The current that would pass through the 30 ohms resistor is 2 A.
<h3>What is electric current?</h3>
Electric current is the rate of flow of electric charge round a conductor.
To calculate the electric current that would pass through the 30 ohms resistor, we use the formula below
Formula:
- I = V/Rt........... Equation 1
Where:
- I = Electric current passing through the 30 ohms resistor
- V = Voltage
- Rt = Total or effective resistance of the resistors.
From the question,
Given:
- V = 100 volts
- Rt = (30+20) ohms (since both resistors are connected in series)
Substitute these values into equation 1
Hence, The current that would pass through the 30 ohms resistor is 2 A.
Learn more about electric current here: brainly.com/question/1100341
Answer:
0.73 m/s
Explanation:
From Newton second law of motion,
I = m(v-u)...................... Equation 1
Where I = Impulse, m = mass of the person, v = final velocity, u = Initial velocity.
make v the subject of the equation
v =(I/m)+u................. Equation 2
Note: u = 0 m/s as the person is falling from an height.
Given: I = 55 Ns, m = 75 kg, u = 0 m/s
Substitute into equation 2
v = 55/75
v = 0.73 m/s
Answer:
L = 0.635m
Explanation:
This problem involves the concept of stationary waves in pipes. For pipes closed at one end,
The frequency f = nv/4L for n = 1,3,5....n
For pipes open at both ends
f = nv/2L for n = 1,2,3,4...n
Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.
The film solution can be found in the attachment below.
Answer:
Due to energy loss while collision ball will not reach to same height while if there is no energy loss then in that case ball will reach to same height
Explanation:
As we know that initially ball is held at height h = 40 cm
So here we can say that kinetic energy of the ball is zero and potential energy is given as

now when strike with the ground then its its fraction of kinetic energy is lost in form of other energies
So the ball will left rebound with smaller energy and hence it will reach to height less than the initial height
While if we assume that there is no energy loss during collision then in that case ball will reach to same height again
Answer:
the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
Explanation:
a) Kinetic energy of block = potential energy in spring
½ mv² = ½ kx²
Here m stands for combined mass (block + bullet),
which is just 1 kg. Spring constant k is unknown, but you can find it from given data:
k = 0.75 N / 0.25 cm
= 3 N/cm, or 300 N/m.
From the energy equation above, solve for v,
v = v √(k/m)
= 0.15 √(300/1)
= 2.598 m/s.
b) Momentum before impact = momentum after impact.
Since m = 1 kg,
v = 2.598 m/s,
p = 2.598 kg m/s.
This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,
u = 2.598 kg m/s / 8 × 10⁻³ kg
= 324.76 m/s.
Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.