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stiks02 [169]
3 years ago
6

Can anyone pls help me out in dis i am struggling in dis!

Physics
1 answer:
goldfiish [28.3K]3 years ago
5 0

Answer:

B

Explanation:

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MATCH THESE ^-^ Match Newton's law with the correct statement.
ikadub [295]
1 and A
2 and B
3 and D
4 and C

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What likely happens to the hart rate as the cardiac mucle gets stronger
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The heart rate will likely decrease. As the cardiac muscle, or heart, gets stronger, it takes less effort to pump more blood. As a result, the heart will probably beat less, decreasing the heart rate. This is why athletes often have lower heart rates than the average person.
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In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l
Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

u=+6.3 m/s is the initial velocity (positive because it is upward)

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

v=+6.3 + (-9.8)(1.7)=-10.4 m/s

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

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3 years ago
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A yellow ball with a mass of 2 kg is rolling across the floor at 3 m's A red ball with a mass of 3 kg is rolling across the
Wewaii [24]
Answer:
The red ball has a greater kinetic energy, because it has a greater mass

Explanation:
Mass is directly proportional to kinetic energy, as one increases, so does the other
6 0
3 years ago
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A bucket of water of mass 10 kg is pulled at constant velocity up to a platform 30 meters above the ground. This takes 10 minute
Rudik [331]

Answer:

W = 2352 J

Explanation:

Given that:

  • mass of the bucket, M = 10 kg
  • velocity of pulling the bucket, v = 3m.min^{-1}
  • height of the platform, h = 30 m
  • time taken, t = 10 min
  • rate of loss of water-mass, m = 0.4 kg.min^{-1}

Here, according to the given situation the bucket moves at the rate,

v=3 m.min^{-1}

The mass varies with the time as,

M=(10-0.4t) kg

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance

∆x =  3∆t meters

So, during this interval change in work done,

∆W = m.g∆x

<u>For work calculation:</u>

W=\int_{0}^{10} [(10-0.4t).g\times 3] dt

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W= 2352 J

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