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tensa zangetsu [6.8K]
3 years ago
8

Zero, a hypothetical planet, has a mass of 5.8 x 1023 kg, a radius of 2.7 x 106 m, and no atmosphere. a 10 kg space probe is to

be launched vertically from its surface. (a) if the probe is launched with an initial kinetic energy of 5.0 x 107 j, what will be its kinetic energy when it is 4.0 x 106 m from the center of zero? (b) if the probe is to achieve a maximum distance of 8.0 x 106 m from the center of zero, with what initial kinetic energy must it be launched from the surface of zero?
Physics
1 answer:
Tom [10]3 years ago
5 0
To answer these questions just use the equations for potential energy using the mass and heights described. the potential energy at the prescribed heights = the initial kinetic energy required to reach that height.

Make sure you calculate the force of gravity on the surface using the radius of the planet.
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Fill in the blank for four and five answer. Question for number 6.
ehidna [41]
Increase .... decrease .... presumably it's the "best shape" for a body which has been formed by the gravitational force
7 0
2 years ago
You accidentally slide a glass of milk off a table that is 0.9 m tall. How long does it take the milk to hit the ground.
labwork [276]

Fine, lets do a retry of this.

Δd = -0.9m

v₁ = 0

v₂ = ?

a = -9.8 m/s²

Δt = ?

We can use the following kinematic equation and solve for Δt.

                             Δd = v₁Δt + 0.5(a)(Δt)²

                             Δd = 0.5(a)(Δt)²

                           2Δd = a(Δt)²

                     √2Δd/a = Δt

√2(-0.9m)/(-9.8 m/s²) = Δt

0.<u>4</u>28571428574048 = Δt

Therefore, it takes 0.4 seconds for the glass to hit the ground, or 0.43s as you said (even though I don't believe it follows significant digit rules)

7 0
3 years ago
215 = 25x – 35, solve for x.
cluponka [151]
Add 35 to 215. then divide by 25. you should get x=10
6 0
2 years ago
Read 2 more answers
The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou
Cerrena [4.2K]

Answer:

a) I = 19.799\,kg\cdot m^{2}, b) T = -3.405\,N\cdot m, c) n_{T} \approx 54.842\,rev

Explanation:

a) The net torque is:

T = I\cdot \alpha

Let assume a constant angular acceleration, which is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}

\alpha = 1.793\,\frac{rad}{s^{2}}

The moment of inertia of the wheel is:

I = \frac{T}{\alpha}

I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }

I = 19.799\,kg\cdot m^{2}

b) The deceleration of the wheel is due to the friction force. The deceleration is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}

\alpha = - 0.172\,\frac{rad}{s^{2}}

The magnitude of the torque due to friction:

T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )

T = -3.405\,N\cdot m

c) The total angular displacement is:

\theta_{T} = \theta_{1} + \theta_{2}

\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}

\theta_{T} = 344.580\,rad

The total number of revolutions of the wheel is:

n_{T} = \frac{\theta_{T}}{2\pi}

n_{T} = \frac{344.580\,rad}{2\pi}

n_{T} \approx 54.842\,rev

5 0
3 years ago
What is an electromagnet ​
damaskus [11]

Answer:

An electromagnet is a magnet that uses electricity.

Explanation:

6 0
3 years ago
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