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Papessa [141]
10 months ago
6

Joanne drives her car at a speed of 20 m/s. when she applied her breaks, a frictional force of 2000 N brought her car to a compl

ete stop in 10 seconds. what is the mass of her car
A) 1000 Kg
B) 1300 Kg
C) 20,000 Kg
D) 800 Kg​
Physics
1 answer:
Papessa [141]10 months ago
8 0

Answer:

A) 1000 kg

Explanation:

vf = vi + at

0 = 20 + (a)(10)

a = -2.0 m/s^2

F = ma

2000 = (m)(2)

m = 1000 kg

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A satellite was in two separate crashes. In both crashes, the satellite had the same mass. Engineers want to know about the spee
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3 years ago
A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e
mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)

200 \times 0.1 L = 53.5 \times (x - 0.6 L)

0.374 L =x - 0.6 L

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7 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
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