Those two objects would “Attract” if one is positive charge and one is negative charge. So, B) should be the answer.

Answer:

980 J, B

Explanation:

Given that.

mass of substance, m = 75 g

initial temperature of system, θ1 = 150° C

final temperature of system, θ2 = 250° C

specific heat capacity, c = 0.13 J/gC

Q = mcΔθ, where

Q = quantity of heat required in J

m = mass of substance in G

c = specific heat capacity of substance in J/gC

Δθ = change in temperature °C

Δθ = θ2 - θ1

Δθ = 250° C - 150° C

Δθ = 100° C

now that we have all our values, what we do next is to substitute and apply all in the initial formula given

Q = mcΔθ

Q = 75 * 0.13 * 100

Q = 7500 * 0.13

Q = 975 J

Thus, we can say they amount of heat required to increase the temperature of 75g of gold, from 150° - 250° is 975 J, which is approximately, 980 J.

Option B

Answer:

Theta1 = 12° and theta2 = 168°

The solution procedure can be found in the attachment below.

Explanation:

The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).

In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.

Bulbs c and b would still be screwed in if they were in to begin with and bulbs A, D, and E. would be unscrewed

Answer:

Energy Flows Quick check answers:

1. Ffd.

2. The kinetic energy decreases, and gravitational potential energy increases.

3. The internal energy of the system increases.

4. KEbox= Etotal-mgh

5. Etotal = 1/2m1(v1)^2+1/2m^2(v2)^2+U