Answer:
<h3>14.97m/s</h3>
Explanation:
Given
Initial velocity of the car u = 8m/s
Distance travelled by the rider S = 40m
Acceleration a = 2m/s²
Required
rider's velocity after the acceleration v
Using the equation of motion
v² = u²+2as
v² = 8²+2(2)(40)
v² = 64+160
v² = 224
v = √224
v = 14.97m/s
Hence the rider's velocity after the acceleration is 14.97m/s
There are no appropriate examples in the list you provided with your question.
Examples of radiation:
... sunshine to tan your skin
... radio energy to bring you the news
... X-ray to check your teeth
... microwave to heat up the meatloaf
... flashlight to see where you're going
... RF energy to get an MRI of your knee
... infrared radiation from the campfire to warm your tootsies
... UHF radio waves to make a call or check Facebook with your smartphone
Answer:
The K.E is maximum when the child is at the vertical position and the P.E is maximum at the extreme deviated position from the vertical.
Explanation:
- A child is swinging on swing up and down has both kinetic and potential energy.
- The total mechanical energy of the system is conserved throughout the system. At any instant the total mechanical energy is given by,
E = K.E + P.E
- The K.E is maximum when the child is at the vertical position.
- The P.E is maximum at the extreme deviated position from the vertical.
- And when K.E is maximum P.E becomes minimum and vice versa as per the law of conservation of energy.
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
