Answer:
She does a work of 689.44 J in the snow.
Explanation:
A force is said to do work when it alters the state of motion of a body. The work of the force on that body will be equivalent to the energy needed to move it.
In other words, Work is a form of energy transmission between bodies. In order to carry out work, a force must be exerted on a body and it must move.
The work is equal to the product of the force times the distance and the cosine of the angle that exists between the direction of the force and the direction that the moving point or object travels:
W= F*d* cos Ф
Work W is measured in joules (J), force is measured in newtons (N), and displacement in meters (m).
In this case:
- F= 180 N
- d=5 m
- Ф= 40 degrees
Replacing:
W= 180 N*5 m* cos 40
Solving:
W= 689.44 J
<u><em>She does a work of 689.44 J in the snow.</em></u>
Answer:
The downwind side of an obstacle such as a ridge. The addition of weight on top of a snowpack, usually from precipitation, wind drifting, or a person. An avalanche that releases from a point and spreads downhill collecting more snow - different from a slab avalanche. Also called a point-release or sluff.
Explanation:
Answer:
Net charge contained in the cubeq= 3.536×10^-6C
Explanation:
Formular for total flux in a cube is given as:
Total flux= E300Acos(180) + E200Acos(0)
Where A is crossectional area
Total flux= A(E200-E300)
Total flux= q/Eo
q= Eo×total flux
q=(8.84×10^-12)×(100)^2×(100-60)
q= 3.536×10^-6C
Answer:
Without any external forces a moving object will continue to move in a straight line. The gravitational force between the two objects will provide the centripetal force to keep the objects moving around one another.
1. satellite in orbit around the earth (motion of earth is negligible)
2. moon in orbit around the earth (center of motion several thousand miles
from center of earth)
3. earth in orbit around sun (center of rotation close to center of sun)
4. binary stars (if masses of stars are equal center of rotation is in middle)
Answer:
c) F = 16000 N
Explanation:
For this exercise we use Newton's second law
F = ma
they tell us that adding the other wagons the acceleration of the locomotive must be maintained
F = m a
by adding the other four wagons
mass = 4 no
therefore to maintain the force you must also raise the same factor
Fe = 4Fo
Fe = 4 4000
F = 16000 N
