Answer:
Explanation:
<u>Center of Gravity
</u>
It refers to a point where all the forces of gravity of a body make a zero total torque. To find the solution we use the fact that the net force acting on the system boat-man is in every moment equal to zero. It's assured by the first Newton’s law, the center of gravity is at rest or in uniform motion in both moments. From an external viewer's point of view, the center of gravity remains unchanged. The formula to compute it is shown below
Originally, the man sits on the stern of the boat. His weight is applied at a distance xm=4.9 m from the pier (assumed as x=0). The boat is assumed to have a uniformly distributed mass applied at its center, i.e. at xb = 4.9 / 2 = 2.45 m. The center of gravity is located originally at
When the man walks to the prow, the boat moves x = 1.2 m from the pier, so its center is located at a distance
The man is located at
The center of gravity is computed now as
Both centers of gravity are equal, thus
Simplifying
Rearranging
Thus
Answer:
79.8 is the answere
I hope this HELP :)
Explanation:
Here, initial velocity is u=18.5m/s and final velocity is v=46.1m/s
Time taken t=2.47s
If a be the acceleration of the car.
Using v=u+at,
46.1=18.5+a(2.47)
⟹ a=11.17m/s
2
If d be the distance traveled by car.
Using formula v
2
−u
2
=2ad,
(46.1)
2
−(18.5)
2
=2(11.17)d
⟹ d=79.8m
Answer:
354.72 m/s
Explanation:
= mass of lead bullet
= specific heat of lead = 128 J/(kg °C)
= Latent heat of fusion of lead = 24500 J/kg
= initial temperature = 27.4 °C
= final temperature = melting point of lead = 327.5 °C
= Speed of lead bullet
Using conservation of energy
Kinetic energy of bullet = Heat required for change of temperature + Heat of melting
Answer:
Explanation:
Momentum before collision = momentum after collision
If
m₁ = 300 kg
v₁= 10 m/s
m₂ =1000 kg
v₂ = 0 as the car was parked
momemum before collision =
m₁ v₁ + v₂ m₂= 3000kgm/sec
momentum after collision= m₁ v₃+ v₄m₂=
300v₃+15m/sx 1000kg
now from the law of conservation of momentum
momentum before collision= momentum after collision
3000kgm/s= 300v₃ +15000kgm/s
-15000+3000= 300v₃
-12000=300v₃
v₃= -40m/s
the velocity of truck after collision is 40m/s and the negative sign indicates that its a recoil velocity.
Answer:
Total mass of combination = 2+3+5 = 10kg.
Acceleration produced = 2m/s^2
hence force =( total mass × acceleration)= (2×10)= 20 N.
Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N
applied force on 2 kg block = 20N
Force between 2 kg and 3 kg block = (20-4) = 16N. ans
Net force on 3 kg block = 3 × 2 =6N.
Applied force on 3 kg block due to 2 kg block = 16N.
hence, force between 3 kg and 5 kg block = (16-6) = 10N .
answers:-
(a) 20 N
(b) 16N
(c) 10 N