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3 years ago

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An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m fr

om the barrier. What minimum coefficient of kinetic friction is required to stop the automobile before it hits the barrier?

1 answer:

AleksAgata [21]3 years ago

5 0

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

$a = \dfrac{400}{2\times 50}$

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

$\mu = \dfrac{a}{g}$

$\mu = \dfrac{4}{9.81}$

μ = 0.408

hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408

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A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pres

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Answer:

a) m = 1.174 grams

b) F_g = 0.01151 N

c) F_c = 1013 N

Explanation:

Given:

- The length of a cube L = 10.0 cm

- The molar mass of air M = 28.9 g/mol

- Pressure of air P = 101.3 KPa

- Temperature of air T = 300 K

- Universal Gas constant R = 8.314 J/kgK

Find:

(a) the mass of the gas

(b) the gravitational force exerted on it

(c) the force it exerts on each face of the cube

(d) Why does such a small sample exert such a great force? (6%)

Solution:

- Compute the volume of the cube:

V = L^3 = 0.1^3 = 0.001 m^3

- Use Ideal gas law equation and compute number of moles of air n:

P*V = n*R*T

n = P*V / R*T

n = 101.3*10^3 * 0.001 / 8.314*300

n = 0.04061 moles

- Compute the mass of the gas:

m = n*M

m = 0.04061*28.9

m = 1.174 grams

- The gravitational force exerted on the mass of gas is due to its weight:

F_g = m*g

F_g = 1.174*9.81*10^-3

F_g = 0.01151 N

- The force exerted on each face of cube is due its surface area:

F_c = P*A

F_c = (101.3*10^3)*(0.1)^2

F_c = 1013 N

- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.

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4.A 31-cm long conducting wire of 9-g carrying 7- A current is placed in a uniform magnetic field. What are the strength and dir

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Answer

Given,

Length of the wire,L = 31 cm = 0.31 m

mass of the wire, m = 9 g = 0.009 Kg

Current in the wire,I = 7 A

Magnetic field strength, B= ?

Equating magnetic force to the weight of the wire.

$BIL = m g$

$B=\dfrac{m g}{IL}$

$B=\dfrac{0.009\times 9.81}{7\times 0.31}$

B = 0.0407 T

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A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/

NemiM [27]

Answer:

a) $F_H=776.952\ N$

b) $F_g=706.32\ N$

c) $v=5.4249\ m.s^{-1}$

d) $KE=1059.48\ J$

Explanation:

Given:

mass of the astronaut, $m=72\ kg$

vertical displacement of the astronaut, $h=15\ m$

acceleration of the astronaut while the lift, $a=\frac{g}{10} =0.981\ m.s^{-2}$

a)

<u>Now the force of lift by the helicopter:</u>

Here the lift force is the resultant of the force of gravity being overcome by the force of helicopter.

$F_H-F_g=m.a$

where:

$F_H=$ force by the helicopter

$F_g=$ force of gravity

$F_H=72\times 0.981+72\times9.81$

$F_H=776.952\ N$

b)

The gravitational force on the astronaut:

$F_g=m.g$

$F_g=72\times 9.81$

$F_g=706.32\ N$

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero, $u=0\ m.s^{-1}$

using equation of motion:

$v^2=u^2+2a.h$

$v^2=0^2+2\times 0.981\times 15$

$v=5.4249\ m.s^{-1}$

c)

Hence the kinetic energy:

$KE=\frac{1}{2} m.v^2$

$KE=0.5\times 72\times 5.4249^2$

$KE=1059.48\ J$

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TiliK225 [7]

Answer:

1-In a uniform electric field, the field lines are straight, parallel, and uniformly spaced this statement is true.

2-Electric field lines near positive point charges radiate outward. this statement is also true.

3-The electric force acting on a point charge is proportional to the magnitude of the point charge. this statement is true as well.

Explanation:

the electric field created by a point charge is defined by E=KQ/r^2 where k is constant, q is magnitude of charge and r is the distance away from the point charge so the electric filed is distance dependent and can not be constant at all distances.

electric field lines near a negative point charge are directed radially inward because negative charge attracts the field and is not clockwise.

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