Answer:
If a body covers equal displacement in equal interval of time,however small may be the body is said to have uniform velocity. If a body covers in equal displacement in equal interval of time is called variable velocity.
Complete solution is given below in attachment
Answer:
A) vectors: veloicty, force
scalar: speed, work
B) t = 1.75 s, C) v = - 17 2 m / s
Explanation:
We answer each part separately
A) A vector magnitude has magnitude and direction instead a scalar magnitude has only magnitude
vector quantities: the speed of a car number is the magnitude and direction is where it goes
Force, the number is the magnitude and above that applies gives direction
Scalar magnitude: how quickly the number of the speedometer of the car
Temperature, work
B) I = 15 m height to the soil and get to calculate time = 0
y = y₀ + v₀ t - ½ g t²
as the ball is loose its initial velocity is zero
0 = 0 +0 - ½ g t²
t =
t = 
t = 1.75 s
C) the velocity to the reach the floor
v = vo - g t
v = 0 - g t
v = - 9.8 1.74
v = - 17 2 m / s
The negative signt iindicates that the speed goes down
Answer:
Explanation:
Use the one-dimensional equation
which says that the final velocity of an object is equal to the object's initial velocity plus its acceleration times time. We are looking for time. That means the equation looks like this:
0 = 30 + (-4)t and
-30 = -4t so
t = 7.5 sec
Answer:
The latent heat of vaporization of water is 2.4 kJ/g
Explanation:
The given readings are;
The first (mass) balance reading (of the water) in grams, m₁ = 581 g
The second (mass) balance reading (of the water) in grams, m₂ = 526 g
The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ
The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ
The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature
Based on the measurements, we have;
The latent heat of vaporization = ΔQ/Δm
∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g
The latent heat of vaporization of water = 2.4 kJ/g