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Ne4ueva [31]
2 years ago
11

A Mercedes-Benz 300SL (m = 1700 kg) is parked on a road that rises 15 degrees above the horizontal. What are the magnitudes of (

a) the normal force and (b) the static frictional force that the ground exerts on the tires?
Physics
1 answer:
kogti [31]2 years ago
4 0

Answer: See below

Explanation:

<u>Given:</u>

Mass of the Mercedes-Benz (m) = 1700 kg

Inclination of the road (θ) = 15.0

<em>The free body diagram is shown in figure attached below</em>

<em />

a) The normal force is equal to the cos component of the weight of the car.

\begin{aligned}&f=m g \cos \theta \\&f=1700 \times 9.81 \times \cos 15 \\&f=16108.74 \mathrm{~N}\end{aligned}

b) The static force will be equal to the weight's sin component.

\begin{aligned}&f=m g \sin \theta \\&f=1700 \times 9.81 \times \sin 15 \\&f=4316.32 \mathrm{~N}\end{aligned}

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d) the sound originates from a vibration.

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The fire alarm goes off, and a 75 kg firefighter slides down a pole with a constant acceleration of a = 6 m/s square. What is th
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Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10
11111nata11111 [884]

Answer:

1.62\times 10^{-8}\ \text{s}

Explanation:

\epsilon_0 = Vacuum permittivity = 8.854\times 10^{-12}\ \text{F/m}

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V = Initial voltage = 100 V

R = Resistance = 1000\ \Omega

Capacitance is given by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}

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V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}

The time taken for the potential difference to reach the required level is 1.62\times 10^{-8}\ \text{s}.

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2 years ago
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