Answer:
The impulse exerted by one cart on the other has a magnitude of 4 N.s.
Explanation:
Given;
mass of the first cart, m₁ = 2 kg
initial speed of the first car, u₁ = 3 m/s
mass of the second cart, m₂ = 4 kg
initial speed of the second cart, u₂ = 0
Let the final speed of both carts = v, since they stick together after collision.
Apply the principle of conservation of momentum to determine v
m₁u₁ + m₂u₂ = v(m₁ + m₂)
2 x 3 + 0 = v(2 + 4)
6 = 6v
v = 1 m/s
Impulse is given by;
I = ft = mΔv = m(
The impulse exerted by the first cart on the second cart is given;
I = 2 (3 -1 )
I = 4 N.s
The impulse exerted by the second cart on the first cart is given;
I = 4(0-1)
I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).
Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.
Answer:
toward the normal
Explanation:
Light travels at different speed in different mediums.
Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.
Or ,
n = c/v.
Light travels at a slower speed in water as compared to air because there are more number of interfering molecules in the path of the light in case of water as compared to liquid.
When a light travels from lower denser medium say water to higher denser medium say water, it bends towards the perpendicular (normal) as its speed reduces in that medium.
Because the temperature remains constant, we can apply Boyle's Law which states that
pV = constant
where
p = pressure
V = volume
Define the two states of the gas.
State 1
Pressure = p₁
Volume = 1000 ml
State 2
Pressure = p₂
Volume = 500 ml
Apply Boyle's law.
1000p₁ = 500p₂
2 = p₂/p₁
By halving the volume, the pressure doubles.
Answer:
The pressure increases by a factor of 2.
Answer:
t_{out} = 
t_{in}, t_{out} = 
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D / 
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D / 
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / 
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =
Answer:
The initial speed of the pelican is 8.81 m/s.
Explanation:
Given;
height of the pelican, h = 5.0 m
horizontal distance, X = 8.9 m
The time of flight is given by;
The initial horizontal speed of the pelican is given by;
X = vₓt
vₓ = X / t
vₓ = 8.9 / 1.01
vₓ = 8.81 m/s
Therefore, the initial speed of the pelican is 8.81 m/s.
