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3 years ago

15

Which statement accurately describes the ultraviolet catastrophe?

1 answer:

serious [3.7K]3 years ago

8 0

Answer:

D

Explanation:

This is a phenomena by a black body. A black body is reckoned to absorb all energy it receives this is because it gives off energy of different frequency ranges hence as the frequency increases the energy rises.

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A car drives 215 km east and then 45 km north. What is the magnitude of the car’s displacement? Round your answer to the nearest

34kurt

Displacement = (straight-line distance between the start point and end point) .

Since the road east is perpendicular to the road north,
the car drove two legs of a right triangle, and the magnitude
of its final displacement is the hypotenuse of the triangle.

Length of the hypotenuse = √ (215² + 45²)

= √ (46,225 + 2,025)

= √ 48,250

= 219.7 miles .

8 0

3 years ago

Read 2 more answers

1. Which of these is not a natural fibre?<br>a. leather<br>b. jute<br>C.Wool<br>d. cotton

Oksanka [162]

Answer:

leather

Explanation:

plz mark as brainliest......hope it helps

6 0

3 years ago

Read 2 more answers

A proposed ocean thermal-energy conversion (OTEC) system is a heat engine that would operate between warm water (16°C) at the oc

LenaWriter [7]

Answer:

$3.46$ %

Explanation:

We have given cooler temperature that is $T_c$ = 6°C =273+6=279 K

And the warm temperature of water that is $T_h=16^{\circ}C=16+273=289\ K$

The maximum possible efficiency is given by $\eta =1-\frac{T_c}{T_h}$

So maximum efficiency $\eta =1-\frac{T_c}{T_h}=1-\frac{279}{289}=0.0346=3.46$ %

6 0

3 years ago

The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it

Keith_Richards [23]

Answer:

The no. of revolutions does the tub turn while it is in motion is = 52.51 revolutions

Explanation:

Given data

$\omega_1$ = 0

$\omega_2$ = 5 $\frac{rev}{sec}$

Time taken = 7 sec

(1). The angular acceleration is given by

$\alpha = \frac{d \omega}{dt}$

$\alpha = \frac{5}{7}$

$\alpha = 0.714 \frac{rev}{s^{2} }$

We know that from the equation of motion

$\omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_1$

$5^{2} = 0 + 2 (0.714) \theta_1$

$\theta_1 = 17.5 \ rev$ -------- (1)

(2). The angular acceleration is given by

$\alpha = \frac{d \omega}{dt}$

$\alpha = - \frac{5}{14}$

$\alpha =$ - 0.357 $\frac{rev}{s^{2} }$

We know that from the equation of motion

$\omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_2$

$0^{2} = 5^{2} + 2 (-0.357) \theta_2$

$\theta_2$ = 35.01 rev ------- (2)

Total no of revolution made by the machine is

$\theta = \theta_1 + \theta_2$

$\theta =$ 17.5 + 35.01

$\theta =$ 52.51 rev

Therefore the no. of revolutions does the tub turn while it is in motion is = 52.51 rev

8 0

3 years ago

Free fall differs from terminal velocity, in that during free fall

viktelen [127]

Answer:

B

Explanation:

8 0

3 years ago