Answer:
7200 N/m
Explanation:
Metric unit conversion
100g = 0.1 kg
5 cm = 0.05 m
50 cm = 0.5 m
As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground
The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m
Let g = 10 m/s2. The change in potential energy can be calculated as the following:
Therefore, as elastic energy is converted to potential energy and work of friction:
Answer:
a = -2.4 m/s²
Explanation:
Given,
The initial speed of the bus, u = 24 m/s
The final speed of bus, v = 12 m/s
Time taken to reach final speed is, t = 5.0 s
The acceleration of the body is given by the change in velocity by time
a = (v - u) / t
= (12 - 24) / 5
= -2.4 m/s²
The negative sign in the acceleration indicates that the bus is decelerating.
Therefore, the acceleration of the bus is, a = -2.4 m/s²
Answer:
Explanation:
If the dragster attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .
So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .
v = u + a t
v₀ = 0 + a tmax
tmax = v₀ / a
The value of tmax is v₀ / a .
Answer:
1.87 A
Explanation:
τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s
d = diameter of copper wire = 2 mm = 2 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
E = magnitude of electric field = 0.01 V/m
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of electron = 9.1 x 10⁻³¹ kg
n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³
= magnitude of current
magnitude of current is given as
= 1.87 A
