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3 years ago

Can anyone answer this question for me please..

Physics

1 answer:

motikmotik3 years ago

8 0

Answer:

You have to sub x in for the number you dont know and then solve for x.

Explanation:

10(5+x)=90

50+10x=90

10x=40

x=4

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When children use the word “doggie” to refer to every four-legged animal they see, __________ occurs.

Oduvanchick [21]

This is known as overextension! (the correct answer is B.)

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4 years ago

The radar system of a navy cruiser transmits at a wavelength of 1.4 cm, from a circular antenna with a diameter of 2.7 m. At a r

Goshia [24]

Answer:

Distance will be 49.34 m

Explanation:

We have given wavelength $\lambda =1.4cm =0.014m$

Diameter of the antenna d = 2.7 m

Range L = 7.8 km = 7800 m

We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D

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3 years ago

Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len

Alenkinab [10]

Answer:

6.75J

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U=1/2KΔx²

U=0.5* 150*0.30^2

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3 years ago

Calculate the change in the energy of an electron that moves from the n = 3 level to the n = 2 level. What type of light is emit

marissa [1.9K]

Answer:

Red light

Explanation:

The energy emitted during an electron transition in an atom of hydrogen is given by

$E=E_0 (\frac{1}{n_2^2}-\frac{1}{n_1^2})$

where

$E_0 = 13.6 eV$ is the energy of the lowest level

n1 and n2 are the numbers corresponding to the two levels

Here we have

n1 = 3

n2 = 2

So the energy of the emitted photon is

$E=(13.6) (\frac{1}{2^2}-\frac{1}{3^2})=1.9 eV$

Converting into Joules,

$E=(1.9 eV)(1.6\cdot 10^{-19} J/eV)=3.0\cdot 10^{-19} J$

And now we can find the wavelength of the emitted photon by using the equation

$E=\frac{hc}{\lambda}$

where h is the Planck constant and c is the speed of light. Solving for $\lambda$ ,

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.0\cdot 10^{-19}}=6.63\cdot 10^{-7} m = 663 nm$

And this wavelength corresponds to red light.

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3 years ago

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3 years ago

Can anyone answer this question for me please..​

Can anyone answer this question for me please..