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Vera_Pavlovna [14]
3 years ago
12

A cart changes its speed from 90 m/s to 100 m/s in 10 seconds. during this interval its acceleration is

Physics
2 answers:
Aliun [14]3 years ago
4 0
The acceleration is 1
vekshin13 years ago
4 0

Answer:

1 \frac{m}{s^{2} }

Explanation:

Hello,I think I can help you with this

the acceleration is the change in the speed, in a given time. it can be calculated

acceleration(a) =\frac{speed_{f} -speed_{i} }{time}\\\\

Step 1

let

\\speed_{f}=100\ \frac{m}{s} \\speed_{i}=90\ \frac{m}{s} \\time=10\ seconds\\

put the values into the equation

acceleration(a) =\frac{100\ \frac{m}{s} -90\ \frac{m}{s} }{10\ sec}\\\\acceleration(a)=\frac{10 \frac{m}{s} }{10\ seconds}\\acceleration(a)=1 \frac{m}{s^{2} }

Have a nice day

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Answer:

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The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
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Read 2 more answers
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