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Lesechka [4]
2 years ago
12

Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu

me that Phobos's orbit is circular. Calculate the mass of Mars. Express your answer in units of kg. (Hint: Use the circular orbit velocity formula ; remember to use units of meters, kilograms, and seconds.) Please round the answer to four significant digits. m
Physics
1 answer:
Elenna [48]2 years ago
4 0

Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)²  = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424  x 10²⁵

= 6.4 x 10²³ kg .

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A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is
valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\  \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}

= 4.7476 m/sec

= 4.75 m/s

6 0
2 years ago
What will be the change in velocity of a 850kg car if a force of 50,000 N
yKpoI14uk [10]

Answer:

29.412m/s

Explanation:

F=ma where F= force, m= mass, and a=acceleration

we also know that,

a = Δv / t where Δv = change in velocity and t = time

thus F = m ( Δv / t)

50000=850(\frac{v}{0.5})

\frac{50000}{1700}= Δv

29.412m/s=Δv

8 0
3 years ago
Question 1 (1 point)
Dmitry [639]

1). trajectory

2). person sitting in a chair

3). 490 meters

4). 65 m/s

5). False.  The projectile's displacement, velocity, and acceleration have vertical and horizontal components, but the projectile doesn't.

6). False

7). The vertical component of a projectile doesn't change due to gravity, but the vertical components of its displacement, velocity, and acceleration do.

The vertical components do NOT equal the horizontal components.

8). Decreasing if you include the effects of air resistance.  Constant if you don't.  Gravity has no effect on horizontal velocity.

9). We can't see the simulation.  But if the projectile doesn't have jets on it, then as it travels upward, its vertical velocity must decrease, because gravity is trying to not let it get away.

10). We can't see the simulation.  But if the projectile is traveling downward, we would call that "falling", and its vertical velocity must increase, because gravity is pulling it downward.

6 0
3 years ago
Read 2 more answers
HEre now can somebody help!! <br> Pic is what i need help with
AleksAgata [21]

Answer:

which of the cars are speeding up: c

which of the cars or slowing down: a

which of the cars are maintaning a constant speed: b

Explanation:

8 0
2 years ago
Read 2 more answers
Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
leonid [27]

Answer:

KE_A=33\ J

KE_B=99\ J

Explanation:

Given:

Let mass of the particle B be, m_B=m

then the mass of particle A, m_A=3m

Energy stored in the compressed spring, E=132\ J

Now when the compression of the particles with the spring is released, the spring potential energy must get converted into the kinetic energy of the particles and their momentum must be conserved.

Kinetic energy:

\frac{1}{2}m_A.v_A^2+\frac{1}{2}m_B.v_B^2=132

3m.v_A^2+m.v_B^2=264 .............................(1)

<u>Using the conservation of linear momentum:</u>

m_A.v_A+m_B.v_B=0

3m.v_A+m.v_B=0 .............................(2)

Put the value of v_A from eq. (2) into eq. (1)

3m\times (\frac{-v_B}{3})^2+m.v_B^2=264

v_B^2=\frac{198}{m}  ...........................(3)

<u>Now the kinetic energy of particle B:</u>

KE_B=\frac{1}{2}\times m_B\times v_B^2

KE_B=\frac{1}{2}\times m\times \frac{198}{m}

KE_B=99\ J

Put the value of v_B^2 form eq. (3) into eq. (1):

v_A^2=\frac{22}{m}

<u>Now the kinetic energy of particle A:</u>

<u />KE_A=\frac{1}{2}m_A.v_A^2<u />

<u />KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}<u />

KE_A=33\ J

6 0
3 years ago
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