5. The table below lists the masses and velocities of several moving things.

What is the relationship between the wavelength of the peak (maximum) of the Gaussian curve and the temperature in degrees Kelvi

Tomtit [17]

Answer:

$\lambda_{max}=\frac{b}{T}$

Explanation:

Wien's displacement law states that the radiation of the black body curve for different temperatures will give peak values at different wavelengths and this wavelength is related inversely to the temperature.

Formally the law of Wien displacement states that the black body's spectral radiation per unit of wavelength, will give peaks at the wavelength of $\lambda_{max}$ which is given by the mathematical expression.

$\lambda_{max}=\frac{b}{T}$

Here, b is proportionality constant with value of $2.897771955...\times 10^{-3} mK$

The wavelength of the peak of the Gaussian curve is inversely related to temperature in degree kelvin.

3 0

2 years ago

A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and

Troyanec [42]

The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

• net vertical force:

∑ F = N - W = 0

• net horizontal force:

∑ F = Fs = m a

where

N = magnitude of normal force

W = car's weight

Fs = mag. of static friction

m = car's mass

a = v ²/R = mag. of the centripetal acceleration

v = car's speed

R = radius of curve

Now,

• compute the car's weight:

W = m g = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

N = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

Fs = 0.5 N = 7350 N

• solve for the (maximum) acceleration:

7350 N = (1500 kg) a → a = 4.9 m/s²

• solve for the (maximum) speed:

4.9 m/s² = v ²/ (35 m) → v ≈ 13 m/s

4 0

3 years ago

A man who works for a moving company is loading a box onto a moving van. He pushes a 200N box up a 5m long ramp. If he pushes wi

OlgaM077 [116]

Answer:

η = 0.667 = 66.7%

Explanation:

The efficiency of the man can be given by the following formula:

η = output/input

where,

η = efficiency of man = ?

output = potential energy gain of the box = Wh

input = work done by man = Fd

Therefore,

$\eta = \frac{Wh}{Fd}$

where,

W = weight of box = 200 N

h = height gained by box = 1 m

F = force exerted by man = 60 N

d = length of ramp = 5 m

Therefore,

$\eta = \frac{(200\ N)(1\ m)}{(60\ N)(5\ m)}$

η = 0.667 = 66.7%

8 0

2 years ago

35. What would a ship's position be if that ship started at

Tcecarenko [31]

Answer:

The final position of the ship after the given time period is 42 km West of B.

Explanation:

Given;

average velocity of the ship, v = 35 km/h

time taken for the ship to reach point D, t = 1.2 hours

The position of the ship after the given time period is calculated as follows;

x = v x t

x = (35 km/h) x 1.2 h = 42 km

x = 42 km West of B.

Therefore, the final position of the ship after the given time period is 42 km West of B.

5 0

2 years ago

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static fricti

vovikov84 [41]

Answer:

$\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

Normal reaction from 40 kg slab on 10 kg block

M × g = 10 × 9.8 = 98 N

Static frictional force = 98 × 0.7 N

Static frictional force = 68.6 N is less than 100 N applied

10 kg block will slide on 40 kg slab and net force on it

= 100 N - kinetic friction

$=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)$

= 100 - 39.2

= 60.8 N

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}$

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}$

$10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}$

$\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }$

Frictional force on 40 kg slab by 10 kg block = 98 × 0.4

Frictional force on 40 kg slab by 10 kg block = 39.2 N

$40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}$

$40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}$

40 kg slab will move with = $0.98 \mathrm{m} / \mathrm{s}^{2}$

$\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}$

3 0

2 years ago