Answer:
a) W = 180.87 J
, b) ΔU = -180.87 J
, c) ΔU = -180.87 J
Explanation:
a) Work is defined as
W = F .ds
Where bold indicates vectors, we can write the scalar product
W = F s cos θ
Where the angle is between force and displacement.
The force of gravity is the weight of the body, which is directed downwards and the displacement thickens the tip of the cliff at the bottom, so that it is directed downwards, therefore the angle is zero degrees
W = y
W = m g y
For this problem we must fix a reference system, from the statement it is established that the system is placed at the base of the cliff, so that final height is zero and the initial height (y₀ = 7.69m)
W = 2.40 9.8 (7.69-0)
W = 180.87 J
b) The potential energy is
U = mg y
The change in potential energy,
ΔU = - U₀
ΔU = mg (- y₀)
ΔU = 2.4 9.8 (0 -7.69)
ΔU = -180.87 J
c) in this case we change the reference system to the height of the cliffs, for this configuration
y₀ = 0
= -7.69 m
ΔU = 2.4 9.8 (-7.69 -0)
ΔU = -180.87 J
<span>(a)
Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get:
18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component.
(b)
To much the same end do we derive the vertical component:
18.0*sin37.5 = v_y = 10.96 ms^-1
Which we then divide by acceleration, a_y, to derive the time till maximal displacement,
10.96/9.8 = 1.12 s
Finally, doubling this value should yield the particle's total time with r_y > 0
<span>2.24 s
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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Answer: 1) It keeps us on the earth so that we can live on the earth and not flying someone else in the atmosphere and space. 2) It maintains the motion of motion of all the planets around the sun and moon around the earth. 3) It pulls all the object towards the earth. 4) The flowing of water in the rivers and seas.
Explanation:
Answer:
The refractive index of fluid 2 is 1.78
Explanation:
Refractive index , n = real depth/apparent depth
For the first fluid, n = 1.37 and apparent depth = 9.00 cm.
The real depth of the container is thus
real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm
To find the refractive index of fluid index of fluid 2, we use the relation
Refractive index , n = real depth/apparent depth.
Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.
So, n = 12.33 cm/6.86 cm = 1.78
So the refractive index of fluid 2 is 1.78