Answer:
(A) The force would be lower on the school day than the weekend.
Explanation:
In a school day the bus is on the road travelling with kids. But in a weekend it is parked. When a object stays still that means the the force which is working on the earth by the bus is equal to the force which works on the bus by the earth. we can understand it clearly by, Newton's second law of motion.
This pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. so as it is when it is balanced it is not moving.
And on a school day the bus is moving so it has a force pulling itself forward . so it means that the force which the bus has is greater than the gravitational force.
We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is
X=0m/s
From the question we are told that
mosquito flying 2 m/s
against a 2 m/s headwind
Generally
The speed over the ground is the Flight Speed minus resistance speed
Generally the equation for the speed over the ground is mathematically given as
X=Flight Speed-resistance speed
Therefore
X=2-2
X=0m/s
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Answer : The mass of a sample of water is, 888.89 grams
Explanation :
Latent heat of vaporization : It is defined as the amount of heat energy released or absorbed when the liquid converted to vapor at atmospheric pressure at its boiling point.
Formula used :
where,
q = heat = 2000 kJ = 
(1 kJ = 1000 J)
L = latent heat of vaporization of water = 
m = mass of sample of water = ?
Now put all the given values in the above formula, we get:
(1 kg = 1000 g)
Therefore, the mass of a sample of water is, 888.89 grams
Answer:
-20.0 m/s and 30.0 m/s
Explanation:
Momentum is conserved:
m (30.0) + m (-20.0) = m v₁ + m v₂
30.0 − 20.0 = v₁ + v₂
10.0 = v₁ + v₂
Since the collision is perfectly elastic, energy is also conserved. Since there's no rotational energy or work done by friction, the initial kinetic energy equals the final kinetic energy.
½ m (30.0)² + ½ m (-20.0)² = ½ mv₁² + ½ mv₂²
(30.0)² + (-20.0)² = v₁² + v₂²
1300 = v₁² + v₂²
We now have two equations and two variables. Solve the system of equations using substitution:
1300 = v₁² + (10 − v₁)²
1300 = v₁² + 100 − 20v₁ + v₁²
0 = 2v₁² − 20v₁ − 1200
0 = v₁² − 10v₁ − 600
0 = (v₁ + 20) (v₁ − 30)
v₁ = -20, 30
If v₁ = -20, v₂ = 30.
If v₁ = 30, v₂ = -20.
So either way, the final velocities are -20.0 m/s and 30.0 m/s.
