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ELEN [110]
3 years ago
15

A cruise ship travels west at 25.0 m/s. John is taking his morning jog on the upper deck and jogs from the port side to the star

board side at 5 m/s. what 8s john velocity with respect to the water?​
Physics
1 answer:
gogolik [260]3 years ago
7 0

Answer:

25 m/sec

Explanation:

If the water is assumed to have zero velocity with respect to the boat, John and travels perpendicular  to the boat traveling direction going from port to starboard, his velocity with respect to the water is the same as the ship, 25.m/sec

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The path a star takes during its lifetime depends on
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c.) Mass (is the answer)

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After the box comes to rest at position x₁, a person starts pushing the box, giving it a speed v₁. When the box reaches position
Vesna [10]

Answer:

W_p = \frac{1}{2}mv_1^2

Explanation:

As we know that box is initially at rest

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v_i = 0

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v_f = v_1

now we know by work energy theorem that work done by all the forces on the box will be equal to the change in kinetic energy

So here we will have

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8 0
3 years ago
An image of the moon is focused onto a screen using a converging lens of focal length f= 34.3 cm. The diameter of the moon is 3.
Rashid [163]

Answer:

The diameter of the moon's image is 0.31 cm.

Explanation:

Given that,

Focal length = 34.3 cm

Diameter of the moon d=3.48\times10^{6}\ m

Mean distance from the earth d=3.85\times10^{8}\ m

At that distance the object is assumes to be at infinity. hence the image will be formed at a distance equal to focal length

So, the image distance is 34.3 cm.

We need to calculate the  diameter of the moon's image

Using formula of magnification

m= \dfrac{image\ distance}{object\ distance}=\dfrac{height\ of\ image}{height\ of\ object}

\dfrac{0.343}{3.85\times10^{8}}=\dfrac{h'}{3.48\times10^{6}}

h'=\dfrac{0.343\times3.48\times10^{6}}{3.85\times10^{8}}

h'=0.0031\ m= 0.31\ cm

Hence, The diameter of the moon's image is 0.31 cm.

7 0
3 years ago
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