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Ludmilka [50]
3 years ago
9

The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that th

e acceleration an object experiences is
Physics
1 answer:
Mars2501 [29]3 years ago
6 0

Answer:

According to Newton's 2nd law

The force acting on a body produces acceleration in its direction which is directly propotional to the force but inversly propotinal to the mass of tbe body.

Explanation:

a = F/m

F = ma

Where( F) is force (m) is mass and (a) is acceleration.

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4 0
3 years ago
A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?
katrin [286]

Answer:

<em>k = 25.18 N/m</em>

Explanation:

<u>Simple Harmonic Oscillator</u>

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}

If the period is known, we can find the value of the constant by solving for k:

\displaystyle k=m\left(\frac{2\pi}{T}\right)^2

Substituting the given values m=5 Kg and T=2.8 seconds:

\displaystyle k=5 \left(\frac{2\pi}{2.8}\right)^2

k = 25.18 N/m

5 0
3 years ago
Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe
vodka [1.7K]
 We will use formula for the orbital velocity of Venus, which is v = 35.02 km/s. 
    An average distance to the Sun ( In kilometers ) is:
    R = 0.723 * 149,579,871 km= 108,150,260 km.
    Than we will calculate the orbital period ( T ).
    v = 2 π R / T
    T = 2 π R / v
    T = 2 * 3.14 * 108,150,260 km / 126,072 km/s
    T = 5389.75 s ≈ <span>224.5 days
    The orbital period of Venus is approximately 224.5 days.</span>
6 0
3 years ago
A police car is traveling due east at a speed of 15.0 m/s relative to the earth. You are in a convertible following behind the p
Neporo4naja [7]

Answer:

A.) 355 m/s

B.) 0.71 m

C.) 500Hz

Explanation:

Given that a police car is traveling due east at a speed of 15.0 m/s relative to the earth. You are in a convertible following behind the police car. Your car is also moving due east at 15.0 m/s relative to the earth, so the speed of the police car relative to you is zero. The siren of the police car is emitting sound of frequency 500 Hz. The speed of sound in the still air is 340 m/s

a.) What is the speed of the sound waves relative to you?

Since the car is moving away from the observer, the relative velocity will be:

Relative velocity = 340 + 15

Relative velocity = 355 m/s

b.) What is the wavelength of the sound waves at your location?

Using the wave speed formula

V = frequency × wavelength

Make wavelength the subject of formula.

Wavelength = Velocity / frequency

Wavelength = 355/500

Wavelength = 0.71 m

c.) What frequency do you detect?

Fo = Fs ( C + V ) / ( C + v )

Fo = Fs

That is, the frequency of the observer will be equal to the frequency of the source.

Therefore, Fo = 500Hz

4 0
2 years ago
Consider two less-than-desirable options. In the first you are driving 30 mph and crash head-on into an identical car also going
Keith_Richards [23]

Answer:

B. The force would be the same in both cases.

Explanation:

As we know that the force is defined as the rate of change in momentum of the system

So here we will have

F = \frac{\Delta P}{\Delta t}

so in both the cases the change in the momentum of the car will be same

\Delta P = m(\Delta v)

also the time interval is given same in both the cases

so here we can say

F = \frac{m\Delta v}{\Delta t}

so force will be same on the car in both the cases

6 0
3 years ago
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