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Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

Answers

Answer 1

Answer:

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

= 1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

Answer 2

Answer:

Final answer:

To calculate the thermal energy dissipated from the brakes of a car, use the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill. The temperature change of the brakes can then be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Explanation:

The thermal energy dissipated from the brakes of a car can be calculated by converting the gravitational potential energy lost by the car into internal energy of the brakes. By using the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill, we can calculate the thermal energy dissipated. From there, the temperature change of the brakes can be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

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Name the four forces in physics?

Answers

Answer:

Gravitational

Electrostatic

magnetic

Frictional

gravitational

electrostatic

magnetic

frictional

hope it helps

pls mark as brainliest

A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student holds a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?A. 30 cm
B. 60 cm
C. 75 cm
D. 90 cm

Answers

B. 60 cm

All parallel light rays are bent through the focal point of a convex lens, so the rays from the flashlight 150 cm above the floor must go through the same point on the principal axis as the rays from the flashlight 120 cm above the floor. The location of the focal point does not change when the position of the object is moved either vertically or horizontally.

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surface of this cube?

Answers

Answer:

The flux through the surface of the cube is $2.314\ Nm^(2)/C$

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = $8.0* 10^(- 2)\ m$

Volume Charge density, $\rho_(v) = 40 nC/m^(3) = 40* {- 9}\ C/m^(3)$

Now,

To calculate the electric flux:

$\phi = (q)/(\epsilon_(o))$ (1)

where

$\phi$ = electric flux

$\epsilon_(o) = 8.85* 10^(- 12)\ F/m$ = permittivity of free space

Volume Charge density for the given case is given by the formula:

$\rho_(v) = (Total\ charge, q)/(Volume of cube, V)$ (2)

Volume of cube, $V = a^(3)$

Thus

$V = (8.0* 10^(- 2))^(3) = 5.12* 10^(- 4)\ m^(3)$

Thus from eqn (2), the total charge is given by:

$q = \rho_(v)V = 40* {- 9}* 5.12* 10^(- 4)$

$q = 2.048* 10^(-11)\ F = 20.48\ pF$

Now, substitute the value of 'q' in eqn (1):

$\phi = (2.048* 10^(-11))/(8.85* 10^(- 12)) = 2.314\ Nm^(2)/C$

An isotropic point source emits light at wavelength 500 nm, at the rate of 185 W. A light detector is positioned 380 m from the source. What is the maximum rate ∂B/∂t at which the magnetic component of the light changes with time at the detector's location?

Answers

Answer:

$(dB)/(dt) = 3.49 *10^(6) \ \ T/s$

Explanation:

Given that

An isotropic point source emits light at a wavelength $\lambda$ = 500 nm

Power = 185 W

Radius = 380 m

Let's first calculate the The intensity of the wave , which is = $(Power )/(Area)$

= $(Power)/(4 \pi r ^2)$

= $(185 \ W)/( 4 \pi (380)^2)$

= $1.0195*10^(-4) \ W/m^2$

Now;

The amplitude of the magnetic field is calculated afterwards by using poynting vector

i.e

$I = ((c)/(2 \mu_0 ))B_(max^2)$

$B_(max^2) = ((2 \mu_0 I)/( c))$

$B_(max^2) = ((2 *4 \pi *10^(-7)*1.0195*10^(-4))/( 3*10^8))$

$B_(max^2) = 8.5409*10^(-19)$

$B_(max) = \sqrt {8.5409*10^(-19)}$

$B_(max) = 9.242*10^(-10)$

The magnetic field wave equation can now be expressed as;

$B = B_(max) sin (kx - \omega t)$

Taking the differentiation

$(dB)/(dt)= - \omega B_(max) \ cos ( kx - \omega t)$

The maximum value ;

$(dB)/(dt) = \omega B _(max)$

where ;

$\omega = 2 \pi f\n\omega = (2 \pi c)/(\lambda)$

then

$(dB)/(dt) = (2 \pi c)/(\lambda) B _(max)$

$(dB)/(dt) = (2 \pi 3*10^8*9.242*10^(-10))/(500*10^(-9))$

$(dB)/(dt) = 3484751.917$

$(dB)/(dt) = 3.49 *10^(6) \ \ T/s$

The maximum rate $(∂B/∂t)$ at which the magnetic component of the light changes with time at the detector's location is approximately $6.8 x 10^9$ Tesla per second (T/s).

To find the maximum rate at which the magnetic component of the light changes with time at the detector's location, you can use the formula for the rate of change of magnetic field due to an electromagnetic wave. The formula is given by:

$∂B/∂t = (2π / λ) * E * c$

Where:

$∂B/∂t$ is the rate of change of the magnetic field.

λ is the wavelength of the light.

E is the electric field strength.

c is the speed of light in a vacuum, approximately $3 x 10^8 m/s.$

You have the wavelength (λ) as 500 nm, which is 500 x 10^-9 meters, and the electric field strength (E) can be calculated using the power (P) and the distance (r) from the source. The power emitted by the source is 185 W, and the distance from the source to the detector is 380 m.

First, calculate the electric field strength (E):

$E = sqrt(P / (2π * r^2))$

$E = sqrt(185 W / (2π * (380 m)^2))$

$E = sqrt(185 W / (2π * 144400 m^2))$

$E ≈ 6.325 x 10^-5 N/C$

Now, you can calculate the rate of change of the magnetic field:

$∂B/∂t = (2π / λ) * E * c$

$∂B/∂t = (2π / (500 x 10^-9 m)) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)$

$∂B/∂t ≈ (3.77 x 10^15 Hz) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)$

$∂B/∂t ≈ 6.8 x 10^9 T/s$

So, the maximum rate at which the magnetic component of the light changes with time at the detector's location is approximately $6.8 x 10^9$ Tesla per second (T/s).

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If i throw up an object up at 31 m/s, how long will it take to get to its highest point

Answers

Answer:

Explanation:

vf=vi+at

vf=31 m/s

vi=0 m/s

a=g=9.8 m/s2

t=?

vf-vi=at

vf-vi/a=t

t=vf-vi/a

t=31 m/s-0/9.8

t=3.16 s

Students have four identical, hollow, uncharged conducting spheres, W, X, Y, and Z.Sphere Z is given a positive charge of +40 C. Sphere Z is touched first to sphere W, then sphere X, and finally to sphere Y. What is the resulting charge on sphere Y?

a. +5 με

b. +10 μC

c. +20 μC

d. +40 με

Answers

Explanation:

because they made contact that means their new force will be the same

Final answer:

Sphere Z is initially charged with +40 C. When it is touched to three other spheres, the charge is evenly distributed among them. The resulting charge on sphere Y is +10 μC.

Explanation:

The initial charge on sphere Z is +40 C. When sphere Z is touched to sphere W, the charge is evenly distributed between the two spheres, resulting in each sphere having a charge of +20 C. Then, when sphere Z is touched to sphere X, the total charge is evenly distributed between all three spheres, resulting in each sphere having a charge of +13.33 C. Finally, when sphere Z is touched to sphere Y, the total charge is evenly distributed between all four spheres, resulting in each sphere having a charge of +10 C. Therefore, the resulting charge on sphere Y is +10 μC (option b).

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