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enyata [817]
3 years ago
10

A runner begins a race from the starting line and accelerates to a speed of 8.9 m/s. If it takes the runner 3 seconds to reach h

is maximum speed, what was his acceleration?
Physics
2 answers:
stellarik [79]3 years ago
4 0

Answer:

i believe its 26.7

Explanation:

if the runner goes 8.9 m/s each second while accelerating for 3 seconds to reach top speed, the top speed would be 26.7 m/s

ale4655 [162]3 years ago
3 0

Answer:

it will be 8.9 over 3 m/s/s or m/s square

Explanation:

acceleration equal to speed over time

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4 years ago
The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a
Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

with m the mass , a the acceleration and \sum\overrightarrow{F} the net force forces that is:

(F-f) (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

(F-f)=ma

solving for a:

a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}

Now let's use the Galileo’s kinematic equation

Vf^{2}=Vo^{2}+2a\varDelta x (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and \varDelta x the time took to achieve that velocity, solving (3) for \varDelta x:

\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}

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8 0
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Water enters a cylindrical tank through two pipes at rates of 250 and 100 gal/min. If the level of the water in the tank remains
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Answer:

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Then, if the level of the water remains constant, this means that the water leaves the tank at a rate of 350gal/min.

We know that the diameter of the pipe is 8 inches, then the area of the pipe is:

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now, the flow can be calculated as:

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if we want to write our velocity in inches per minute, then we need to write the entering flow in cubic inches:

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Then the water that leaves the tank must be the same amoun, we have:

Q = 80,850 in^3/min. = v*A = v*50.24in^2

v =  (80,850in^3/min)/50.24in^2 = 1609.3 in/min.

The velocity of the flow leaving the tank is 1609.3 in/min.

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