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jolli1 [7]
4 years ago
13

The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact wit

h the ground. Determine the weight of the automobile.
Physics
1 answer:
Stels [109]4 years ago
7 0

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, P=2.2\times 10^5\ Pa

Area of each tire, A=0.023\ m^2

Area of 4 tires,  A=0.092\ m^2

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

P=\dfrac{F}{A}

F=P\times A

F=2.2\times 10^5\times 0.092

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

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Answer:240 meters

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An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.
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Answer:

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Explanation:

From the question we are told that

Electric field of intensity E= 3.25 kN/C

Rectangle parameter Width W=0.350 m  Length L=0.700 m

Angle to the normal \angle=30.5 \textdegree

Generally the equation for Electric flux at parallel to the yz plane \varphi_1 is mathematically given by

\varphi_1=EA cos theta

\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0

\varphi_1= 796.25 N m^2/C

Generally the equation for Electric flux at parallel to xy  plane \varphi_2 is mathematically given by

\varphi_2=EA cos theta

\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90

\varphi_2= 0 N m^2/C

Generally the equation for Electric flux at angle 30 to x plane \varphi_3 is mathematically given by

\varphi_3=EA cos theta

\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5

\varphi_3=686.072219  N m^2/C

\varphi_3=686.1  N m^2/C

7 0
3 years ago
Two manned satellites approach one another at a relative velocity of v = 0.150 m/s, intending to dock. The first has a mass of m
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Answer:

= - 0.41m/s

Explanation:

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Velocity of the second satellite

V_2 = \frac{2m_1}{m_1 + m_2} u

V_2 = \frac{2  \times 4 \times 10^3}{4 \times 10^3 + 7.5  \times 10^3} \\\\= 0.10435

Final velocity = V(1) - V(2)

V = -0.30435 - 0.10435\\\\= -0.4087

≅ -0.41m/s

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Answer:

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Explanation:

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What is the oscillation period of your eardrum when you are listening to the A4 note on a piano (frequency 440 Hz)?
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