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gregori [183]
3 years ago
11

a runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her

big push to the finish ?
Physics
2 answers:
Rama09 [41]3 years ago
5 0
Acceleration = velocity/time
A= 3.5m/s/15s
A= 0.23m/s^2
goblinko [34]3 years ago
4 0

Answer:

0.027m/s²

Explanation:

Acceleration (in its simpler form) is defined as the change in velocity divided by the interval of time in which occurred that change. Mathematically:

a=\frac{v_{2}-v_{1} }{t}

Where:

v_{1}: initial velocity

v_{2}: final velocity

t: time needed to that increase in velocity

Plugging the given values in the equation:

a=\frac{3.5m/s-31m/s}{15s}

a=\frac{0.4m/s}{15s}

<em>a= 0.027m/s²</em>

So, her acceleration during her big push was 0.027m/s²

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Ian walks 2km to his best friends house, the. Walks .5km to the library. Then makes a 2.5 km walk home. The entire walk took 3 h
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Read 2 more answers
Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i
Rashid [163]

Answer:

<em>a) 0.72 V</em>

<em>b) 19.2 mA</em>

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = N_{p} = 500 turns

input voltage = V_{p} = 120 V

number of secondary turns = N_{s} = 3 turns

output voltage = V_{s} = ?

using the equation for a transformer

\frac{V_{s} }{V_{p} }  = \frac{N_{s} }{N_{p} }

substituting values, we have

\frac{V_{s} }{120 }  = \frac{3 }{500} }

500V_{p}  = 120*3\\500V_{p} = 360

V_{p} = 360/500 =<em> 0.72 V</em>

<em></em>

b) by law of energy conservation,

I_{P}V_{p} = I_{s}V_{s}

where

I_{p} = input current = ?

I_{s} = output voltage = 3.2 A

V_{s} = output voltage = 0.72 V

V_{p} = input voltage = 120 V

substituting values, we have

120I_{p} = 3.2 x 0.72

120I_{p} = 2.304

I_{p}  = 2.304/120 = 0.0192 A

= <em>19.2 mA</em>

<em></em>

c) power input = I_{p} V_{p}

==> 0.0192 x 120 = <em>2.304 Watts</em>

7 0
3 years ago
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