Distance, luminosity, brightness, radius, chemical composition and temperature
Answer: Graph C is the correct option
Explanation:
The question is incomplete, please remember to submit the whole question :)
However, the rest of the question with its corresponding figures is below:
Assume that positive momentum is to the right, which of the following best represents the momentum of the cart of mass m as a function of time before and after the collision?
The initial momentum 
of the cart with mass 
(before the collision) is:
(1) Note the negative sign indicates the direction of cart's velocity (to the left, as seen in the first image attached)
On the other hand, the final momentum 
of both carts (after the inelastic collision) is:
(2)
So, according to this, the correct graph tha best represents the situation is C. Since before the collision the momentum is negative, then both carts slow down after the collision (
), and taking into account the linear momentum is directly proportional to the velocity (although is in the positive direction) is less than .
Answer:
2.5 ft
Explanation:
minimum size of the mirror should be one half the person's height.
For this woman of 5 feet height, she would therefore need a mirror of length 2.5 feet
5ft ÷ 2 ft (half) = 2.5ft OR half of 5t is 2.5ft
Answer: wavelength=velocity×period
Explanation:the relation between velocity, wavelength and period is
Wavelength=velocity×period
This question is incomplete, the complete question is;
A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick. What is the maximum charge?
(The dielectric constant of mica is 5.4, and its dielectric strength is 1.00×10⁸ V/m)
Answer: the maximum charge q is 716.85 μF
Explanation:
Given data;
with = 3.0 cm = 0.03
breathe = 5.0 m
Area = 0.03 × 5 = 0.15 m²
dielectric strength E = 1.00 × 10⁸
∈₀ = 8.85 × 10⁻¹²
constant K = 5.4
maximum charge = ?
the capacitor C = KA∈₀ / d
q = cv so c = q/v
now
q/v = KA∈₀ / d
q = vKA∈₀/d = EKA∈₀
we substitute
q = (1.00 × 10⁸) × 5.4 × 0.15 × 8.85 × 10⁻¹²
q = 716.85 × 10⁻⁶ F
q = 716.85 μF
the maximum charge q is 716.85 μF
