Answer:
If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2
Explanation:
We know that in a simple harmonic oscillator the maximum speed is given by
= 
Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to 
of the oscillation .
Since 
= 2
Where 
is the maximum speed when frequency is doubled .
The speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.
<h3>
Speed of the satellite</h3>
v = √(GM/r)
where;
- G is universal gravitation constant
- M is mass of Earth
- r is radius of the satellite
v = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/3.57 x 6.37x 10³)
v = 1.32 x 10⁵ m/s
Thus, the speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.
Learn more about speed of satellite here: brainly.com/question/22247460
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When an unbalanced force acts on an object the change in the object state of rest or motion depends on the size and direction of the force.
If a body is at state of rest or motion, when an unbalanced external force acts on it, its starts moving in the direction of force and magnitude of its velocity or acceleration depends on the magnitude of force applied.
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
Light is refracted when it crosses the interface from air to glass in which it moves more slowly.
Since the light speed changes at the interface, the wave length of the light must change too. The wave length decreases as the light enter the medium and the light wave changes direction.
