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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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A truck initially traveling at a speed of 22 m/s increases at a constant rate of 2.4 m/s^2 for 3.2s. What is the total distance
FinnZ [79.3K]

Answer:

82.7 m

Explanation:

u= 22m/s

a= 2.4 m/s^2.

t= 3.2 secs

Therefore the distance travelled can be calculated as follows

S= ut + 1/2at^2

= 22 × 3.2 + 1/2 × 2.4 × 3.2^2

= 70.4 + 1/2×24.58

= 70.4 + 12.29

= 82.7 m

Hence the distance travelled by the truck is 82.7 m

6 0
3 years ago
Find the volume of a rectangular prism that is 8m long, 4m wide, and 300cm high
konstantin123 [22]
  • L=8m
  • B=4m
  • H=300cm=3m

\\ \bull\tt\longmapsto Volume=LBH

\\ \bull\tt\longmapsto Volume=8(4)(3)

\\ \bull\tt\longmapsto Volume=96m^3

6 0
3 years ago
Read 2 more answers
Ran 300 meter in 40 seconds, what is the speed?
nignag [31]

Answer:

7.5

Explanation:

Speed= distance/ time

Speed= 300m/40sec

Speed= 7.5m/s

5 0
3 years ago
if heat energy is transferred from direct contact between a warm and a cold object it has been transferred by what ?
Katarina [22]

When heat energy is transferred from direct contact between a warm and a cold object , it is known as heat transfer by conduction.

In conduction, the heat transfer takes place within an object or between two objects in contact until the temperature becomes uniform. this kind of heat transfer continues until the temperature at two ends between which the heat transfer take place , becomes equal.  Heat transfer takes place from point at high temperature to point at lower temperature.

5 0
3 years ago
Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri
miv72 [106K]

Answer:

T’= 4/3 T  

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

      T = m_A a

Axis y

     N- W_A = 0

Body B

Vertical axis

     W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

    T = m_A a

    W_B –T = M_B a

We solve this system of equations

     m_B g = (m_A + m_B) a

    a = m_B / (m_A + m_B) g

In this initial case

     m_A = M

     m_B = M

     a = M / (1 + 1) M g

     a = ½ g

Let's find the tension

    T = m_A a

    T = M ½ g

    T = ½ M g

Now we change the mass of the second block

    m_B = 2M

    a = 2M / (1 + 2) M g

    a = 2/3 g

We seek tension for this case

    T’= m_A a

    T’= M 2/3 g

   

Let's look for the relationship between the tensions of the two cases

   T’/ T = 2/3 M g / (½ M g)

   T’/ T = 4/3

   T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer  e

4 0
3 years ago
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