Answer:
The final velocity of the car is 26.65 m/s.
Explanation:
Given;
acceleration of the racecar, a = 6.5 m/s²
initial velocity of the car, u = 0
time of motion, t = 4.1 s
The final velocity of the car is given by;
v = u + at
where;
v is the final velocity of the car
suvstitute the givens
v = 0 + (6.5)(4.1)
v = 26.65 m/s.
Therefore, the final velocity of the car is 26.65 m/s.
Answer:
PE = 44.1 J
Explanation:
Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)
We have:
- 1 kilograms = 1000 grams.
We convert it using a rule of 3, replacing, simplifying units and solving:
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Earth's gravity is known to be 9.8 m/s², so we have:
Data:
- m = 0.3 kg
- g = 9.8 m/s²
- h = 15 m
- PE = ?
Use formula of potencial energy:
Replace and solve:
Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.
The potential energy of the volleyball is <u>44.1 Joules.</u>
Greetings.
Answer:
T = 4 sec / 2 = 2 sec period of revolution
S = 2 pi R = 2 * pi * 1.75 m = 11 m
V = S / T = 11 m / 2 sec = 5.5 m/s speed of object
In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.
Answer:
ω = 12.023 rad/s
α = 222.61 rad/s²
Explanation:
We are given;
ω0 = 2.37 rad/s, t = 0 sec
ω =?, t = 0.22 sec
α =?
θ = 57°
From formulas,
Tangential acceleration; a_t = rα
Normal acceleration; a_n = rω²
tan θ = a_t/a_n
Thus; tan θ = rα/rω² = α/ω²
tan θ = α/ω²
α = ω²tan θ
Now, α = dω/dt
So; dω/dt = ω²tan θ
Rearranging, we have;
dω/ω² = dt × tan θ
Integrating both sides, we have;
(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ
This gives;
-1[(1/ω_o) - (1/ω)] = t(tan θ)
Thus;
ω = ω_o/(1 - (ω_o × t × tan θ))
While;
α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²
Thus, plugging in the relevant values;
ω = 2.37/(1 - (2.37 × 0.22 × tan 57))
ω = 12.023 rad/s
Also;
α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²
α = 8.64926751525/0.03885408979 = 222.61 rad/s²
