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
Answer:
Explanation: The Sun is directly overhead at solar noon at the Equator on the equinoxes, at the Tropic of Cancer (latitude 23°26′11.2″ N) on the June solstice and at the Tropic of Capricorn (23°26′11.2″ S) on the December solstice.
Answer:
Part a)
Part b)
Here final kinetic energy is more than the initial kinetic energy
This increase in kinetic energy is due to spring connected between them as the spring energy is converted into kinetic energy of two blocks
Explanation:
Part a)
As we know that there is no external force on the system of two gliders
So here we can use momentum conservation for two gliders
So we will have
Part b)
now we will have
initial kinetic energy of both gliders is given as
Final kinetic energy of two gliders
so here final kinetic energy is more than the initial kinetic energy
This increase in kinetic energy is due to spring connected between them as the spring energy is converted into kinetic energy of two blocks
A calculator must be used. To put your calculator in degree mode, press the MODE button and select degree, the press the 2nd button then MODE again. For most TI calculators, press the 2nd button then press the cos button then enter the value 0.34. This will give you an answer of 70.123 (when you round to 3 decimal places).
The answer is C
m = mass of the partner which the cheerleader lifts = 59.6 kg
h = height to which the partner is lifted by the cheerleader = 0.749 m
g = acceleration due to gravity = 9.8 m/s²
work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.
W = work done by the cheerleader in lifting the partner
PE = potential energy gained
so W = PE
potential energy is given as
PE = mgh
hence
W = mgh
inserting the values in the above formula
W = 59.6 x 9.8 x 0.749
W = 437.5 J
this is the work done in lifting the partner once.
the cheerleader does this 30 times , hence the total work done is given as
W' = 30 W
W' = 30 x 437.5
W' = 13125 J
