Q3. (a) 0m/s, as they are asking for initial velocit.
(b)(i) The paper has a large surface area or weighs less than the coin,thus,falls smoothly.
(ii)The coin has more mass than the paper.
(c) They fall at the same acceleration and hit the bottom at the same time.
Their mass doesn't matter in the vacuum.
Answer:
I = 215.76 A
Explanation:
The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.
Fm = u_o*I^2*L/2*π*d
where I is the current in each rod, d = 0.0082 m is the distance 27rId
between each, L = 0.85 m is the length of each rod.
Fm = 4π*10^-7*I^2*1.1/2*π*0.0083
The mass of each rod is m = 0.0276 kg
F_m = mg
4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8
I = 215.76 A
note:
mathematical calculation maybe wrong or having little bit error but the method is perfectly fine
Answer:
4.37 * 10^-4 J
Explanation:
Energy stored :
mgΔl / 2
m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m
Δl = mgl / πr²Y
Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m
Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10
Δl = 98 / 3.5 * π * 10^6
Δl = 0.00000891267
Energy stored :
mgΔl / 2
(10 * 9.8 * 0.00000891267) / 2
= 0.00043672083 J
4.37 * 10^-4 J
Answer:
147 J
Explanation:
The energy transferred to potential energy is :
U = m * g * h = (5 kg) * (9.8 m/s^2) * (3 m) = 147 J
1.
m = mass of Mr. Ure = 65 kg
g = acceleration due to gravity = 9.8 m/s²
force of earth's gravity on Mr. Ure is given as
F = mg
F = 65 x 9.8
F = 637 N
2.
F = force of gravity on car = 3050 N
m = mass of the car = ?
g = acceleration due to gravity = 9.8 m/s²
force of gravity on car is given as
F = mg
3050 = m (9.8)
m = 3050/9.8
m = 311.22 kg
3.
m = mass of Mr. Rees = 90 kg
g = acceleration due to gravity = 9.8 m/s²
force of earth's gravity on Mr. Rees is given as
F = mg
F = 90 x 9.8
F = 882 N