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3 years ago

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A space probe is directly between two moons of a planet. If it is twice as far from moon A as it is from moon B, but the net for

ce on the probe is zero, what can be said about the relative masses of the moons? a. Moon A is twice as massive as moon B. b. Moon A has the same mass as moon B. c. Moon A is four times as massive as moon B. d. Moon A is half as massive as moon B.

1 answer:

Dennis_Churaev [7]3 years ago

3 0

Answer:

c. Moon A is four times as massive as moon B

Explanation:

Let's assume the:

mass of the object = $m\,kilogram$

mass of the moon A = $M_A\,kilogram$

mass of the moon B = $M_B\,kilogram$

distance between the center of masses of the object and moon B = $r\,meters$

According to the given condition the object is twice as far from moon A as it is from moon B

∴distance between the center of masses of the object and moon B = $2r\,meters$

<u>As we know, gravitational force of attraction is given by:</u>

$F=G\frac{m_1.m_2}{r^2}$

<em>According to the condition</em>

Force on m due to $M_B=$ Force on m due to $M_A$

$G\frac{m.M_A}{(2r)^2} =G\frac{m.M_B}{(r)^2}$

$\frac{M_A}{4r^2} =\frac{M_B}{r^2}$

$M_A=4M_B$

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A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and

Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

$W=mg$

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

$B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N$

(c) 1524 N

The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

$B=\rho_a V g$

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

4 0

3 years ago

Answer the question with step.

Maru [420]

Answer:

f1/f2 =W1/W2 = 1/3

.0 f2 = 3f1

As ,

1/F= 1/f1 +1/f2

...1/40 = 1/f1 - 1/3f1

f1=> 80/3 cm

... f2 = 2f1 = 3 x 80/3 = 80 cm

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3 years ago

Mechanical energy is conserved in the presence of which of the following types of forces?magnetic

Tju [1.3M]

<h2>Answer: electrostatic and gravitational force </h2><h2 />

Mechanical energy remains constant (conserved) if only <u>conservative forces</u> act on the particles.

In this sense, the following forces are conservative:

-Gravitational

-Elastic

-Electrostatics

While the Friction Force and the Magnetic Force are not conservative.

According to this, mechanical energy is conserved in the presence of electrostatic and gravitational forces.

7 0

3 years ago

Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki

irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

$a=1 m/s^2$

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

$a=-1 m/s^2$ is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

$T=-\frac{u}{a}=-\frac{10}{-1}=10 s$

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

$a=-1 m/s^2$ is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

$s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m$

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3 years ago

An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (

Crazy boy [7]

Answer:

Acceleration, $a=9.36\times 10^{18}\ m/s^2$

Explanation:

It is given that,

Speed of electron, $v=8\times 10^6\ m/s$

Charge on an electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

Magnetic field, $B=5.5i-3.7j$

Magnitude, $|B|=\sqrt{5.5^2+(-3.77)^2}=6.66\ T$

Magnetic force is given by :

$F=qvB$

Also, F = ma

$a=\dfrac{qvB}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 8\times 10^6\times 6.66}{9.1\times 10^{-31}}$

$a=9.36\times 10^{18}\ m/s^2$

So, the acceleration of the electron is $9.36\times 10^{18}\ m/s^2$ . Hence, this is the required solution.

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3 years ago