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Molodets [167]
3 years ago
10

Solar eclipses ___.

Physics
2 answers:
Lelechka [254]3 years ago
4 0

Answer:

Happen when Earth passes between the sun and the moon

Explanation:

Hopes that help

andreev551 [17]3 years ago
3 0

Answer:

Happen when Earth passes between the sun and the moon

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HELP ME ASAP I WILL MARK YOU THE BRAINLIEST NO LINKS
ValentinkaMS [17]

Answer:

rolling friction is the lady in the forest

static friction is the man pushing the fridge

sliding friction is the snow kid

and fluid fraction is the bird I'm pretty sure

4 0
3 years ago
Read 2 more answers
Diagnostic ultrasound of frequency 4.50 MHz is used to examine tumors in soft tissue. (a) What is the wavelength in air of such
mafiozo [28]

(a) 7.62 \times 10^{-5} m is the wavelength in air of such a sound wave.

(b) 3.33 \times 10^{-4}\ m is the wavelength of this wave in tissue.

<u>Explanation:</u>

Frequency and wavelength can be related by the equation,

              Velocity = Wavelength x Frequency

              v=\lambda \times f

where,

v - velocity of light for all EM (electromagnetic) waves in vacuum

Given:

f - 4.50 MHz = 4.50 \times 10^{6} \mathrm{Hz}

a) To find the wavelength in air

We know,

Speed of sound in air = 343 m/s

Apply given frequency and speed of sound in air, we get

        \lambda=\frac{v}{f}=\frac{343}{4.5 \times 10^{6}}=76.2 \times 10^{-6}=7.62 \times 10^{-5}\ \mathrm{m}

b) If the speed of sound in tissue is 1500 m/s, find the wavelength of this wave in tissue

Speed of sound in tissue, v = 1500 m/s

        \lambda=\frac{v}{f}=\frac{1500}{4.5 \times 10^{6}}=333.33 \times 10^{-6}=3.33 \times 10^{-4} \mathrm{m}

4 0
3 years ago
A thin, circular hoop with a radius of 0.22 m is hanging from its rim on a nail. When pulled to the side and released, the hoop
Alex73 [517]

Answer:

Period of oscillation = 1.33 seconds

Explanation:

The period of oscillation is given by:

T = 2π√[I/(MgL)] 

for I = 2MR² and L = R,

Given: L = 0.22m = R

T = 2π√[2R/g] 

T = 2 × 3.142 Sqrt[( 2 × 0.22)/ 9.8]

T = 6.284 Sqrt(0.44/9.8)

T = 6.284 Sqrt(0.0449)

T = 6.284 × 0.2119

T = 1.33 sec

6 0
4 years ago
Betelgeuse is 100,000 times more luminous than our sun, which means that it releases an estimated 3.846 x 1031 W of luminous lig
denis-greek [22]

Answer:

5.4  × 10⁸ W/m²

Explanation:

Given that:

The Power (P) of Betelgeuse is estimated to release 3.846 × 10³¹ W

the mass of the exoplanet = 5.972 × 10²⁴ kg

radius of the earth = 1.27 × 10⁷ m

half the distance (i.e radius r ) = 7.5  × 10¹⁰ m

a) What is the intensity of Betelgeuse at the "earth’s" surface?

The Intensity of  Betelgeuse  can be determined by using the formula:

Intensity \ I = \frac{P}{4 \pi r^2}

I = \frac{3.846*10^{31}}{4 \pi (7.5*10^{10})^2}

I = 544097698.8 W/m²

I = 5.4  × 10⁸ W/m²

8 0
3 years ago
Read 2 more answers
Please help with this
sertanlavr [38]

Answer:I have to say 56

Explanation: because it is going up by 8

6 0
3 years ago
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