A typical wheel used on a roller coaster is constructed by taking an aluminum hub and bonding a polyurethane tire to the hub's outside diameter. This entire “wheel assembly” is then connected to the axle through a bearing.
Answer:
a) V = -0.227 mV
b) V = -0.5169 mV
Explanation:
a)
Inside a sphere with a uniformly distributed charge density, electric field is radial and has a magnitude
E = (qr) / (4πε₀R³)
As we know that
V = -
By solving above equation, we get
V = (-qr²) / (8πε₀R³)
When
R = 1.81 cm
r = 1.2 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-2.80 × 10⁻¹⁵ × (1.2 × 10⁻²)²) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²)³)
V = -2.27 × 10⁻⁴ V
V = -0.227 mV
b)
When
r = R
R = 1.81 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-qR²) / (8πε₀R³)
V = (-q) / (8πε₀R)
V = (-2.80 × 10⁻¹⁵) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²))
V = -5.169 × 10⁻⁴ V
V = -0.5169 mV
The problem describes the relationship of "bulb a" and "bulb b" to be in connected in series. When the switch is open then no current can flow, on the other hand, when it is closed, current will pass through.
When only "bulb a" is connected to the battery then more current is flowing to "bulb a" causing it to be bright.
Closing the switch would mean that "bulb b" is already included in the circuit and the battery will push small current to flow around the whole circuit. The more bulbs are connected, the harder for the current to flow because the resistance will be very high.
So the light of "bulb a" will be dimmer.
Answer:
W = 68.6 N
Explanation:
Given that,
Force acting on a right side = 30 N
Mass of the object, m = 7 kg
The force of friction acting on the car = 3 N
We need to find the weight of the object.
The weight of an object is given by :
W = mg
Substitute all the values,
W = 7 × 9.8
W = 68.6 N
So, the weight of the object is 68.6 N.
