Hi there!
We can use Newton's Second Law:
∑F = net force (N)
m = mass (kg)
a = acceleration (m/s²)
We are given the mass and acceleration, so:
∑F = 20 · 2 = <u>40 N</u>
Answer:
I'm pretty sure its B and C
Explanation:
B bc the weight is gravitational pull x mass so when the object has same mass the weight is smaller on moon
C bc mass is the same - you can't change it
Answer:
North pole
Explanation:
According to the law of magnetism:
<em>Unlike poles attract while like poles repel</em>
Since the south pole of the steel is brought near the nail, and the nail is meant to attract the steel magnet, the nail domain realigns itself to produce a pole opposite to the pole of the steel magnet brought near it.
Since the North pole is the opposite of the south pole, the North pole will be at the pointed end of the nail so that it can attract the steel magnet.
Answer:
ωB = 300 rad/s
ωC = 600 rad/s
Explanation:
The linear velocity of the belt is the same at pulley A as it is at pulley D.
vA = vD
ωA rA = ωD rD
ωD = (rA / rD) ωA
Pulley B has the same angular velocity as pulley D.
ωB = ωD
The linear velocity of the belt is the same at pulley B as it is at pulley C.
vB = vC
ωB rB = ωC rC
ωC = (rB / rC) ωB
Given:
ω₀A = 40 rad/s
αA = 20 rad/s²
t = 3 s
Find: ωA
ω = αt + ω₀
ωA = (20 rad/s²) (3 s) + 40 rad/s
ωA = 100 rad/s
ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s
ωB = ωD = 300 rad/s
ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s
Answer: zero.
Justification:
The downward velocity of the sky diver just before starting to fall is zero, assuming that the helicopter is not moving but just hovering.
Before starting to fall, the velocity of the skydiver is the same of the helicopter, which is zero. It is only, once she jumps out of the helicopter that her weight is not supported by the helicopter and so the gravitational force of the Earth attracts the skydiver and she starts to gain velocity at a rate equal to g (around 9.81 m/s²).
