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3 years ago

Find the acceleration produced.

Physics

2 answers:

AlladinOne [14]3 years ago

7 0

Answer:

<h2>Given </h2>

force = 0.8 N

mass = 1.2 kg

force

<h2>Formula used </h2>

$\boxed{force = m.a}$

<h2>solution </h2>

According to Newton's second law of motion

$\boxed{ \bold \pink {\underline{f =m.a}}}$

where

f= force

m= mass

a= acceleration

substitute the value in the above formula

0.8 = 1.2 × accleration

accleration= 0.8/1.2

acceleration= 0.66 m/sec²

krok68 [10]3 years ago

4 0

Answer:

0.67m/s²

Explanation:

Given parameters:

Mass of toy = 1.2kg

Force applied = 0.8N

Unknown:

Acceleration = ?

Solution:

According to newton's second law of motion;

Force = mass x acceleration

Now,

Acceleration = $\frac{Force}{mass}$

Acceleration = $\frac{0.8}{1.2}$ = 0.67m/s²

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A circuit has an AC voltage source in series with a 50 ohm resistor and a 113 mH inductor. The frequency is 100 cycles/sec, and

user100 [1]

Answer:

Explanation:

The rms voltage = 140/√2 = 140/1.414 = 99 V.

Reactance of inductor = wL = 2 X 3.14 X 100 X 113 X 10⁻³ =70.96 ohm.

Total resistance in terms of vector = 50+70.96j

j is imaginary unit number

Magnitude of this resistance = √ 50² + 70.96² = 86.80 ohm

current in resistance (rms) ( I ) = 99/86.80 = 1.14 A.

Power dissipated in resistor = I² R = 1.14 X 1.14 X 50 = 65 W( approx)

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4 years ago

a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&

Katen [24]

Answer:

a) T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

sin 30 = $\frac{T_x}{T}$

cos 30 = $\frac{T_y}{T}$

Tₓ = T sin 30

T_y = T cos 30

Y axis

T_y -W = 0

T cos 30 = mg (1)

X axis

Tₓ = m a

they relate it is centripetal

a = v² / r

we substitute

T sin 30 = m $\frac{v^2}{r}$ (2)

a) we substitute in 1

T = $\frac{mg }{cos 30}$

T = $\frac{ 0.2 \ 9.8}{cos \ 30}$

T = 2.26 N

b) from equation 2

v² = $\frac{T \ sin 30 \ r}{m}$

If we know the length of the string

sin 30 = r / L

r = L sin 30

we substitute

v² = $\frac{ T \ sin 30 \ L \ sin 30}{m}$

v² = $\frac{TL \ sin^2 30}{m}$

For the problem let us take L = 1 m

let's calculate

v = $\sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }$

v = 1.68 m / s

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3 years ago

Light of wavelength 550 nm falls on a

Brums [2.3K]

Answer:

The first diffraction maximum fringe will be at approximately 2.7 meters from the central maximum.

Explanation:

We can describe single slit diffraction phenomenon with the equation:

$a\sin\theta=m\lambda$ (1)

with θ the angular position of the minimum of order m respect the central maximum, a the slit width and λ the wavelength of the incident light. Because the distances between the first minima and the central maximum ( $y_{m}$ ) are small compared to the distance between the screen and the slit (x), we can approximate $\sin\theta\approx\tan\theta=\frac{y}{x}$ , using this on (1):

$a\frac{y_{m}}{x}=m\lambda$

solving for y

$y_{m}= \frac{mx\lambda}{a}$

Note that $y_{m}$ is the distance between a minimum and the central maximum but we need the position of a maximum not a minimum, here we can use the fact that a maximum is approximately between two minima, so the first diffraction maximum fringe is between the minima of order 1 and 2, so we should find $y_{1}$ , $y_{2}$ add them and divide by two:

$y_{1}= \frac{(1)(10.0m)(550\times10^{-9}\,m)}{3.00\times10^{-6}\,m}$

$y_{1}= 1.8 m$

$y_{2}= \frac{(2)(10.0m)(550\times10^{-9}\,m)}{3.00\times10^{-6}\,m}$

$y_{1}= 3.6 m$

$maximum = \frac{1.8+3.6}{2}=2.7m$

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3 years ago

Why does a pencil in a glass container filled with water appear to be bent?

Bogdan [553]

The phenomenon is called "refraction". It's the process of
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from water into air.

Your brain doesn't know that the light waves bent toward a
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ALWAYS does ... in the direction the light waves are coming FROM.

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4 years ago

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Answer:

A pie chart.

Explanation:

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