A cat lifts up her kitten by its neck.
A bird carries a worm to its nest.
A horse pulls a wagon along a road.
Explanation:
Work is defined as the product between the force applied to an object and the distance the object has been moved:
therefore, work is non-zero only if the object has been moved by a distance different from zero. This is exactly what happens in the three following scenarios:
A cat lifts up her kitten by its neck.
A bird carries a worm to its nest.
A horse pulls a wagon along a road.
In all these cases, an object (the kitten, the nest and the wagon) are moved through a certain distance, so work is done. On the contrary, in the other two examples:
A dog rubs his back against a large tree
A goat butts its head against the barn wall.
No object is moved, so d=0 in the formula and no work is done.
The given question is incomplete. The complete question is as follows.
A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.
A) If the net force acting on the particle is N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.
Enter your answers numerically separated by commas.
Explanation:
The given data is as follows.
Q =
E = 1300 N/C in the +x direction
B = 1.02 T in the +z direction
and, in the +x direction
Also,
= qE - qvB
Now, we will calculate the value of v as follows.
v =
=
v = 458.507 m/s
Using the value for velocity, we need to know which direction it's going.
You know +x direction for E, +z direction for B and +x for .
Using the right hand rule where:
your right thumb goes toward the , then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.
Thus, we can conclude that = 0, -(458.507), 0.
Answer:
The net force = 0
Explanation:
The given information includes;
The mass of the crate = 250 kg
The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)
In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.
The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate
The weight of the crate, ↓ = 250 kg × 9.81 m/s² = 2,452.5 N
The force the helicopter should provide to just lift the crate, ↑ = The weight of the crate = 2,452.5 N
The net force, =
↑ -
↓ = 2,452.5 N - 2,452.5 N = 0
The net force = 0.