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2 years ago

A mass of 9.2 kg is accelerated at 2.1 m/s2. What is the force?

Physics

1 answer:

Zielflug [23.3K]2 years ago

6 0

the standard formula for force is f = ma, so multiply the mass and the acceleration and you're good

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A race car makes three and half laps around a circular track in 4.5 min. What is the car’s angular speed?

slavikrds [6]

Answer:

the angular speed is around 45

Explanation:

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2 years ago

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(340-331)/0.6 = temp

9/0.6

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2 years ago

2) How could a motorcycle (a vehicle with less mass) make a van (a

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2 years ago

Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i

Rashid [163]

Answer:

a) 0.72 V

b) 19.2 mA

c) 2.304 Watts

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 = 0.72 V

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

120 $I_{p}$ = 3.2 x 0.72

120 $I_{p}$ = 2.304

$I_{p}$ = 2.304/120 = 0.0192 A

= 19.2 mA

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

==> 0.0192 x 120 = 2.304 Watts

7 0

2 years ago

A car traveling at 27.4 m/s hits a bridge abutment. A passenger in the car, who has a mass of 65.0 kg, moves forward a distance

Minchanka [31]

Answer:

$F=43570.9N$

Explanation:

We can calculate the acceleration experimented by the passenger using the formula $v_f^2=v_i^2+2ad$ , taking the initial direction of movement as the positive direction and considering it comes to a rest:

$a=\frac{v_f^2-v_i^2}{2d}=\frac{-v_i^2}{2d}$

Then we use Newton's 2nd Law to calculate the force the passenger of mass m experimented to have this acceleration:

$F=ma=\frac{-mv_i^2}{2d}$

Which for our values is:

$F=\frac{-(65kg)(27.4m/s)^2}{2(0.56m)}=43570.9N$

6 0

3 years ago