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3 years ago

14

A child has a toy car on a horizontal platform. The car starts from rest and reaches a maximum speed in 4 s. If the mass of the

car is
0.1 kg and engine has an effective pull of 0.4 N Find the acceleration of the car.

2 answers:

dmitriy555 [2]3 years ago

7 0

Answer:

a=4m/s²

Explanation:

F=ma

0.4=0.1a

Morgarella [4.7K]3 years ago

3 0

Answer:

a=4m/s

Explanation:

F=ma

0.4=0.1a

$\frac{0.4}{0.1} = \frac{0.1}{0.1}$

a =4m/ s

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An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic

mezya [45]

Answer:

d. 332 V

Explanation:

Given;

number of turns in the wire, N = 40 turns

area of the coil, A = 0.06 m²

magnitude of the magnetic field, B = 0.4 T

frequency of the wave, f = 55 Hz

The maximum emf induced in the coil is given by;

E = NBAω

Where;

ω is angular velocity = 2πf

E = NBA(2πf)

E = 40 x 0.4 x 0.06 x (2 x π x 55)

E = 332 V

Therefore, the maximum induced emf in the coil is 332 V.

The correct option is "D"

d. 332 V

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3 years ago

You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin

faust18 [17]

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance $L=0.450H=0.45\times 10^{-3}H$

Current in the inductor $i=1.90mA=1.90\times 10^{-3}A$

Voltage v = 13 volt

Inductive reactance of the circuit $X_l=\frac{v}{i}$

$X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm$

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3 years ago

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

emmasim [6.3K]

• Net vertical force on the block:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

(<em>n</em> = magnitude of normal force, <em>w</em> = weight)

<em>n</em> = <em>w</em> = <em>m g</em>

(<em>m</em> = mass, <em>g</em> = 9.8 m/s²)

<em>n</em> = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ <em>F</em> = -<em>f</em> = <em>m a</em>

(<em>f</em> = mag. of friction, <em>a</em> = acceleration)

We have <em>f</em> = <em>µ</em> <em>n</em> = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) <em>a</em>

<em>a</em> = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆<em>x</em> in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆<em>x</em>

∆<em>x</em> = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

8 0

3 years ago