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VikaD [51]
2 years ago
13

The four tires of an automobile are inflated to a gauge pressure of 2.02×10^5 Pa. Each tire has an area of 217 cm^2 in contact w

ith the ground. Determine the weight of the automobile.
Physics
1 answer:
Nezavi [6.7K]2 years ago
5 0

The weight of the automobile is 17,533.6 N.

<h3>Weight of the automobile</h3>

The weight of the automobile is calculated as follows;

P = F / A

F = (W/4)

P = (W/4) / A

P = W/4A

W = 4AP

where;

  • P is pressure
  • A is area

W = (4)(217 x 10⁻⁴ m²)(2.02 x 10⁵ Pa)

W = 17,533.6 N

Thus, the weight of the automobile is 17,533.6 N.

Learn more about weight here: brainly.com/question/2337612

#SPJ1

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Lenz's law

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A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how
GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

x=V_{o}cos\theta t   (1)

Where:

V_{o}=48.1 m/s is the cannonball's initial velocity

\theta=0 because we are told the cannonball is shot horizontally

t is the time since the cannonball is shot until it hits the ground

y-component:

y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}   (2)

Where:

y_{o}=1.5m  is the initial height of the cannonball

y=0  is the final height of the cannonball (when it finally hits the ground)

g=9.8m/s^{2}  is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it X_{max} and this occurs when y=0.

So, firstly we will find the time with (2):

0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2   (3)

Rearranging the equation:

0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m   (4)

-(4.9 m/s^{2})t^2+(48. 1 m/s)  t+1.5 m=0   (5)

This is a <u>quadratic equation</u> (also called <u>equation of the second degree</u>) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} (6)

Where:

a=-4.9 m/s^{2}

b=48.1 m/s

c=1.5 m

Substituting the known values:

t=\frac{-48.1 \pm \sqrt{48.1^{2}-4(-4.9)(1.5)}}{2(-4.9)} (7)

Solving (7) we find the positive result is:

t=9.847 s (8)

Substituting this value in (1):

x=(48.1 m/s)cos(0\°) (9.847 s)   (9)

x=473.640 m  This is the horizontal distance the cannonball traveled before it landed on the ground.

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WITCHER [35]
Measurement=<span>the size, length, or amount of something

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