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morpeh [17]
3 years ago
5

Show with proof, the planet which an astronaut would have a greater weight if Planet A has twice the mass and twice the radius o

f Planet B.
Physics
1 answer:
meriva3 years ago
4 0

Answer:

First let's look at the gravitational formula by newton:

f = g \times  \frac{m1 \times m2}{ {r}^{2} }

Now the f and g should be capital but that's not possible with the equation system. But we can see that the force the astronaut is pulled at is dependent on the mass and the distance if we assume his mass stays the same (mass and weight aren't the same also g is a constant). If a planet has twice the radius the force will by four times as weak because of the ^2. This is not compensated by the twice as big mass. Therefore the astronaut will have a higher force and thus a higher weight on planet B

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An ideal gas is enclosed in a cylinder that has a movable piston on top. The piston has a mass m and an area A and is free to sl
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given mass of piston \left ( m\right )

no. of moles =n

Given Pressure remains same

Temperature changes from T_1 to T_2

Work done\left ( W\right ) is given by=\int_{V_1}^{V_2}PdV

W=P\left ( V_2-V_1\right )

also PV_1=nRT_1

PV_2=nRT_2

W=nR\left ( T_2-T_1\right )

4 0
3 years ago
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
Lubov Fominskaja [6]

Answer:

The moment of inertia is I=\frac{M}{12} a^{2}

Explanation:

The moment of inertia is equal:

I=\int\limits^a_b {r^{2} } \, dm

If r is -\frac{a}{2}

and dm=\frac{M}{a} dr

I=\int\limits^a_b {r^{2}\frac{M}{a}  } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}

I=\frac{M}{a} \int\limits^a_b {r^{2}  } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\  I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}

7 0
3 years ago
A car accelerates from rest at a constant rate of 2m/s2 for 5s. what is the speed of a car at the end of that time?
Fiesta28 [93]
Good afternoon.


We have:

\mathsf{V_0 = 0}\\ \mathsf{a = 2 \ m/s^2}\\ \mathsf{t = 5 \ s}

The function of velocity:

\mathsf{V = V_0+at}\\ \\ \mathsf{V = 0 + 2t}\\ \\ \mathsf{V = 2t}


For t = 5 s:

\mathsf{V = 2\cdot 5}\\ \\ \boxed{\mathsf{V = 10 \ m/s}}
4 0
3 years ago
A toy plane is flying in a horizontal circle by being attached to a 0.75 meter string. The plane has a mass of 101.7 grams and m
nadezda [96]

Answer:

F = 2,894 N

Explanation:

For this exercise let's use Newton's second law

      F = m a

The acceleration is centripetal

     a = v² / r

Angular and linear variables are related.

     v = w r

Let's replace

     F = m w² r

The radius r and the length of the rope is related

    cos is = r / L

    r = L cos tea

Let's replace

    F = m w² L cos θ

Let's reduce the magnitudes to the SI system

     m = 101.7 g (1 kg / 1000g) = 0.1017 kg

      θ = 5 rev (2π rad / rev) = 31,416 rad

     w =  θ / t

     w = 31.416 / 5.1

     w = 6.16 rad / s

     F = 0.1017 6.16² 0.75 cos  θ

     F = 2,894 cos  θ

The maximum value of F is for  θ equal to zero

     F = 2,894 N

7 0
3 years ago
Which statement is true?
Maurinko [17]
The correct answer is<span> B.The speed of sound in air is directly proportional to the temperature of the air.

When the temperature increases so does the speed of sound. Sound is faster by </span>0.60 m/s for every higher degree in air temperature because the air density is reduced and the sound can travel faster.
8 0
3 years ago
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