The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of
1 answer:
You might be interested in
The coefficient of friction is missing and it has a value of μ = 0.4
Answer:
a = 3.924 m/s²
Explanation:
I've attached the kinematic free body diagram.
Taking the sum of all upward and downward forces,
EFy = 0;
N1 + N2 - m_p•g - m_v•g = 0
N1 + N2 = m_p•g + m_v•g
Where;
N1 and N2 are the normal reactions at the wheels
m_p is the mass of the driver
m_v is the mass of the vehicle
g is the acceleration due to gravity.
Plugging in the relevant values in the question,we obtain;
N1 + N2 = (385 + 75) x 9.81
N1 + N2 = 4512.6N - - - (eq1)
Now, taking sum of all horizontal forces;
EFx = (m_p + m_v) x a
So,
μ(N1 + N2) = (mp + mv) x a
Thus,
0.4(N1 + N2) = (385 + 75)a
0.4(N1 + N2) = 460a
N1 + N2 = 1150a
From eq(1),N1 + N2 = 4512.6N
Thus,
1150a = 4512.6N
a = 4512.6/1150
a = 3.924 m/s²
Therefore, the acceleration, a = 3.924 m/s²
Answer:
35
Explanation:
We are given that
Initial voltage,
Final voltage,
Number of tuns in primary coil of the transformer,
Rms current,
We have to find the number of turns are there on the secondary coil.
We know that
Using the formula
Hence, there are number of turns on the secondary coil=35