Answer:
Option C
100 J
Explanation:
Kinetic energy, KE is given by
where m is the mass and v is the velocity
Substituting 50 Kg for mass, m and 2 m/s for velocity v then we obtain
Therefore, the child's kinetic energy is equivalent to 100 J
Answer:
Hope it helps for you :)))))
Answer:
The instantaneous speed of the object after the first five seconds is 12.5 m/s.
(C) is correct option.
Explanation:
Given that,
An object starts at rest. Its acceleration over 30 seconds.
We need to calculate the instantaneous speed of the object after the first five seconds
We know that,
Area under the acceleration -time graph gives speed.
According to figure,
Hence, The instantaneous speed of the object after the first five seconds is 12.5 m/s.
Answer:
Option C. 210 J.
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 0.75 Kg
Height (h) = 12 m
Velocity (v) = 18 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Total Mechanical energy (ME) =?
Next, we shall determine the potential energy of the plane. This can be obtained as follow:
Mass (m) = 0.75 Kg
Height (h) = 12 m
Acceleration due to gravity (g) = 9.8 m/s²
Potential energy (PE) =?
PE = mgh
PE = 0.75 × 9.8 × 12
PE = 88.2 J
Next, we shall determine the kinetic energy of the plane. This can be obtained as follow:
Mass (m) = 0.75 Kg
Velocity (v) = 18 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.75 × 18²
KE = ½ × 0.75 × 324
KE = 121.5 J
Finally, we shall determine the total mechanical energy of the plane. This can be obtained as follow:
Potential energy (PE) = 88.2 J
Kinetic energy (KE) = 121.5 J
Total Mechanical energy (ME) =?
ME = PE + KE
ME = 88.2 + 121.5
ME = 209.7 J
ME ≈ 210 J
Therefore, the total mechanical energy of the plane is 210 J.
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
