URGENT!! An astronaut on the International Space Station is doing a spacewalk to fix a solar panel that has malfunctioned. While
2 answers:
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Answer:
Check attachment for solution and diagrams
Explanation:
Given that,
Mass of crate m=63kg
Distance travelled d=40m
Horizontal force Fx=130N
Angle the force applied on cord makes with horizontal is θ=23°.
The weight of the crate is given by
W=mg
W=63×9.81
W=618.03N
Horizontal force Fx=130N
Resolving the applied Force F to the horizontal will give
Fx=FCos θ
F=Fx/Cos θ
F=130/Cos23
F=141.2N
a. Check attachment for model diagram
b. Check attachment for free body diagram
c. Check attachment for pictorial representation
d. Work done by gravitational force.
We, know that the body did not move upward, then the distance d=0
Work done is given as
W=F×d
So, d=0
W=F×0
W=0J
So, no work is done by gravity
e. Normal force?
Using newton law of motion
ΣFy = may
Since the body is not moving upward, then ay=0m/s²
N+141.2Sin23-618.03=0
N=618.03-141.2Sin23
N=562.86N
f. Work done by normal force.
The body is not moving upward, then the distance is zero
d=0
Work done by normal=normal force × distance
Wn=562.86×0
Wn=0J
No work is done by the normal force
g. Frictional force?
Since the coefficient of kinetic friction is zero, then the surface is frictionless
So, no frictional force is acting on the body
Fictional force is given as
Fr=μk•N
Given that, μk=0
Fr=0×562.86
Fr=0N
d. Work done by frictional force?
Since the frictional force Is zero, then, no work is done by friction
W(friction ) = frictional force × d
Here, the body moved a distance of 40m
W(fr)=0×40
W(fr)=0J
No work is done by friction
I. Work done by exerted force
The horizontal component of the exerted force is 130N and the body traveled a distance of 40m
Then, work done is given as
Workdone=force ×distance
Work done=130×40
W=5200J
W=5.2KJ
h. Net workdone?
Since no work is lost by friction, then, the net work done is equal to the work done by the exerted force.
Went, = work done by force exerted - work done by friction
Wnet=5200-0
Wnet, =5200
Wnet=5.2KJ
Answer:
The correct statement is "The electric field is directed toward the electron and has a magnitude of ".
Explanation:
According to Coulomb's law, the magnitude of the electric field due to a static point charge q at a point r distance away from it is given by
- k is the Coulmob's constant.
The direction of the electric field along the line joining the charge and the point where electric field is to be found and it is directed from positive charge to negative charge.
Conventionally, we assume a positive test charge placed at the point where electric field is to be found, the test charge has very small charge such that its charge does not affect the electric field due to the given charge.
The charge on the electron = -e.
The electric field due to an electron is given by
The direction of this electric field is from positive test charge, placed at the point where electric field is to be found, towards the electron along the line joining the two.
Thus, the correct statement is "The electric field is directed toward the electron and has a magnitude of ".