Answer:
1. Energy = 2880 Joules.
2. Energy = 60 Joules.
3. Quantity of charge = 120 Coulombs.
Explanation:
Given the following data;
1. Voltage = 12 Volts
Current = 0.5 Amps
Time, t = 8 mins to seconds = 8 * 60 = 480 seconds
To find the energy;
Power = current * voltage
Power = 12 * 0.5
Power = 6 Watts
Next, we find the energy transferred;
Energy = power * time
Energy = 6 * 480
Energy = 2880 Joules
2. Charge, Q = 4 coulombs
Potential difference, p.d = 15V
To find the total energy transferred;
Energy = Q * p.d
Energy = 4 * 15
Energy = 60 Joules
3. Voltage = 6 Volts
Current = 1 Amps
Time = 2 minutes to seconds = 2 * 60 = 120 seconds
To find the quantity of charge;
Quantity of charge = current * time
Quantity of charge = 1 * 120
Quantity of charge = 120 Coulombs
Answer:
c. natural force or pull toward the earth
Explanation:
Gravity of the earth is the force of attraction that it naturally possesses to attract any mass.
An apple falls on the earth due to this force of gravity.
The force of gravity near to the surface of the earth is given as:
where:
m= mass of the body
g= acceleration due to gravity
The variation of the gravitational force with height is given as:
where:
where
R = radius of the earth
An interesting problem, and thanks to the precise heading you put for the question.
We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.
Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m
Assume g = -9.81 m/s^2
initial velocity, v m/s (to be determined)
Solution:
(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.
substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s
(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s
Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m
(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m
(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)
vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)
Answer:
1) black body radiation
2) Electromagnetic radiation through the window
Explanation:
Blackbody radiation is the radiation produced by heated objects, particularly from a blackbody. The car is heated by sunshine. Actually, everything is heated by sunshine, and the car being a blackbody object absorbs all radiation (visible light, infrared light, ultraviolet light, etc.) that falls on it. This also means that it will also radiate at all frequencies that heat energy produces in it.
Radiation from the Sun is absorbed, through the glass, by the inside of the car. That is why it helps to have tinted glass or use a screen on the windows while it is parked.