Answer:
2.78 cm³/s
Explanation:
From the question,
Q = v/A'.................... Equation 1
Where Q = Average flow rate, A' = inverse of Area, v = velocity of the car.
Given: v = 100 km/h, A' = 10 km/L
Substitute this value into equation 1
Q = 100/10
Q = 10 L/h.
Now, we convert L/h to cm³/s.
Since,
1 L = 1000 cm³, and
1 h = 3600 s
Therefore,
Q = 10(1000/3600) cm³/s
Q = 2.78 cm³/s
Answer:
Explanation:
It is given that,
Mass of the loop, m = 10 g = 0.01 kg
Resistance of the loop, R = 0.02 ohms
Dimension of square loop, 7 cm × 7 cm
Area of the loop,
At time t = 0 s, the magnetic field increases from 0 to 1 T in 0.01 s
(a) Due to change in magnetic field, an emf is induced in the loop. Using the formula for induced emf as :
Now using Ohm's law to find the induced current in the loop. It is given by :
I =24.5 A
(b) A magnetic force acting on the loop is given by :
F = 0.8575
Since,
v = 0.8575 m/s
Hence, this is the require solution.
Given that,
Bulk modulus
Change in volume
Edge
The volume of the cube at the ocean's surface will be
Where, a = edge
Put the value into the formula
We need to calculate the change in pressure
Using formula of pressure
Put the value into the formula
The pressure increases by for every meter of depth
We need to calculate the depth
Using formula for depth
Hence, The depth is 9.75 m.
Answer:
The frequency of the sound wave in the helium in the stopped pipe is 749.25 Hz.
Explanation:
Given that,
Length = 1.00 m
Temperature T = 20°C
Speed of sound = 999 m/s
We need to calculate the frequency of the sound wave in the helium in the stopped pipe
Using formula of frequency
Put the value into the formula
Hence, The frequency of the sound wave in the helium in the stopped pipe is 749.25 Hz.