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Sauron [17]
4 years ago
13

I need help with this problem.

Physics
1 answer:
tamaranim1 [39]4 years ago
7 0

Answer:

Explanation:

Line CD

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Forces are exerted I believe : all of the above
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Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea
melisa1 [442]

Answer:

m = 1.26*10²⁵ kg.

Explanation:

Assuming that the mass of the stone is much smaller than the mass of the planet, we can get the mass, applying the Universal Law of  Gravitation to both masses, as follows:

Fg = G* ms* mp / rp²

Now, if we apply Newton's 2nd Law to the mass of the stone, we can get the gravitational acceleration, as follows:

Fg = ms*a = ms*g ⇒ g = G*mp / rp²

First of all, we need to get the value of g.

Assuming that this acceleration is constant, we can appy the kinematic equations to this situation.

We know that the stone is thrown upward with an initial velocity vo = 15 m/s.

At the highest point in the trajectory, just before of changing direction, the stone comes momentarily to a stop.

At this point, applying the definition of acceleration, we can write:

vf = vo -g*t ⇒ 0 = vo -gt ⇒ g = vo/t (1)

We have the total time since the stone was thrown upwards, not the one used for the upward trajectory.

It can be showed, using the expression for the displacement (which is the same in both directions) that the time used for going up, it's the same used to go down, so the time that we need to put in (1). is just the half of the total time.

So, replacing in (1) we get the value of g, as follows:

g = 15 m/s / 4.5 s = 3.33 m/s²

Now, we can replace this value in the equation that gives us g based in the Universal Law of Gravitation, as follows:

g=G*mp / rp² (2)

Before solving for mp, however, we need to get the value of the radius of the planet.

Assuming that it's a perfect sphere, we can get this value from the value of the circumference at the planet's equator:

rp = 2*π*rp / 2*π ⇒ rp = 1.0*10⁵ km / 2*π = 15,915 km.

With this value for  rp, we can solve (2) for mp, as follows:

mp= g*rp² / G = 3.33 m/s² * (15,915 km)² / 6,67*10⁻¹¹ N.m²/kg²

mp = 1.26*10²⁵ kg.

8 0
3 years ago
A cylinder of mass 250 kg and radius 2.60 m is rotating at 4.00 rad/s on a frictionless.
aleksandrvk [35]

Answer:

The angular momentum of a cylinder, when it is rotating with constant angular velocity is Lini =Iωi

. When two cylinders are added to the rotating cylinder, which are identical in their dimensions, the moment of inertia of the entire system increases (since mass increases). The final moment of inertia will be 3I

Since friction exist, all the cylinders start rotating with same angular velocity, the new angular velocity can be calculated using conservation of angular momentum

Thus, Iωi =3Iωf ⟹ωf =ωi/3 = 0.33ωi

8 0
3 years ago
Activity that uses 150 calories of energy per day, or 1,000 calories per week, describes ___________.
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Moderate physical activity
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4 years ago
A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruis- ing speed
Nataly_w [17]

Answer: P=5573.43\ W

Explanation:

Given

Mass of the elevator is M=650\ kg\\\

Time period of ascension t=3\ s

cruising speed v=1.75\ m/s

Distance moved by elevator during this time

Suppose Elevator starts from rest

\Rightarrow v=u+at\\\Rightarrow 1.75=0+a(3)\\\Rightarrow a=0.583\ s

Distance moved

\Rightarrow h=ut+0.5at^2\\\Rightarrow h=0+0.5\times 0.5833\times (3)^2\\\Rightarrow h=2.62\ m

Gain in Potential Energy is

\Rightarrow E=mgh\\\Rightarrow E=650\times 9.8\times 2.62\\\Rightarrow E=16,720.3\ N

Average power during this period is

\Rightarrow P=\dfrac{E}{t}\\\\\Rightarrow P=\dfrac{16,720.3}{3}\\\\\Rightarrow P=5573.43\ W

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3 years ago
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