Answer:
3.75 m/s south
Explanation:
Momentum before collision = momentum after collision
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
Since the car and truck stick together, v₁ = v₂.
m₁ u₁ + m₂ u₂ = (m₁ + m₂) v
Given m₁ = 1500 kg, u₁ = -15 m/s, m₂ = 4500 kg, and u₂ = 0 m/s:
(1500 kg) (-15 m/s) + (4500 kg) (0 m/s) = (1500 kg + 4500 kg) v
-22500 kg m/s = 6000 kg v
v = -3.75 m/s
The final velocity is 3.75 m/s to the south.
Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer:
Aluminum
Explanation:
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Answer with Explanation:
We are given that
Force acts on a pebble=
N
Position vector=
m
a.We have to find the resulting torque on the pebble about origin.
Torque=
Substitute the values then we get
Torque=
N-m
By using 
b.
Torque about point (2,0,-3)
N-m
Answer: The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center. So it would be D.
