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3 years ago

Select the correct answer.

Physics

1 answer:

Pachacha [2.7K]3 years ago

8 0

Answer:

Explanation:

It's c the last one u see

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What equation can be used to calculate either wavelength, frequency or speed?

lakkis [162]

Answer:

speed = wavelength * frequency

Explanation:

Thenks and mark me brainliest :)

7 0

3 years ago

. Are there human health risks from exposure (or overexposure) to X-rays? If so:

algol [13]

<h3>yes, there are human health risks caused by exposure to x rays</h3>

describe the main risk (s) ::: - may cause damage to cells in the body

- large doses may lead cells to be cancerous

- causes mutations in the DNA

describe common sources of exposure ( or overexposure ) :::

- natural background radiation e.g. - cosmic rays

- medical radiation e.g. - CT scans & medical x - rays

- man - made radiation e.g. - airport security scanners

describe a technology, device, or behavior that can be used for protection, or to decrease exposure :::

- x - ray beam filters

- digital x - ray detectors

- technologies reduce radiation doses in radiography &

mammography

<em>hope that helps !!</em>

3 0

3 years ago

Read 2 more answers

If something is 50% efficient, how many joules of wasted energy will there be if 750J of energy is put in?

evablogger [386]

Answer:

375J of wasted energy

Explanation:

Efficiency=work output/work input×100%

50=w.o/750×100

750×50=100w.o

w.o=750×50/100

w.o=75×5

w.o=375J

Wasted energy=750J-375J

=375J

Aliter

If the machine is 50% efficient then it is also 50% inefficient

50%×750J=375J

5 0

3 years ago

Solve for v<br> when<br> d = 10 and<br> t=5

VashaNatasha [74]

Answer:

asfsdfasasfsda

Explanation:

7 0

3 years ago

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A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab

Drupady [299]

Answer

According the conservation of energy

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2}I\omega_i^2+0 = mgh + 0$

I for ball = $\dfrac{2}{3}mr^2$

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2} \dfrac{2}{3}mr^2\omega_i^2= mgh$

$\omega_i = \dfrac{v_i}{r}$

$v_i^2+\dfrac{2}{3}r^2(\dfrac{v_i}{r})^2 = 2gh$

$v_i^2+\dfrac{2}{3}v_i^2 = 2gh$

$v_i^2+[1+\dfrac{2}{3}]=2gh$

$v_i^2\dfrac{5}{3}=2gh$

$v_i=\sqrt{\dfrac{6gh}{5}}=\sqrt{\dfrac{6\times 9.8\times 5}{5}}$

$v_i = 7.67\ m/s$

a) $\omega_i = \dfrac{v_i}{R}$

$\omega_i = \dfrac{7.67}{0.113}$

$\omega_i =67.87\ rad/s$

b) $K_{rot} = \dfrac{1}{2}\dfrac{2}{3}mR^2\omega_i^2$

$K_{rot} = \dfrac{1}{3}\times 0.426\times 0.112^2\times 67.87^2$

$K_{rot} = 8.205\ J$

4 0

3 years ago