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2 years ago

How far does a car travel in 10 s if its initial velocity is 3 m/s and its acceleration

Physics

1 answer:

victus00 [196]2 years ago

4 0

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A blackbody curve relates _____. intensity of radiation to energy intensity of radiation to wavelength wavelength to energy surf

neonofarm [45]

A blackbody curve represents the relation between <u>intensity of radiation with wavelength.</u>

Here in this curve we can see that all ideal blackbody radiates almost all wavelength of radiations and these radiations are of different intensity.

here intensity will be maximum for a given wavelength of radiation and the relation of this wavelength with the temperature of the object is given by Wein's law

It is given by

$\lambda = \frac{b}{T}$

now if we increase the temperature the maximum intensity for which wavelength is given will shift to the left.

Using this all we can also compare the temperature of two blackbody for which radiation graph is given to us.

5 0

3 years ago

Read 2 more answers

A circuit supplied with 110V carries 5 Amps current. Calculate the Resistance value of the circuit?

IgorC [24]

Answer:

The value is $R = 22 \ \Omega$

Explanation:

From the question we are told that

The voltage is $V = 110 \ V$

The current is $I = 5 \ A$

Generally the resistance value is mathematically represented as

$R = \frac{V}{I}$

=> $R = \frac{110}{5}$

=> $R = 22 \ \Omega$

6 0

2 years ago

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

Two moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2250 J of heat is added to the gas, and 870 J of work i

Lapatulllka [165]

Answer: The final temperature is 470K

Explanation: Using the relation;

Q= ΔU +W

Given, n = 2mol

Initial temperature T1= 345K

Heat =Q= 2250J

Workdone=W=-870J(work is done on gas)

T2 =Final temperature =?

ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

5 0

2 years ago

This diagram shows half of the magnetic field of a bar magnet. Carefully draw in the other

alukav5142 [94]

Answer:

Explanation:

5 0

2 years ago