Answer:
Twice as fast
Explanation:
Solution:-
- The mass of less massive cart = m
- The mass of Massive cart = 2m
- The velocity of less massive cart = u
- The velocity of massive cart = v
- We will consider the system of two carts to be isolated and there is no external applied force on the system. This conditions validates the conservation of linear momentum to be applied on the isolated system.
- Each cart with its respective velocity are directed at each other. And meet up with head on collision and comes to rest immediately after the collision.
- The conservation of linear momentum states that the momentum of the system before ( P_i ) and after the collision ( P_f ) remains the same.
- Since the carts comes to a stop after collision then the linear momentum after the collision ( P_f = 0 ). Therefore, we have:
- The linear momentum of a particle ( cart ) is the product of its mass and velocity as follows:
m*u - 2*m*v = 0
Where,
( u ) and ( v ) are opposing velocity vectors in 1-dimension.
- Evaluate the velcoity ( u ) of the less massive cart in terms of the speed ( v ) of more massive cart as follows:
m*u = 2*m*v
u = 2*v
Answer: The velocity of less massive cart must be twice the speed of more massive cart for the system conditions to hold true i.e ( they both come to a stop after collision ).
Answer:
The speed of the ball when it reaches equilibrium position is 3.31 m/s
Explanation:
Given;
mass of the object, m = 0.25 kg
initial displacement of the object, h₁ = 0.56 m
spring constant, k = 105 N/m
displacement at equilibrium position, h₂ = 0
initial velocity of the object, v₁ = 0
velocity of the object at equilibrium position = v₂
The change in gravitational potential energy at the equilibrium position is given as;
ΔP.E = mg(h₂ - h₁)
The change in kinetic energy of the object at the equilibrium position is given as;
ΔK.E = ¹/₂m(v₂² - v₁²)
Apply the principle of conservation of mechanical energy;
ΔK.E + ΔP.E = 0
¹/₂m(v₂² - v₁²) + mg(h₂ - h₁) = 0
¹/₂m(v₂² - 0) + mg(0 - h₁) = 0
¹/₂mv₂² - mgh₁ = 0
¹/₂mv₂² = mgh
¹/₂v₂² = gh
v₂² = 2gh
v₂ = √2gh
v₂ = √(2 x 9.8 x 0.56)
v₂ = 3.31 m/s
Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s
Explanation:
It is given that,
A nerve signal travels 150 meters per second. It is the speed of the nerve signal. We need to convert the number of kilometers that the nerve signal will travel in the same time.
We know that,
1 kilometer = 1000 meter
1 hour = 3600 seconds
So, the nerve signal will travel at the rate of 540 km/h. Hence, this is the required solution.
Infra-red radiation has a slightly longer wavelength than red light ...
the longest wavelength that is visible to the human eye.
Although infrared can't be detected by the eye, we do have other
receptors that CAN detect it. They are some of the nerve endings
in our skin. When infrared radiation hits these, it produces the feeling
of warmth. Infrared radiation carries heat energy.
First question:
The magnitude of current flowing in a circuit is described in
units of Amperes. The device used to measure it is an Ampmeter,
or Ammeter.
Second question:
This question is so absurd that it should not be dignified with an
answer. Although 'E' is often used as the symbol for Electromotive
force, potential difference, and voltage, there's certainly no rule.
Anyone is free to use 'M', 'Q', 'Θ', or 'Щ' to denote voltage when
they write electrical formulas, just as long as they make sure to
explain the meaning of whatever symbols they use.
