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kotegsom [21]
3 years ago
11

What is simple harmonic motion ​

Physics
2 answers:
Irina-Kira [14]3 years ago
5 0
Simple harmonic motion is defined as a periodic motion of a point along a straight line. It’s acceleration is always towards a fixed point in that line and is proportional to its distance from that point.
ozzi3 years ago
5 0
“a special type of periodic motion, where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position.”
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What is the momentum of a 5 kg object that has a velocity of 1.2 m/s? 3.8 kg • m/s 4.2 kg • m/s 6.0 kg • m/s 6.2 kg • m/s
Gnesinka [82]

Answer:

Your answer will be 6.0kg•m/s

Explanation:

In the given question all the required details d given. Using these information's a person can easily find the momentum of the object. In the question it is already given that the mass of the object is 5 kg and the velocity at which it is traveling is 1.2 m/s.We know the equation of finding momentum asMomentum = mass * velocity                   = 5 * 1.2                    = 6So the momentum of the object is 6 Newton.

4 0
3 years ago
Read 2 more answers
Which of the following statements about electromagnetic radiation it true? A.electromagnetic waves with long wavelength are more
mina [271]

Answer:

Electromagnetic radiation is an electric and magnetic disturbance traveling through space at the speed of light (2.998 × 108 m/s). It contains neither mass nor charge but travels in packets of radiant energy called photons, or quanta.

7 0
2 years ago
a golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0. if the hang time of the golf ball is 33.4 s
Kobotan [32]
Solving this using the time, we know that range = horizontal velocity x time of flight 

since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have 

range = 36 cos 28 x 3.44 s = 109.3 m

6 0
2 years ago
A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.
alex41 [277]

The magnitude of the unknown height of the projectile is determined as 16.1 m.

<h3>Magnitude of the height</h3>

The magnitude of the height of the projectile is calculated as follows;

H = u²sin²θ/2g

H = (36.6² x (sin 29)²)/(2 x 9.8)

H = 16.1 m

Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.

Learn more about height here: brainly.com/question/1739912

#SPJ1

3 0
1 year ago
A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea
viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}                           (1)

time taken = 5 years

We know that :

time = \frac{distance}{speed}

5 = \frac{L''}{v}                                                      (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}

Solving the above eqn:

t'' = 20.56 years

4 0
3 years ago
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