A change in an object's speed has a(n) _________ effect on its kinetic energy than a change in its mass = <span>A greater effect.</span>
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
Answer:
3.33 seconds
Explanation:
We can use the velocity formula [ v = u + at ] to solve.
Find the value "u".
135km/h -> 135km*1000m/3600s -> 37.5m/s
Find the value "v".
75km/h -> 75km*1000m/3600s -> 20.83m/s
Keep in mind we are dealing with "deceleration" so when we input 5.0m/s into the formula, it will be a negative value.
Now, find "t" which is the value we aren't given with the values we're given in the question.
20.83 = 37.5 - 5t
-16.67 = -5t
3.33 = t
Best of luck!
Answer:
a. 0.199 ms b. 5.03 kHz c. 0.1 mJ
Explanation:
a. The period of oscillation of an L-C circuit is T = 2π√(LC) where L = inductance = 20 mH = 20 × 10⁻³ H and C = capacitance = 0.005 mF = 5 × 10⁻⁶ F.
So, T = 2π√(LC)
= 2π√(20 × 10⁻³ H × 5 × 10⁻⁶ F)
= 2π√(100 × 10⁻¹¹)
= 2π√(10 × 10⁻¹⁰)
= 2π(3.16 × 10⁻⁵)
= 19.87 × 10⁻⁵
= 1.987 × 10⁻⁴ s
= 1.99 × 10⁻⁴ s
= 0.199 × 10⁻³ s
= 0.199 ms
b. frequency , f = 1/T where T = period = 0.199 × 10⁻³ s.
So, f = 1/0.199 × 10⁻³ s
= 5.03 × 10³ Hz
= 5.03 kHz
c. The electromagnetic energy E = 1/2Li² where L = inductance = 20 × 10⁻³ H and i = current = 100 mA = 0.1 A
So, E = 1/2Li²
= 1/2 × 20 × 10⁻³ H × (0.1 A)²
= 0.1 × 10⁻³ J
= 0.1 mJ
Answer:
6) 2.6 m/s, 31°
7) 9.2 m/s
8) 1.2 s
Explanation:
I'll do #6, #7, and #8 as examples. You can solve #9 using the equation from #7, and #10 using the equation from #8.
6) Take north to be +y and east to be +x.
Given:
vₓ = 2.2 m/s
vᵧ = 1.3 m/s
Find: v
v² = vₓ² + vᵧ²
v² = (2.2 m/s)² + (1.3 m/s)²
v ≈ 2.6 m/s
θ = atan(vᵧ / vₓ)
θ = atan(1.3 / 2.2)
θ ≈ 31°
7) Given:
Δy = -4.3 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-4.3 m)
v ≈ 9.2 m/s
8) Given:
Δy = -6.7 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-6.7 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 1.2 s