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3 years ago

The planet whose distance from the sun equals 1 a.u. is: venus earth mars jupiter

Physics

2 answers:

son4ous [18]3 years ago

7 0

The answer is Earth. hope this helped

Katarina [22]3 years ago

7 0

earth is the correct answer

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An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago

Which of the following causes an object to move in a curved path?

Brrunno [24]

Answer:

A. centripetaal force

3 0

3 years ago

Read 2 more answers

An oil droplet is sprayed into a uniform electric field of adjustable magnitude. The 0.11 g droplet hovers

ohaa [14]

Answer:

The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.

F = N e E where E is the value of the field and N e the charge Q

M g = N e E and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check: N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13 electrons

7 0

2 years ago

The sound produced by the loudspeaker in the drawing has a frequency of 11999 Hz and arrives at the microphone via two different

const2013 [10]

The speed at which sound travels through the gas in the tube is 719.94m/s

<u>Explanation:</u>

Given:

Frequency, f = 11999Hz

Wavelength, λ = 0.03m

Velocity, v = ?

Sound speed in the tube is calculated by multiplying the frequency v by the wavelength λ.

As the sound loudness changed from a maximum to a minimum, then we know the sound interference in the case changed from constructive interference (the two sound waves are in phase, i.e. peaks are in a line with peaks and so the troughs), to a destructive interference (peaks coinciding with troughs). The least distance change required to cause such a change is a half wavelength distance, so:

λ/2 = 0.03/2

λ = 0.06m

We know,

v = λf

v = 0.06 X 11999Hz

v = 719.94m/s

Therefore, the speed at which sound travels through the gas in the tube is 719.94m/s

3 0

3 years ago

A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement

mamaluj [8]

Answer:

sup qwertyasdfghjk

Explanation:

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3 years ago