Electrons move from the sweater to the balloon. The sweater becomes positively charged, while the balloon becomes negatively charged.
Explanation:
- Sweater is a conductive material, which means it readily gives away its electrons.
- Consequently, when you rub a balloon on Sweater, this causes the electrons to move from the Sweater to the balloon's surface.
- The rubbed part of the balloon now acquired a negative charge. Objects made of rubber, such as the balloon, are basically electrical insulators, meaning that they resist electric charges flowing through them.
- This is why only part of the balloon may have a negative charge (where the wool rubbed it) and the rest may remain neutral after electrostatic process.
Answer:
Rise in level of fluid is 0.11 m
Rise in level of fluid in case of mercury is 0.728 cm or 7.28 mm
Solution:
As per the question:
Density of oil, 
Change in Pressure in the tank, 
Density of the mercury, 
Now,
To calculate the rise in the level of fluid inside the manometer:
We know that:
1 mmHg = 133.332 Pa
Thus

Also,

where
g = acceleration due to gravity
h = height of the fluid level

h = 0.11 m
Now, if mercury is used:


h = 0.00728 m = 7.28 mm
Answer:
The answer is below
Explanation:
The amplitude decreases by 2% during each oscillation. Hence the decrease in amplitude can be represented by an exponential decay in the form:
y = abˣ; where x ad y are variables, a is the initial value and b is the factor.
Let y represent the amplitude after x oscillations. Since the initial amplitude is 10 cm, hence:
a = 10 cm, b = 2% = 0.02.
Therefore:
y = 10(0.02)ˣ
The amplitude after 25 oscillations is gotten by substituting x = 25 into the equation. Hence:
y = 10(0.02)²⁵
y= 3.355 * 10⁻⁴² cm
The amplitude after 25 oscillations is 3.355 * 10⁻⁴² cm
Answer:
V = 5.67 V
Q' = 2.09 x 10^-4 C
Explanation:
Potential difference, V = 5.67 V
Charge, Q = 3.49 x 10^-5 C
dielectric constant, K = 5.99
Capacitance of the capacitor
C = Q / V = 3.49 x 10^-5 / 5.67 = 6.15 x 10^-6 F
Now the dielectric is inserted, so,
C' = K x C = 5.99 x 6.15 x 10^-6 = 3.687 x 10^-5 F
After insertion of dielectric, as the battery is connected, so potential difference remains same.
Charge, Q' = C' x V = 3.687 x 10^-5 x 5.67 = 2.09 x 10^-4 C
Answer:
60 kg
80 kg
Explanation:
Work is equal to the change in energy.
W = ΔE = E − E₀
Let's start with block B. The work done by the tension force is equal to the change in energy. Initially, the block has potential energy. Finally, the block has kinetic energy.
W = ΔE
FΔy = ½ mv² − mgh
T (-2.0 m) = ½ m (6.00 m/s)² − m (10 m/s²) (2.0 m)
T (-2.0 m) = m (-2 m²/s²)
T = m (1 m/s²)
Now let's look at block A. The work done by tension and against friction is equal to the change in energy. Initially, the block has no energy. Finally, it has both kinetic and potential energy.
W = ΔE
Fd = ½ mv² + mgh − 0
(T − Nμ) (2.0 m) = ½ (4.00 kg) (6.00 m/s)² + (4.00 kg) (10 m/s²) (⅗ × 2.0 m)
(T − Nμ) (2.0 m) = 120 J
T − Nμ = 60 N
Draw a free body diagram of block A and sum the forces in the perpendicular direction to find the normal force N.
N = mg cos θ
N = (4.00 kg) (10 m/s²) (⅘)
N = 32 N
Substitute:
T − 32μ = 60 N
If μ = 0, then T = 60 N and m = 60 kg.
If μ = ⅝, then T = 80 N and m = 80 kg.