The horizontal component of the displacement is 566.9 m
Explanation:
The horizontal (x-) and vertical (y-) components of a vector on the Cartesian plane can be found as follows:
where
v is the magnitude of the vector
is the angle representing the direction of the vector, measured as above the x-axis.
In this problem, we have:
v = 740 m (magnitude of the vector)
(direction of the vector)
Therefore, the two components are
Learn more about vector components:
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A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. The reflecting object would be 3000 m away .
Phenomenon of hearing back our own sound is called an echo. It is due to successive reflection of sound waves from the surfaces or obstacles of large size. To hear an echo, there must be a time gap of 0.1 second in original sound and the reflected sound.
Given
time = 20 μs = 20 * s
let distance to the reflecting surface be = x
total distance travelled by pulse will be = 2x
speed = 3.0 × m/s
distance = speed * time
2x = 3.0 × * 20 *
x = 3000 m
The reflecting object would be 3000 m away
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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
