Ask question

Ask question

All categories

2 years ago

12

Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net

gravitational force on a third mass would be zero.

1 answer:

kotykmax [81]2 years ago

7 0

Answer:

0.29D

Explanation:

Given that

F = G M m / r2

F = GM(6m) / (D-r)2

G Mm/r2 = GM(6m) / (D-r)2

1/r2 = 6 / (D-r)2

r = D / (Ö6 + 1)

r = 0.29 D

See diagram in attached file

You might be interested in

The force against a solid object moving through a gas or a liquid

boyakko [2]

This drag force is always opposite to the object's motion, and unlike friction between solid surfaces, the drag force increases as the object moves faster.

3 0

2 years ago

A 40kg load is raised to a height of 25m. If the operation requires 1 min, find the power required

OlgaM077 [116]

Answer:

163.33 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 40 Kg

Height (h) = 25 m

Time (t) = 1 min

Power (P) =..?

Next, we shall determine the energy. This can be obtained as follow:

Mass (m) = 40 Kg

Height (h) = 25 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 40 × 9.8 × 255

E = 9800 J

Finally, we shall determine the power. This can be obtained as illustrated below:

Time (t) = 1 min = 60 s

Energy (E) = 9800 J

Power (P) =?

P = E/t

P = 9800 / 60

P = 163.33 Watts

Thus, the power required is 163.33 Watts

8 0

2 years ago

Which equation is used to determine the density of a substance?

Mademuasel [1]

The answer is D=M/V hope it helps!!

3 0

3 years ago

Read 2 more answers

A 35 N force makes a 10 degree angle with the positive x-axis. What is the magnitude of the vertical component of the force?

Snowcat [4.5K]

Answer:

6.07 N

Explanation:

Given that,

Force, F = 35 N

It makes 10 degree angle with the positive x-axis.

We need to find the magnitude of the vertical component of the force. It can be given by :

$F_y=F\sin\theta\\\\=35\times \sin(10)\\\\=6.07\ N$

So, the magnitude of the vertical component of the force is 6.07 N.

5 0

2 years ago

Read 2 more answers

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

as

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

5 0

2 years ago