Answer:
This is an example of Inelastic colission
Explanation:
Step one:
given:
mass of moose m1 = 620 kg
mass of train m2= 10,000kg
Initial velocity of moose u1= 0 m/s
Initial velocity of train v1 = 10m/s
combined velocity of the system is given as v
Applying the conservation of momentum equation we have
m1u1+ m2u1= (m1+m2)V
substitutting we have
620*0+10000*10= (620+10000)V
100000= 10620V
divide both sides by 10620
V = 100000/10620
V=9.41m/s
The velocity of the moose after impact is 9.41m/s
Answer:
45.864N
Explanation:
Using the formula
F = nR
n is the coefficient of friction
R is the normal reaction
R = mg
F = nmg
F = 0.78 * 6 * 9.8
F = 45.864N
Hence the amount of force with which the hand pushes the box is 45.864N
Answer:
The correct answer is:
(a) 84.240 kg
(b) 24.038 m
Explanation:
The given values are:
Force,
F = 81.0 N
Distance,
S = 13.0 m
Time,
t = 5.20 s
As we know,
The acceleration of mass will be:
⇒
On substituting the given values, we get
⇒
⇒
⇒
(a)
The mass of the block will be:
⇒
On substituting the given values, we get
⇒
⇒
(b)
The final velocity after a given time i.e.,
t = 5.00 s
⇒
On substituting the values, we get
⇒
⇒
In time, t = 5.00 s
The distance moved by the block will be:
⇒
On putting the values, we get
⇒
⇒
Answer:
1.52g
Explanation:
Given parameters:
Force = 125N
Mass combined = 82kg
Unknown:
Acceleration of the bicycle = ?
Solution:
From Newton second law of motion suggests that:
Force = mass x acceleration
Acceleration = = = 1.52g