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3 years ago

A blackbody curve relates the wavelength of emitted light to _____. intensity color frequency speed

Physics

2 answers:

valina [46]3 years ago

7 0

Answer: A

Explanation: I got it right on my test

Brainliest?

Kitty [74]3 years ago

4 0

The best answer would be intensity. A black body curve relates to the wavelength of the emitted light to its intensity. To expands the thought, the curve is referring to the power that is being transferred per unit area.

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When the object is at half its amplitude from equilibrium, is the magnitude of its acceleration at half its maximum value?

Nezavi [6.7K]

We have F = kx or ma = kx where m and k are constants. Therefore, if x is halved, a must be halved too.

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4 years ago

Which of the following is a densit-independent factor

Gnom [1K]

Density-dependent factors operate only when the population density reaches a certain level.

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3 years ago

Which represents the net force

ipn [44]

Answer:

The net force is 500N downwards

Explanation:

When Haley is trying to pull an object upward. The below forces are acting on the object.

Fp = 5500N

Fg = 6000N

because the force of gravity is more than the force of the pull.

Fnet = Fg - Fp = 6000N - 5500N = 500N

And, the direction of the resultant force is the direction of the larger force.

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3 years ago

Use the values from PRACTICE IT to help you work this exercise. Suppose the same two vehicles are both traveling eastward, the c

Mariulka [41]

Answer:

A. $v_{3}=12.17m/s$

B. $v_{car}=6.3m/s\\v_{truck}=-6.3m/s$

C. ΔK $=-4.13x10^3J$

Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

$v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg$

A. Since the two vehicles become entangled the final mass is:

$m_{3}=102kg+103kg=205kg$

From linear momentum we got that:

$p_{1}=p_{2}$

$m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}$

$v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}$

$v_{3}=12.17m/s$

B. The change in velocity of both vehicles are:

For the car

$v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s$

For the truck

$v_{truck}=12.17m/s-18.5m/s=-6.3m/s$

C. The change in kinetic energy is:

ΔK= $K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})$

ΔK= $\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J$

ΔK $=-4.13x10^{3}J$

6 0

3 years ago

The time, t, required to drive a fixed distance varies inversely as the speed, r. It takes 3 hr at a speed of 14 km/h to driv

kenny6666 [7]

Answer:

time taken with speed 23 km/h will be 1.8 hours or 1 hour 48 minutes

Explanation:

Given:

Time is inversely proportional to the speed

mathematically,

t ∝ (1/r)

let the proportionality constant be 'k'

thus,

t = k/r

therefore, for case 1

time = 3 hr

speed = 14 km/hr

3 = k/14

also,

for case 2

let the time be = t

r = 23 km/h

thus,

we have

t = k/23

on dividing equation 2 by 1

we get

$\frac{t}{3}=\frac{k/23}{k/14}$

$t=\frac{14\times3}{23}$

t = 1.8 hr = or 1 hour 48 minutes ( 0.8 hours × 60 minutes/hour = 48 minutes)

4 0

3 years ago