Answer:
the distance traveled by the car is 42.98 m.
Explanation:
Given;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
the braking force applied to the car, f = 5620 N
time of motion of the car, t = 2.5 s
The decelaration of the car is calculated as follows;
-F = ma
a = -F/m
a = -5620 / 2500
a = -2.248 m/s²
The distance traveled by the car is calculated as follows;
s = ut + ¹/₂at²
s = (20 x 2.5) + 0.5(-2.248)(2.5²)
s = 50 - 7.025
s = 42.98 m
Therefore, the distance traveled by the car is 42.98 m.
Answer:
W = Fd = KE =1/2mv²
Explanation:
not sure if that's what your looking for but i'm pretty sure this is it.
Answer:
915 Hz
Explanation:
The observed frequency from a sound source is given as
f₀ = f [(v + v₀)/(v+vₛ)]
where
f₀ = observed frequency of the sound by the observer = ?
f = actual frequency of the sound wave = 983 Hz
v = actual velocity of the sound waves = 343 m/s
vₛ = velocity of the source of the sound waves = 55.9 m/s
v₀ = velocity of the observer = 28.4 m/s
f₀ = 983 [(343+28.4)/(343+55.9)]
f₀ = 915.2 Hz = 915 Hz
Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
Answer:
5 ms-2
Explanation:
F = ma
F = 100N
m = 20kg ( you should make sure the unit is kg before you answer the question)
100 = 20a
a = 100÷ 20
a = 5 ms-2
