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Illusion [34]
3 years ago
13

Note: Take East as the positive direction. A(n) 81 kg fisherman jumps from a dock into a 128 kg rowboat at rest on the West side

of the dock. If the velocity of the fisherman is 4.5 m/s to the West as he leaves the dock, what is the final velocity of the fisherman and the boat? Answer in units of m/s.
Physics
1 answer:
Blizzard [7]3 years ago
6 0

Answer:

final velocity of fisherman will be 1.744 m /sec in west direction

Explanation:

We have given mass of the fisherman m_1=81kg

Velocity of fisherman v_=-4.5m/sec ( As east direction is positive direction )

Mass of the rowboat m_2=128kg

As the rowboat is at rest so v_2=0m/sec

Now according to conservation of momentum

m_1v_1+m_2v_2=(m_1+m_2)v

81\times -4.5+128\times 0=(81+128)\times v

v=-1.744m/sec

So final velocity of fisherman will be 1.744 m /sec in west direction

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Define force and provide an example​
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Answer:

force-strength,power or energy as an attribute of motion, movement or action. Example: Frictional force.

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A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hi
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Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g

U is the speed = 300m/s

H is the maximum height = 78.4m

g is the acceleration due to gravity = 9.8m/s²

Substitute into the fromula;

R = 300√2(78.4)/9.8

R = 300 √(16)

R = 300*4

R = 1200m

Hence the projectile travelled 1200m before hitting the ground

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2 years ago
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I believe the answer is a

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3 years ago
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What is the resultant of the vectors shown?
ad-work [718]

Answer:

Option B

Explanation:

Option A is the wrong answer because the horizontal vector is in the opposite direction.

Option C is the wrong answer as the horizontal vector is in the opposite direction and all the vectors are connected head to tail [of the arrows] [Triangle law of vector addition]

Option D is the wrong answer as the horizontal vector is in the opposite direction.

5 0
2 years ago
(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i
PilotLPTM [1.2K]

Answer:

Part a)

a_c = 1.67 \times 10^{-10} m/s^2

Part b)

v = 2.18 \times 10^5 m/s

Explanation:

Time period of sun is given as

T = 2.60 \times 10^8 years

T = 2.60 \times 10^8 (365 \times 24 \times 3600) s

T = 8.2 \times 10^{15} s

Now the radius of the orbit of sun is given as

R = 3.00 \times 10^4 Ly

R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m

R = 2.84 \times 10^20 m

Part a)

centripetal acceleration is given as

a_c = \omega^2 R

a_c = \frac{4\pi^2}{T^2} R

a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})

a_c = 1.67 \times 10^{-10} m/s^2

Part b)

orbital speed is given as

v = \frac{2\pi R}{T}

v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}

v = 2.18 \times 10^5 m/s

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3 years ago
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