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Vilka [71]
2 years ago
5

A cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5 m/s2 for 8.0

seconds?
A. 36 m/s
B. 9 m/s
C. 72 m/s
D. 15 m/s​
Physics
1 answer:
Katarina [22]2 years ago
6 0

Answer:

V = 15m/s

Explanation:

Given the following data;

Initial velocity = 3m/s

Time = 8secs

Acceleration = 1.5m/s²

To find the final velocity, we would use the first equation of motion;

V = U + at

Substituting into the equation, we have

V = 3 + 1.5*8

V = 3 + 12

V = 15m/s

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the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is
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Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

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Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

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

Answer:

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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}

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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}

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You are trying to convert mass to volume. That ain't working
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40/5.79 = 6.9 meters per second approximately.

3 0
2 years ago
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