The correct answer is:
D. Electromagnetic waves.
The arrows represent electromagnetic waves.
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
Answer:
The correct option is (B).
Explanation:
The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,
It is mentioned that, an asteroid with an orbital period of 8 years. So,
So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.
Answer:
740 N
Explanation:
We are given that
Radius,r=0.3 m
Torque,
We have to find the magnitude of the static frictional force.
According to question
Torque by engine=Torque by static friction
Hence, the magnitude of static frictional force=740 N
