When a warm air mass catches up with a cold air mass, forming a warm front, the warm air slides over the top of the cold air, be
1 answer:
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Answer:
A) F = - 8.5 10² N, B) I = 21 N s
Explanation:
A) We can solve this problem using the relationship of momentum and momentum
I = Δp
in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction
v₀ = 8.50 m / s
v_f = -8.50 m / s
F t = m v_f -m v₀
F =
let's calculate
F =
F = - 8.5 10² N
B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations
v² = v₀² - 2g (y- y₀)
as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0
v =
calculate
v =
v = 14 m / s
to calculate the momentum we use
I = Δp
I = m v_f - mv₀
when it hits the ground its speed drops to zero
we substitute
I = 1.50 (0-14)
I = -21 N s
the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is
I = 21 N s
Answer : The correct option is, (B) -5448 kJ/mol
Explanation :
First we have to calculate the heat required by water.
where,
q = heat required by water = ?
m = mass of water = 250 g
c = specific heat capacity of water =
= initial temperature of water = 293.0 K
= final temperature of water = 371.2 K
Now put all the given values in the above formula, we get:
Now we have to calculate the enthalpy of combustion of octane.
where,
= enthalpy of combustion of octane = ?
q = heat released = -81719 J
n = moles of octane = 0.015 moles
Now put all the given values in the above formula, we get:
Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.
Answer:0
Explanation:
Given
circumference of circle is 2 m
Tension in the string
In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is