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nika2105 [10]
3 years ago
6

An automobile having a mass of 2000 kg deflects its suspension springs 0.02 m under static conditions. Determine the nafural fre

quency of the automobile in the vertical direction by assuming damping to be negligible.
Physics
1 answer:
alexandr1967 [171]3 years ago
7 0

Answer:Frequency = 3.525 Hertz

Explanation:In static equilibrium, kd =mg

Where k= effective spring constant of the spring.

mg= The weight of the car.

d= static deflection.

Therefore, w =SQRTg/d

w = SQRT 9.81/0.02

w= 22.15 rad/sec

Converting to Hertz unit for frequency

1 rad/s = 0.1591

22.15rad/s=?

22.15 × 0.1591= 3.525 hertz

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Help yet again :) A hockey player is skating on the ice at 15km/h. He shoots the puck at 138 km/h according to a radar gun on th
IceJOKER [234]

Answer:

speed of puck acc. to the radar gun = 138 km/h

speed of player = 15 km/h

since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,

speed of puck = speed of player + speed of puck acc. to player

138 = 15 + speed of puck acc. to player

speed of puck acc. to player = 138 -15

speed of puck acc. to player = 123 km/h

Brainly this answer if you think it deserves it

7 0
3 years ago
A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction
lara31 [8.8K]
<h3>Answer</h3><h3>7 Ns</h3><h3>Explanation</h3>

Given in the question,

mass of foul ball = 0.140 kg

initial speed with which ball was hit with the bat = 30 m/s

final speed  = 40 m/s

According to the scenario the whole scene is making a right angle triangle

So, to the solve the question we will use pythagorus theorem

<h3>Hypotenuse² = base² + height²</h3>

Here,

Hypotenuse= Magnitude of impulse

Base = 1st change of momentum

height = 2nd change of momentum

 

1st impulse (1st change of momentum)

p = m(1)v(1) = (0.14 kg)(40.0 m/s) = 5.6 kg m / s = 5.6 N s

2nd impulse (2nd change of momentum)

p = m(2)v(2) = (0.14 kg)(30.0 m/s) = 4.2 kg m / s = 4.2 N s

Magnitude of impulse (hypotenuse of triangle)

impulse² = (5.6)² + (4.2)²

impulse² = 31.36 + 17.64

impulse² = 49

impulse² = √49

impulse = 7.0 N s

7 0
3 years ago
Equal masses are suspended from two separate wires made of the same material. The wires have identical lengths. The first wire h
choli [55]
D. I think is the correct answer
6 0
3 years ago
A weight lifter does 700J of work on a weight that he lifts in 3.1s. What is the power with
IgorLugansk [536]

Power = Work done/Time taken

So, keeping this in mind,we can solve it as follows:

= 700/3.1

= 7000/31

= 225.80 W

8 0
2 years ago
two billiard balls moving along the same line hit each other head-on. each has a mass of 0.220 kg; one has an initial velocity o
Tems11 [23]

Hi there!

Since the collision is elastic, we must also satisfy the following condition:

Ei = Ef, or:

KEi = KEf

Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.

mv1 + mv2 = mvf1 + mvf2

0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)

0.2882/0.220 = vf1 + vf2

1.31 = vf1 + vf2

Now, we can express this as a conservation of energy:

1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²

Plug in values and simplify:

0.403315 = 1/2m(vf1² + vf2²)

Simplify further:

3.6665 = vf1² + vf2²

Use the equation derived from momentum above and solve for one variable:

vf2 = 1.31 - vf1

Plug in this expression for vf2:

3.6665 = vf1² + (1.31 - vf1)²

Expand:

3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²

Simplify:

1.9504 = -2.62vf1 + 2vf1²

Solve for vf1 using a graphing calculator:

vf1 = -0.53 m/s or 1.84 m/s; we must figure out which one is correct.

Since v1 is heading to the right initially with a velocity of 1.84 m/s, we know that the ball's velocity could not have stayed the same in both magnitude and direction, so the final velocity must be -0.53 m/s.

Now, we can solve for the velocity of the other ball (initial of 0.53 m/s):

vf2 = 1.31 - (-0.53) = 1.84 m/s.

Now, you could have also made the connection that when two balls of the SAME MASS experience an ELASTIC collision, the velocities are simply "exchanged" from one to another. I just used this more "extensive" method to prove this.

7 0
3 years ago
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