A train starts from rest. It acquires a speed of 25 m/s after
1 answer:
Given parameters:
Initial speed of train = 0m/s
Final speed = 25m/s
time = 20s
Unknown:
Total distance covered = ?
Solution:
Speed is a scalar quantity that has magnitude and no direction.
Mathematically;
Speed =
distance = speed x time
Input the parameters and solve for distance;
Distance = 25 x 20 = 500m
The distance covered is 500m
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Answer:
D. The period would decrease by sqrt (2)
Explanation:
The period of a mass-spring system is given by:
where
m is the mass
k is the spring constant of the spring
If the spring constant is doubled,
k' = 2k
So the new period will be
So the correct answer is
D. The period would decrease by sqrt (2)
Answer:
a. The ratio of their resistance is 2783:64
b. The ratio of their energy is 4:23
c. The charge on the first bulb is 5.75 C
The charge on the second bulb is C
Explanation:
The voltage on one of the electric bulbs, V₁ = 40 volts
The power rating of the bulb, P₁ = 230 w
The voltage on the other electric bulbs, V₂ = 110 volts
The power rating of the bulb, P₂ = 40 w
a. The power is given by the formula, P = I·V = V²/R
Therefore, R = V²/P
For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96
The resistance of the second bulb, R₂ = 110²/40
The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64
∴ The ratio of their resistance, R₂:R₁ = 2783:64
b. The energy of a bulb, E = t × P
Where;
t = The time in which the bulb is powered on
∴ The energy of the first bulb, E₁ = 230 w × t
The energy of the second bulb, E₂ = 40 w × t
The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23
∴ The ratio of their energy, E₂:E₁ = 4:23
c. The charge on a bulb, 'Q', is given by the formula, Q = I × t
Where;
I = The current flowing through the bulb
From P = I·V, we get;
I = P/V
For the first bulb, the current, I = 230 w/40 V = 5.75 amperes
The charge on the first bulb per second (t = 1) is therefore;
Q₁ = 5.75 A × 1 s = 5.75 C
The charge on the first bulb, Q₁ = 5.75 C
Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = C
The charge on the second bulb, Q₂ = C.
d. The question has left out parts
Answer:
The friction coefficient's minimum value will be "0.173".
Explanation:
The given query seems to be incomplete. Below is the attached file of the complete question.
According to the question,
(a)
The net friction force's magnitude will be:
⇒
(b)
For m₃,
⇒
Or,
⇒
