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trasher
3 months ago
13

A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa

lly by the two tires. If the pressure in each tire is 7.60 x 10^5 Pa, what is the area of contact between each tire and the ground?
Physics
1 answer:
Keith_Richards [3.2K]3 months ago
7 0

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

Given data:

Person's weight = 625 N

Bike's weight = 98 N

Pressure per tire = 7.60 x 10⁵ Pa

Find: Contact area per tire

Total system weight = 625 + 98 = 723 N

Let F represent the force supported by each tire

2F = 723 N

Therefore, F = 361.5 N

Using the formula F = P × A

A = \dfrac{F}{P}

A = \dfrac{361.5}{7.60 \times 10^5}

Contact area, A = 4.76 x 10⁻⁴ m²

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A string is stretched by two equal but opposite forces f newton each what is tension in string
Maru [3345]

The string does not experience any force of tension, as it balances two forces acting in the same direction. Hence, the tension is zero.

Explanation:

If tension existed in the string, it would mean that two equal but opposite forces are exerting pull in contrary directions.

When a force of f newtons is applied from the right and another force of f newtons from the left, the resulting action occurs through one force. Because there is action on the same string in opposing directions, the tension in the string can only be equal to the magnitude of the string itself.

Therefore, the string indeed has no tension since it is dealing with two forces acting in the same direction. Thus, the tension is zero.

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3 months ago
50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his
Softa [3030]

At time t, the ball's horizontal and vertical velocities can be represented as

v_x=v_{xi}

v_y=v_{yi}-gt

However, since the ball is thrown horizontally, we have v_{yi}=0. The horizontal and vertical positions at time t are

x=v_{xi}t

y=20.0\,\mathrm m-\dfrac g2t^2

The ball travels a distance of 22 m horizontally from the throw point, thus

22\,\mathrm m=v_{xi}t

With this, we determine that the time for the ball to reach the ground is

t=\dfrac{22\,\rm m}{v_{xi}}

When it touches down, y=0 and

0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2

\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}

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2 months ago
Suppose you are designing an amplifier and loudspeaker system to use at a rock concert. You want to make it as loud as possible.
Softa [3030]
The response is outlined below. Audio power amplifiers are present in various sound systems, including those for sound reinforcement, public addresses, home audio, and musical instrument amplifiers like those for guitars. This component is the final electronic element in the audio playback chain before signals reach the loudspeaker. To achieve the loudest possible sound, it is essential to maximize output while maintaining high input and low output impedance.
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2 months ago
A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a const
Keith_Richards [3271]

a) -1.54 m/s^2

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

mg=m\frac{v^2}{r}where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top g=9.8 m/s^2

= loop radius

vBy solving for v,

r=1200 m

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived fromv_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

v_A=550 km/h =152.8 m/s

Given the plane's deceleration is consistent, we can obtain it using the following equation:

s=\pi r=\pi(1200)=3770 m

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2- The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B:N=m\frac{v_B^2}{r}

v_B

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B:

s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m

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