A 2-kg toy car accelerates from 5 to 10 m/s.
1 answer:
<h3><u>Full Question:</u></h3>
A 2-kg toy car accelerates from 5 to 10 m/s in 2 seconds. Find the applied force.
<h2><u>Answer:</u></h2>Force applied is .
Force is defined as something which changes or tends to change the velocity of a body. And acceleration is defined as the rate of change of velocity of a body per time. Both of them are vector quantities.
The initial velocity of the body = 5 m/s.
The final velocity of the body = 10 m/s.
So the change in velocity = 10 - 5 m/s = 5m/s.
Time period = 2 seconds.
So acceleration ==
The force is defined as the product of mass and acceleration of the body.
Mass of body = 2 kg.
So, force applied = . =
.
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Answer:
E = 1.655 x 10⁷ N/C towards the filament
Explanation:
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E = [/tex]
where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is
8.85 x 10⁻¹².
Putting the given values in the equation given above
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Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s