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3 years ago

If an object in space is giving off a frequency of 10^13 wavelength of 10^-6 what will scientist be looking for?

Physics

1 answer:

aev [14]3 years ago

5 0

Answer:

The scientist will be looking for the velocity of the wave in air which is equivalent to 10^7m/s

Explanation:

If an object in space is giving off a frequency of 10^13Hz and wavelength of 10^-6m then the scientist will be looking for the velocity of the object in air.

The relationship between the frequency (f) of a wave, the wavelength (¶) and the velocity of the wave in air(v) is expressed as;

v = f¶

Given f = 10^13Hz and ¶ = 10^-6m,

v = 10¹³ × 10^-6

v = 10^7 m/s

The value of the velocity of the object in space that the scientist will be looking for is 10^7m/s

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Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?

Stolb23 [73]

Answer:

$200 cal/^{\circ}C$

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

$Q=C\Delta T$

where

Q is the heat supplied

C is the heat capacity of the object

$\Delta T$ is the change in temperature

In this problem we have:

$Q=10,000 cal$ is the energy supplied

$\Delta T=75C-25C=50C$ is the change in temperature of the object

Therefore, the heat capacity of the object is:

$C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C$

6 0

3 years ago

A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres

Lemur [1.5K]

Answer:

$V_2 = 0.125 m^3$

Work done = = 5 kJ

Explanation:

Given data:

volume of nitrogen $v_1 = 0.08 m^3$

$P_1 = 150 kPa$

$T_1 = 200 degree celcius = 473 Kelvin$

$P_2 = 80 kPa$

Polytropic exponent n = 1.4

$\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}$

putting all value

$\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}$

$\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS$

polytropic process is given as

$P_1 V_1^n = P_2 V_2^n$

$150\times 0.08^{1.4} = 80 \times V_2^{1.4}$

$V_2 = 0.125 m^3$

work done $= \frac{P_1 V_1 -P_2 V_2}{n-1}$

$= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}$

= 5 kJ

4 0

3 years ago

A man can jog 10 miles in 90 minutes. What’s his speed in mph?

Yuki888 [10]

Answer:

hi there!

the correct answer to this question is: 6.67 mph

Explanation:

you convert minutes to hours

10 miles * 60 mins / 90 mins

7 0

3 years ago

The work-energy theorem states that the work done on an object is equal to a change in which quantity?

Fynjy0 [20]

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

8 0

2 years ago

What is gathering info using your senses?

stiks02 [169]

Observations is the answer.

6 0

3 years ago

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