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eimsori [14]
3 years ago
8

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

a gazelle assumes an acceleration of 4.2 m/s2m/s2 for 6.5 ss , after which the gazelle continues at a steady speed. What is the gazelles top speed?
Physics
1 answer:
Novay_Z [31]3 years ago
5 0

Answer:

v = 27.3 m/s

Explanation:

Given that

Acceleration ,a= 4.2 m/s²

Time ,t= 6.5 s

Lets take the maximum speed gain by Thomson's= v

We know that ,if acceleration is constant then the speed v is given as

v= u +  a t

v=final speed

u=initial speed

a=acceleration

t=time

Here the initial speed of Thomson's ,u = 0 m/s

Now by putting the values in the above equation we get

v= 0 + 4.2 x 6.5 m/s

v = 27.3 m/s

Therefore the maximum speed gain by Thomson will be 27.3 m/s.

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Skater 1 has a mass of 105.0 kg and a velocity of 2.0 m/s to the left. He
mart [117]

The final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.

<h3>What is velocity?</h3>

Velocity can be defined as the ratio of the displacement and time of a body.

To calculate the final velocity of Skater 1 we use the formula below.

Formula:

  • mu+MU = mv+MV............ Equation 1

Where:

  • m = mass of the first skater
  • M = mass of the second skater
  • u = initial velocity of the first skater
  • U = initial velocity of the second skater
  • v = final velocity of the first skater
  • V = final velocity of the second skater.

make v the subject of the equation.

  • v = (mu+MU-MV)/m................ Equation 2

Note: Let left direction represent negative and right direction represent positive.

From the question,

Given:

  • m = 105 kg
  • u = -2 m/s
  • M = 71 kg
  • U = 5 m/s
  • V = -3.4 m/s.

Substitute these values into equation 2

  • v = [(105×(-2))+(71×5)-(71×(-3.4))]/105
  • v = (-210+355+241.4)/105
  • v = 386.4/105
  • v = 3.68 m/s
  • v ≈ 3.7 m/s

Hence, the final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.

Learn more about velocity here: brainly.com/question/25749514

7 0
1 year ago
An object in free fall is dropped from a building. Its starting velocity is 0 m/s. Ignoring the effects of air resistance, what
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Answer:

30m/s^2

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Acceleration=Final Velocity-Initial Velocity/Time

10m/s^2= Final Velocity-0m/s/3

30m/s^2= Final Velocity

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Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.
Ksivusya [100]

|v| =\sqrt{ G \cdot M / r}, where

  • M the mass of the planet, and
  • G the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em>  (see below) relates net force the object experiences, \Sigma F to its orbit velocity v and its mass m required for it to stay in orbit :

\Sigma F = m \cdot v^{2} / r <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force \Sigma F acting on the soccer ball shall equal to its weight, W = m \cdot g where g the gravitational acceleration constant. Thus

\Sigma F = W = m \cdot g <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for v; note how the mass of the soccer ball, m, cancels out:

m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0) <em>(equation 3)</em>

<em>Equation 4 </em> gives the value of gravitational acceleration, g, a point of negligible mass experiences at a distance r from a planet of mass M (assuming no other stellar object were present)

g = G \cdot M/ r^{2} <em>(equation 4)</em>

where the universal gravitation <em>constant</em> G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}

Thus

\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}

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