d.

Let V be the potential difference of the battery.

The energy stored in the capacitor is given by

U = Q^2/2C   ...(1)

Now the battery is disconnected, it means the charge is constant.

the separation between the plates is doubled.

The capacitance of the parallel plate capacitor is inversely proportional to the distance between the plates.

C' = C/2

the new energy stored

U' = Q^2 /  2C'

U' = Q^2/C = 2 U

The energy stored in the capacitor is doubled the initial amount.

8 0
3 years ago
On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan
frutty [35]

Answer:

The distance in kilometers is 4012 ×10^{-6} km.

Explanation:

We know that the conversion of 1 millimeters is equal to 10^{-3} meter. And then the conversion of 1 meter is equal to 10^{-3} km. Then the conversion of 1 millimeter to km will be

1 mm = 10^{-3} m

1 m = 10^{-3} km

So, 1 mm = 10^{-3}×10^{-3} km = 10^{-6} km.

As here the the distance is 4012 mm, then the distance in km will be

4012 mm = 4012 ×10^{-6} km.

So the distance is 4012 ×10^{-6} km.

5 0
3 years ago
calculate the average speed of an athele who runs a distance of 100m in 16 s and an additional of 400m in a 44s
Delvig [45]

Explanation:

speeds = distance/time

=100/16

=6.25m/s

second speed is;

400/44 =9.09

av. speed = total speed /n

= (6.25+9.09)/2

=7.67 m/s

5 0
3 years ago
28. Ken and Musa shared a cake such that Ken got twice the size
ratelena [41]

Answer:

Musa = \frac{1}{3}

Ken = \frac{2}{3}

Explanation:

Given

Ken = 2 * Musa --- Ken's share

Required

The fraction each got

Since they both shared a cake, we have:

Ken + Musa = 1

Substitute: Ken = 2 * Musa

2 * Musa+ Musa = 1

Factorize

Musa(2+ 1)= 1

Musa(3)= 1

Divide both sides by 3

Musa = \frac{1}{3}

Recall that: Ken = 2 * Musa

Ken = 2 * \frac{1}{3}

Ken = \frac{2}{3}

3 0
3 years ago
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