In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.
Answer:
Explanation:
Given that,
The mass of a Hubble Space Telescope, 
It orbits the Earth at an altitude of 
We need to find the potential energy the telescope at this location. The formula for potential energy is given by :
Where
is the mass of Earth
Put all the values,
So, the potential energy of the telescope is 
.
Answer:
u = 10.63 m/s
h = 1.10 m
Explanation:
For Take-off speed ..
by using the standard range equation we have
R = 9.1 m
θ = 26º,
Initial velocity = u
solving for u
u = 10.63 m/s
for Max height
using the standard h(max) equation ..
h = 1.10 m
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
For the answer to this question,
Thalia must consider the weight of the object and the mass of the sculpture. Weight and mass are different things. She should also consider the time on how long it will take to move it and where she'll move it.
