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nordsb [41]
3 years ago
6

A 20.0-kg mass is traveling to the right with a speed of 3.00 m/s on a smooth horizontal surface when it collides with and stick

s to a second 20.0-kg mass that is initially at rest but is attached to a light spring with force constant 170 N/m . Find the frequency of the subsequent oscillations.
Express your answer with the appropriate units. Find the amplitude of the subsequent oscillations.

Express your answer with the appropriate units.
Physics
1 answer:
anygoal [31]3 years ago
4 0

Answer:

a) 0.328 Hz b) 73 cm (0.73 m)

Explanation:

Using law of conservation of momentum:

M1U1 + M2U2 = V(M1+M2) since the mass stuck together. where M1 is the mass 20 kg travelling with the U1  speed 3m/s and M2 is the 20kg at rest with U2 = 0 and V is the final speed of the mass after collision.

substitute the values into the equation

20 × 3 + 20 ×0 = V (20+20)

V = 60 / 40 = 1.5 m/s

The period of the oscillation can be calculated with the formula:

T = 2π√ ( m/k)  where T is the period in seconds and F is the frequency of the oscillation

substitute the values into the formula

T = 2×3.142×√ (40/170) = 3.05s

T = 1/F

F = 1/T = 1/ 3.05 = 0.328Hz

b) The amplitude of the oscillation can calculated using the formula

V = x√(K/m) where x is the amplitude in cm and V is the maximum speed that caused the displacement (amplitude) x

substitute the values into the formula

1.5 = x × √ (170/40) = x × 2.062

divide both side by 2.062

x = 1.5 / 2.062 = 0.73m = 0.73 cm

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san4es73 [151]

Answer:

BC and DE

Explanation:

In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.

Area under AB, d_1=\dfrac{1}{2}\times 2\times 10=10\ m

Area under BC, d_2=\dfrac{1}{2}\times 2\times 4=4\ m

Area under CD, d_3=\dfrac{1}{2}\times 2\times 7=7\ m

Area under DE, d_4=\dfrac{1}{2}\times 2\times 4=4\ m

Area under EF, d_5=\dfrac{1}{2}\times 2\times 3=3\ m

So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".

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Arada [10]

Answer : The final temperature is, 25.0^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ice = 2.09J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ice = 50 g

m_2 = mass of water = 200 g

T_f = final temperature = ?

T_1 = initial temperature of ice = -15^oC

T_2 = initial temperature of water = 30^oC

Now put all the given values in the above formula, we get:

50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC

T_f=25.0^oC

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aleksley [76]

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A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the
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Explanation:

Given that,

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

We need to find his displacement from the starting position.

We know that,

Displacement = shortest path covered

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For direction,

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Hence, this is the required solution.

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