Different types of Questions not many though

[hackeract]

an elevator of mass M is suspended from a vertical cable. When the elevator is accelerating downward with an acceleration magnitude of 5.8 m/s^2, the tension in the cable is 3644N, what is the mass of the elevator?

I had a few ideas..
T - mg = ma
T = m(g + a)
T/(g+ a) = m

Or i was thinking 3644N = Kg/M/s^2, you could just divide by 5.8 and the M/s^2 cancel out, leaving M which is = 628kg

..

what is the moons orbital speed?

Mass Earth 6 x 10^24 kg
Mass Moon 7.36 x 10^22 kg

Radius Earth 6400km
Radius Moon 1740km

Mooms orbital radius 3.84 x 10^8

for this i was thinking a = v²/r, and plug that into Newton's second law.. but i don't think that's right

A 24kg mass sits on a frictionless plane. Calculate the force required to give an acceleration of 5 m/s^2 up the plane
(mass = 24kg, Angle=37)


and

a speeding motorist passes a stopped police car. At the moment he passes the police car begins accelerating at a constant rate of 4.4 m/s². the motorist unaware that he is being chased, continues at constant speed until the police car catches him 12s later. how fast was the motorist going?

 

[jdstokes]

an elevator of mass M is suspended from a vertical cable. When the elevator is accelerating downward with an acceleration magnitude of 5.8 m/s^2, the tension in the cable is 3644N, what is the mass of the elevator?

I had a few ideas..
T - mg = ma
T = m(g + a)
T/(g+ a) = m

Or i was thinking 3644N = Kg/M/s^2, you could just divide by 5.8 and the M/s^2 cancel out, leaving M which is = 628kg

..

what is the moons orbital speed?

Mass Earth 6 x 10^24 kg
Mass Moon 7.36 x 10^22 kg

Radius Earth 6400km
Radius Moon 1740km

Mooms orbital radius 3.84 x 10^8

for this i was thinking a = v²/r, and plug that into Newton's second law.. but i don't think that's right

A 24kg mass sits on a frictionless plane. Calculate the force required to give an acceleration of 5 m/s^2 up the plane
(mass = 24kg, Angle=37)


and

a speeding motorist passes a stopped police car. At the moment he passes the police car begins accelerating at a constant rate of 4.4 m/s². the motorist unaware that he is being chased, continues at constant speed until the police car catches him 12s later. how fast was the motorist going?

You're close in the first question, except the acceleration is down, meaning that the net force is negative. $T - mg = - ma$ so $m = \frac{T}{g-a}$. The simplest way to find the orbital speed is to equate the gravitational force $\frac{GMm}{r^2}$ with the centripetal force, so you get $v=\sqrt{\frac{GM}{r}}$. Identify all the forces acting on the block (there are 3) and decompose them into perpendicular components. You'll find it easier if you work in a perpendicular coordinate system where the $x$-axis is inclined 37 degrees from the horizontal, since this coordinate system has one of its axes oriented in the direction of the block's motion. For last question, write down the positions of the objects as functions of time, using the police car as the origin. Equate the the positions and solve for time.

 

[Tide]

The fact that the Moon's mass was provided suggests you might want to consider that the center of mass of the Earth-Moon system may not coincide with the center of the Earth!

 

[hackeract]

You're close in the first question, except the acceleration is down, meaning that the net force is negative. $T - mg = - ma$ so $m = \frac{T}{g-a}$. The simplest way to find the orbital speed is to equate the gravitational force $\frac{GMm}{r^2}$ with the centripetal force, so you get $v=\sqrt{\frac{GM}{r}}$. Identify all the forces acting on the block (there are 3) and decompose them into perpendicular components. You'll find it easier if you work in a perpendicular coordinate system where the $x$-axis is inclined 37 degrees from the horizontal, since this coordinate system has one of its axes oriented in the direction of the block's motion. For last question, write down the positions of the objects as functions of time, using the police car as the origin. Equate the the positions and solve for time.

Im still confused on the orbital question..

would it look something like

v= √(6.67 x 10^-11)(7.36 x 10^22)
_________________________
1740km

im not sure what exactly what information to plug in... and it sucks for the last 2 questions we havn't covered how to find them like that... this is only first tri however..