Answer:
420000N
Explanation:
Given parameters:
Mass of the train = 5.6 x 10⁵kg
Acceleration = 0.75m/s²
Unknown:
Resultant force = ?
Solution:
According to newton's second law, force is the product of mass and acceleration;
Force = mass x acceleration
Resultant force that acts on the train is given below;
Force = 5.6 x 10⁵kg x 0.75m/s² = 420000N
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (1,127 kg) x (6 m/s² forward)
= (1,127 x 6) newtons forward
= 6,762 newtons forward
______________________________
Momentum = (mass) x (speed)
= (69 kg) x (6 m/s)
= 414 kg-m/s
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
Complete question is;
A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?
Answer:
v = 12 m/s
Explanation:
We are given;
Initial velocity; u = 0 m/s (because ship starts from rest)
Acceleration; a = 4 m/s²
Time; t = 3 s
To find velocity after 3 s, we will use Newton's first equation of motion;
v = u + at
v = 0 + (4 × 3)
v = 12 m/s
Weight = (mass) x (gravity)
Weight = (7.0 kg) x (gravity)
On Earth, where (gravity) is roughly 10 N/kg . . .
Weight = (7.0 kg) x (roughly 10 N/kg)
Weight = roughly 70 Newtons
That's <em>B </em>on Earth.
It would be some other number on other bodies.
