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LenKa [72]
2 years ago
11

1. A car’s velocity increases from 4.0 m/s to 36 m/s over a 4.0 second time interval. What is its average acceleration?

Physics
1 answer:
Finger [1]2 years ago
7 0

Explanation:

36-4/4= 9 m/squared. meter per squared.

acceleration unit is meter per second Square.equation is velocity by time.for average final(36) minus initial(4)

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Imagine you could travel to the moon where the acceleration due to gravity is 1.6 m/s^2. What would be the period of
VladimirAG [237]

Answer:

4.9612 s

Explanation:

Applying,

T = 2π√(L/g)............... Equation 1

Where T = period of the pendulum, L = Lenght of the pendulum, g = acceleration due to gravity of the moon, π = pie.

From the question,

Given: L = 1 m, g = 1.6 m/s²

Constant: π = 3.14

Substitute these values into equation 1

T = 2×3.14×√(1/1.6)

T = 6.28√(0.625)

T = 6.28×0.79

T = 4.9612 s

6 0
3 years ago
Which of the following are events?
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8 0
3 years ago
Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

6 0
3 years ago
What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?
EleoNora [17]
Force = mass x acceleration
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force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN

i hope this is right (^^)
4 0
2 years ago
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