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2 years ago

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The (nonconservative) force propelling a 1.50 103-kg car up a mountain road does 5.00 106 J of work on the car. The car starts f

rom rest at sea level and has a speed of 25.0 m/s at an altitude of 1.90 102 m above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are non conservative forces.

1 answer:

kipiarov [429]2 years ago

4 0

Answer:

car. The car starts from rest at sea level and has a speed of 29.0 m/s at an altitude of 2.20x10^2 m above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are nonconservative forces.

Explanation:

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pitot tube on an airplane flying at a standard sea level reads 1.07 x 105 N/m2. What is the velocity of the airplane?

Allushta [10]

Answer:

V_infinty=98.772 m/s

Explanation:

complete question is:

The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?

<u>solution:</u>

<u>given:</u>

<em>p_o=1.07*10^5 N/m^2</em>

<em>ρ_infinity=1.23 kg/m^2</em>

<em>p_infinity=1.01*10^5 N/m^2</em>

p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2

V_infinty^2=9756.097

V_infinty=98.772 m/s

8 0

2 years ago

Norma kicks a soccer ball with an initial velocity of 10.0 meters per second at an angle of 30.0°. If the ball moves through the

enyata [817]

<u>Answer</u>

27.7

<u>Explanation</u>

The ball was hit at an angle of 30°, with the horizontal at a speed of 10 m/s. We have to find the horizontal component of speed.

cosx = adjacent/hypotenuse

cos 30 = adjacent / 10

adjacent = 10 cos30

= 8.66 m/s ⇒ This is the horizontal speed.

Now find the horizontal distance.

Distance = speed × time

= 8.66 × 3.2

= 27.71

Answer to the nearest tenth = 27.7

4 0

2 years ago

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A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it

vredina [299]

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

$F = ma$ ...(I)

Using formula of friction force

$F= \mu m g$ ....(II)

Put the value of F in the equation (II) from equation (I)

$ma=\mu mg$ ....(III)

$a = \mu g$

Put the value in the equation (III)

$a=0.24\times9.8$

$a=2.352\ m/s^2$

We need to calculate the distance,

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}2.352\times(3.0)^2$

$s=10.584\ m\ approx\ 11\ m$

Hence, The distance is 11 m.

3 0

2 years ago

A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the

Elis [28]

Answer:

a) $R = 171$ μΩ

b) $E = 1.7 *10^{-4} V/m$

c) $R_{2} = 1.16 *10^{-4}$ Ω

here * stand for multiplication

Explanation:

length of cylinder = 1.5 m

radius of cylinder = 1.1 cm

resistivity depends on the distance x from the left

$p(x)=a+bx^2$ ............(i)

using equation

$R = \frac{pl}{a}$

let dR is the resistance of thickness dx

$dR =\frac{p(x)dx}{a}$

where p(x) is resistivity l is length

a is area

$\int\limits^R_0 {dR} =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx \\$ .........................(2)

after integration

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$ ...............(3)

it is given $p(0) = a = 2.25 * 10 ^{-8}$ Ωm

$p(L) = a + b(L)^2$ = $8.5 * 10 ^{-8}$ Ωm

$8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\$

(here * stand for multiplication )

on solving we get

$b = 2.78* 10^{-8}$ Ωm

put each value of a and b and r value in equation 3rd we get

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$

$R = 1.71 * 10^{-4}$ Ω

$R = 171$ μΩ

FOR (b)

for mid point x = L/2

E = p(x)L

for x = L/2

$p(L/2) = a+b(L/2)^2$

for given current I = 1.75 A

so electric field

$E = \frac{[a+b(L/2)^2]I }{\pi r^2}$

by substitute the values

we get;

$E = 1.7 *10^{-4} V/m$

(here * stand for multiplication )

c ).

75 cm means length will be half

that is x = L/2

integrate the second equation with upper limit L/2

Let resistance is $R_{1}$

so after integration we get

$R_{1} = \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}$

substitute the value of a , b and L we get

$R_{1} = 5.47 * 10 ^{-5}$ Ω

for second half resistance

$R_{2} = R- R_{1}$

$R_{2} = 1.7 *10^{-4} -5.47 *10^{-5}$

$R_{2} = 1.16 *10^{-4}$ Ω

(here * stand for multiplication )

5 0

2 years ago

A spring has a spring constant of 48 N/m. The end of the spring hangs 8 m above the ground. How much weight can be placed on the

Setler79 [48]

The answer is 96 N .....................................

7 0

2 years ago

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