For some reason it's still not correct. The only units that I converted were the km -> m. Is there anything else that I needed to convert? OH and my answer should be in joules.
A spacecraft of mass 3120 kg is in a circular orbit a distance 1650 km above the surface of Mars. How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 4740 km above the surface?
So I have work = change in energy
so
work=...
Q:
Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: a 2.50-kg stone thrown upward from the ground at 15.0 m/s returns to the ground in 7.00 s; the circumference of Mongo at the equator is 1×10^5 km; and...
When in orbit, a communication satellite attracts the Earth with a force of 16.7 kN and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - 1.43*10^11 J. Find the satellite's altitude above the Earth's surface. The radius of the Earth is 6.38*10^6...
Nope still not working. using the formula for F that I gave earlier, and that
changeinlength = F(originallength)/(crosssectionalarea)(modulus)
I get that:
change in length = r(omega^2)m(orig length) / AY where Y is the modulus.
so I use the following values:
R = 14m
omega =...
ok so to find the tensile force, I will use the centripetal force. so F=(mv^2)/R. And v=R(omega). So F=R(omega)^2(m). But to find this force wouldn't I need to know the radius of the rotation?
An amusement park ride consists of airplane-shaped cars attached to steel rods. Each rod has a length of 14.0 m and a cross-sectional area of 0.000780m^2. When operating, the ride has a maximum angular speed of 8.00 rev/min. How much is the rod stretched?
Take the Young's modulus for the rod...
One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle theta with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.370 .
What is the maximum value the angle theta can...
Seesaw: simple statics problem?
Q:
A uniform seesaw is of length 3.90 m and weighs 235 N. The seesaw is pivoted about a point on its bottom surface and the location of this pivot along the length of the seesaw can be adjusted as far as a distance 0.170 m from the center of the seesaw. Little...