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Bezzdna [24]
3 years 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:
kozerog [31]3 years ago
7 0

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

given,                                  

weight of the person = 625 N

weight of the bike = 98 N        

Pressure on each Tyre = 7.60 x 10⁵ Pa

Area of contact on each Tyre = ?          

total weight of the system = 625 + 98

                                             = 723 N

Let F be the force on both the Tyre

F + F = W                                    

2 F  = 723                                    

F = 361.5 N                                

F = P A                                            

A = \dfrac{F}{P}                          

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

A = 4.76 x 10⁻⁴ m²

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allochka39001 [22]

Answer:

ni = 2.04e19

Explanation:

we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have

n = p = ni

from intrinsic carrier concentration

\sigma = n\left | e \right | \mu_e + n\left | e \right | \mu_h

\sigma = ni\left | e \right | \mu_e  + ni\left | e \right | \mu_h

\sigma = ni \left | e \right | ( \mu_e + \mu_h)

1.7 = ni * 1.6*10^{-19} * (.35 + .17)

ni = 2.014 *10^{19} m^{-3}

ni = 2.04e19

5 0
2 years ago
1. Identify which of the following will not increase the current induced in a wire loop moving through a magnetic field. a. incr
djyliett [7]

Answer:

Rotating the loop until it is perpendicular to the field  

Explanation:

Current is induced in a conductor when there is a change in magnetic flux.

The strength of the induced current in a wire loop moving through a magnetic field can be increased or decreased by the following methods:

By increasing the strength of the magnetic field there will be increased in the induced current. If the strength of the magnetic field is decreased then there is a decrease in induced current.    

By increasing the speed of the wire there will be increased in the induced current. When the speed of the wire is decreased then there is a decrease in induced current.

By increasing the number of turns of the coil the strength of the induced current can be increased. when there is less number of turns in coils then there is a decrease in induced current.  

Rotating the loop until it is perpendicular to the field will not increase the current induced in a wire loop moving through a magnetic field.

Therefore, the option is (c) is correct.

4 0
3 years ago
A ball is thrown straight up. If the launch velocity is 15 m/s, at what SPEED will the ball return to the thrower’s hand?
IgorC [24]

Answer:

Vf = 15 m/s

Explanation:

First we consider the upward motion of ball to find the height reached by the ball. Using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = -9.8 m/s² (negative sign for upward motion)

h = height =?

Vf = Final Velocity = 0 m/s (Since, ball momentarily stops at highest point)

Vi = Initial Velocity = 15 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (15 m/s)²

h = (-225 m²/s²)/(-19.6 m/s²)

h = 11.47 m

Now, we consider downward motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height = 11.47 m

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

2(9.8 m/s²)(11.47 m) = Vf² - (0 m/s)²

Vf = √(224.812 m²/s²)

<u>Vf = 15 m/s</u>

8 0
3 years ago
Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h
vodka [1.7K]

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

R_A=2R_B

Also Potential of both sphere is same

V_A=V_B

V=\frac{kQ}{R}

thus

k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}

\frac{Q_A}{Q_B}=\frac{R_A}{R_B}

\frac{Q_A}{Q_B}=\frac{2}{1}=2

\frac{Q_B}{Q_A}=\frac{1}{2}

(b)Ratio of \frac{E_B}{E_A}

Electric Field is given by E=\frac{kQ}{R^2}

thus E_A=\frac{kQ_A}{R_A^2}----1

E_B=\frac{kQ_B}{R_B^2}----2

Divide 2 by 1

\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}

\frac{E_B}{E_A}=\frac{1}{2}\times 4=2

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