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xeze [42]
3 years ago
12

A rocket of mass m is to be launched fromplanet X, which has a mass M and a radius R. What is the minimum speed that the rocket

must have for it to escape into space?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

The minimum speed = \sqrt{\frac{2GM}{R} }

Explanation:

The minimum speed that the rocket must have for it to escape into space is called its escape velocity. If the speed is not attained, the gravitational pull of the planet would pull down the rocket back to its surface. Thus the launch would not be successful.

The minimum speed can be determined by;

                      Escape velocity = \sqrt{\frac{2GM}{R} }

where: G is the universal gravitational constant, M is the mass of the planet X, and R is its radius.

If the appropriate values of the variables are substituted into the expression, the value of the minimum speed required can be determined.

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A Carnot heat engine has an efficiency of 0.400. If it operates between a deep lake with a constant temperature of 298.0k and a
tatuchka [14]

Answer:

496.7 K

Explanation:

The efficiency of a Carnot engine is given by the equation:

\eta = 1 - \frac{T_H}{T_L}

where:

T_H is the temperature of the hot reservoir

T_C is the temperature of the cold reservoir

For the engine in the problem, we know that

\eta = 0.400 is the efficiency

T_C = 298.0 K is the temperature of the cold reservoir

Solving for T_H, we find:

\frac{T_C}{T_H}=1-\eta\\T_H = \frac{T_C}{1-\eta} =\frac{298.0}{1-0.400}=496.7 K

6 0
3 years ago
There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl
kaheart [24]

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28  × 10⁵km

Orbital period  T = 28.9days

T = 28.9  × 24 × 60 × 60

   = 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s

7 0
3 years ago
Read 2 more answers
A sort of "projectile launcher" is shown below. A large current moves in a closed loop composed of fixed rails, a power supply,
Zarrin [17]

Answer:

wallah i don't understand anything with my stoopid brain

Explanation:

3 0
3 years ago
A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th
son4ous [18]

Answer:

Part a)

v = 7.57 km/h

Part b)

\theta = 67.5 degreeNorth of East

Explanation:

Speed of train towards East = 60 km/h

displacement towards East is given as

d_1 = 40 km

now it turns towards 50 degree East of North

so its distance is given as

d_2 = 20 km(sin50 \hat i + cos50\hat j)

d_2 = 15.3 \hat i + 12.8 \hat j

then finally it moves towards west for 50 min

d_3 = -50 \hat i

Now the total displacement of the train is given as

d = d_1 + d_2 + d_3

d = (40 + 15.3 - 50)\hat i + 12.8 \hat j

d = 5.3\hat i + 12.8 \hat j

now total time duration of the motion is given as

T = 40 min + 20 min + 50 min

T = 1.83 h

now average velocity is given as

v_{avg} = \frac{5.3\hat i + 12.8\hat j}{1.83}

v_{avg} = 2.89\hat i + 6.99\hat j

Part a)

magnitude of the average velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{2.89^2 + 6.99^2}

v = 7.57 km/h

Part b)

Direction of the velocity is given as

tan\theta = \frac{v_y}{v_x}

tan\theta = \frac{6.99}{2.89}

\theta = 67.5 degreeNorth of East

6 0
4 years ago
An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis.
Galina-37 [17]
Vx=cos60(4)
x-component of velocity
If you think about it, it makes a right triangle when you combine all the different types of forces together such as v, vx and vy. Then, you can use trigonometry and soh cah toa in order to figure out vx.
8 0
3 years ago
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