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Rus_ich [418]
3 years ago
14

A planet is orbiting its star at a maximum radius of 1.2 x 10^7 meters. If the mass of the star is 2 x 10^30 kg, what is the tim

e period of the orbit?
22.6 seconds


A. 80 seconds


B. 22.6 days



C.80 days


D.22.6 minutes
Physics
1 answer:
kirill115 [55]3 years ago
5 0







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Find the momentum (in kg.m/s) of a helium nucleus having a mass of 6.68 x 10^-27 kg that is moving at 0.386c.
igomit [66]

Answer:

7.73544\times 10^{-21}kgm/sec

Explanation:

We have given mass of helium nucleus m=6.68\times 10^{-27}kg

Velocity of helium nucleus v=0.386c=0.386\times 3\times 10^{8}m/sec=1.158\times 10^{8}m/sec

Momentum of the helium nucleus is given by P=mv where m is mass and v is velocity

So P=mv=6.68\times 10^{-27}\times 1.158\times 10^{8}=7.73544\times 10^{-21}kgm/sec

6 0
3 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
Getyour pointsssssssssssssssssssssss
Maurinko [17]

Answer:

thankkkk youuuu soo muchhh

3 0
3 years ago
Read 2 more answers
There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.
Sloan [31]

Via half-life equation we have:


A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }


Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes.  By plugging those values in we get:

A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g


There is 6.25 grams left of Ra-229 after 12 minutes.

4 0
4 years ago
Please help I’ll give brainliest
BigorU [14]

Answer:

The answer are as followed: B is the correct answer, hope that helps :)

8 0
3 years ago
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