Answer:
Explanation:
A) <em>Large</em>: As she opens her parachute, she begins to displace a large volume of air. This leads to a Large air resistance
B) <em>increase, weight</em>: As she falls, the air resistance force <u><em>increases</em></u>. Now there is a force acting in opposite directions to her <u><em>weight.</em></u>
C)<em>Weight, Decelerate:</em> The skydiver has only the downward force of her <u><em>weight</em></u> pulling down on her, so she starts to <u><em>decelerate</em></u>
D) <em>Weight, Upward, Resultant</em>:
Her <u><em>weight </em></u>is now equal to the <u><em>upward </em></u>force from the ground. Her <u><em>resultant </em></u>force is then zero
E) <em>Increases, same, constant, resultant, terminal</em>:
As she accelerates faster, the air resistance force <u><em>increases</em></u>. It is now the <u><em>same </em></u>as her weight. She now moves at a <u><em>constant</em></u> speed because the <u><em>resultant </em></u>force acting on her is zero. She is now at her <u><em>terminal </em></u>velocity.
F) <em>Increases, same, constant, terminal</em>:
As she decelerates, the air resistance force on her parachute <u><em>increases </em></u>until it is the <u><em>same</em></u> as her weight. She is now moving with a <u><em>constant </em></u>speed until she hits the ground - a new slower <u><em>terminal</em></u> velocity
Answer:
Total momentum before collision
P1 =.4 * 3.5 = 1.4 ignoring units here
Total momentum after collision
P2 = .6 * V - .4 * .7 = .6 V - .28
.6 V = 1.4 + .28 momentum before = momentum after
V = 2.8 cm/sec
In 5 sec V moves 2.8 cm/sec * 5 sec = 14 cm
Answer:
q = 3.87 x 10⁵ C
Explanation:
given,
Electric field, E = 8.60 x 10¹ = 86 N/C
radius of earth, R = 6371 Km = 6.371 x 10⁶ m
Coulomb constant, K = 9 x 10⁹ N · m²/C²
Charge on the earth = ?
the electric field at the point


inserting all the values

q = 3.87 x 10⁵ C
The electric charge on the earth is equal to 3.87 x 10⁵ C
Answer:
Explanation:
Given that,
The mutual inductance of the two coils is
M = 300mH = 300 × 10^-3 H
M = 0.3 H
Current increase in the coil from 2.8A to 10A
∆I = I_2 - I_1 = 10 - 2.8
∆I = 7.2 A
Within the time 300ms
t = 300ms = 300 × 10^-3
t = 0.3s
Second Coil resistance
R_2 = 0.4 ohms
We want to find the current in the second coil,
The same induced EMF is in both coils, so let find the EMF,
From faradays law
ε = Mdi/dt
ε = M•∆I / ∆t
ε = 0.3 × 7.2 / 0.3
ε = 7.2 Volts
Now, this is the voltage across both coils,
Applying ohms law to the second coil, V=IR
ε = I_2•R_2
0.72 = I_2 • 0.4
I_2 = 0.72 / 0.4
I_2 = 1.8 Amps
The current in the second coil is 1.8A
Work= force*distance
Work= x*12
Force= mass*acceleration
Force= 5 kg*6
Force= 40 N
Work= 40×12
Work= 480 J (joules)
I think this is it