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Tomtit [17]
3 years ago
12

A compact car has a mass of 1300 kg . When empty, the car bounces up and down on its springs 1.5 times per second. What is the c

ar's oscillation frequency while carrying four 78 kg passengers?
Physics
1 answer:
Snezhnost [94]3 years ago
7 0

Answer:

0.76 oscillations per second

Explanation:

Time period = 2\pi \sqrt{m/k} where k is the spring constant and m is the mass attached to the spring.

Calculating the spring constant:

1/1.5=2\pi\sqrt{1300/k};

k=3.68*10^4

When the car is loaded, its mass increases to 1300+ 4*78 = 1612

Calculating the new time period:

T=2\pi \sqrt{1612/36800}=1.1315s

Frequency=1/T;

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Hello,help me with this out please i need it hurry but please ensure your answer is correct..I attach here with my question.
lilavasa [31]

Answer:

.409 N

Explanation:

For this to balance, the moments around the fulcrum must sum to zero.

On the left you have   .21   ( is that down? I will assume it is)

      Counterclockwise moments :

        .21 * 40     +  1.0 * 20    

     Clockwise moments :

        .5 * 20     +     F * 45

these moments must equal each other

.21*40 + 1 *20   =  .5 * 20 + F * 45

   F = .409 N

7 0
2 years ago
When a cup is placed on a table, which force prevents the cup from falling to the ground?
zhuklara [117]

Answer:

B. normal force

Explanation:

Because there is no frictional or resistance force. However gravitational force is applied downroad from the center of the cup thus the contact force that is perpendicular to the surface that an object contacts which is the normal force exerted upward from the table that prevents an object from falling.

6 0
3 years ago
Please help me , thank you
mash [69]

Answer:

Since the reading wasn't specified, it would be most likely A

Explanation:

A is the most similar to a protoplanetary disk, so it'd be A most likely

5 0
3 years ago
A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

\dfrac{dz}{dt}=0.65\ ft/s

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
When an object is balanced about a pivot, the total clockwise moment must be equal to the total __________ __________. What two
Keith_Richards [23]

Answer:

When an object is balanced, about a pivot, the total clockwise moment must be equal to the total anticlockwise moment about that pivot.

Hope that helps.

5 0
3 years ago
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