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4 years ago

1.

Physics

2 answers:

xz_007 [3.2K]4 years ago

5 0

Answer: A. All of the answers are correct.

Anton [14]4 years ago

5 0

Answer:

A. All are correct

Explanation:

When a force is applied on an object, that object is accelerated. Acceleration is a change in the objects speed

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If the fan blades rattle while the fan is turning, you can stop the noise by

daser333 [38]

I beleve the answer is D).

3 0

4 years ago

A box is 300 kg & it covers the area of 150 m2 find the pressure exerted by the box in the ground.

fiasKO [112]

Answer:

If we assume that the box is a cube with 150 m² surface area

Area of 1 face = 150 m² / 6 = 25 m²

The box will exert a downward force of

Force = mass * acceleration due to gravity

Force = 300 kg * 9.81 m/s² N

Force = 2943 N

The box exerts a pressure ( force / unit area) = 2943 N / 25 m²

Pressure = 117.7 N/m²

Thank you, if you found this helpful please comment.

7 0

3 years ago

A short term affect of anorexia might include wight loss

Serggg [28]

Is there a question....

3 0

3 years ago

A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1

Pie

Answer:

(a) E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3

(b) $I=0.0085\ A$

Explanation:

Given:

radius if the coil, $r=0.0395\ m$

no. of turns in the coil, $n=520$

variation of the magnetic field in the coil, $B=(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4$

resistor connected to the coil, $R=560\ \Omega$

(a)

we know, according to Faraday's Law:

$emf=n.\frac{d\phi}{dt}$

where:

$d \phi=$ change in associated magnetic flux

$\phi= B.A$

where:

A= area enclosed by the coil

Here

$A=\pi.r^2$

$A=\pi\times 0.0395^2$

$A=0.0049\ m^2$

$\therefore \phi=((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049$

So, emf:

$emf= 520\times \frac{d}{dt} [((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049]$

$emf= 520\times 0.0049\times \frac{d}{dt} [(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)]$

$emf= 2.548\times [0.012+(13.8\times 10^{-5})t^3)]$

$emf= 0.0306+3.516\times 10^{-4}\ t^3$

(b)

Given:

$t_0=5.25\ s$

Now, emf at given time:

$emf=4.7755\times 10^{-2}\ V$

∴Current

$I=\frac{emf}{R}$

$I=\frac{4.7755\times 10^{-2}}{560}$

$I=8.5\times 10^{-5} A$

6 0

4 years ago

a voltmeter and an ammeter are used to respectively monitor the voltage across the current through the resistor

dusya [7]

When a current is established in a closed conducting loop , we use ammeter. Due to this resultant potential developed detected by voltmeter.

When current flows through loop , it gets conducted.
Due to this Electric field is generated . This happens due to induced magnetic field.
The free electrons were at rest until current is there , still ammeter detects some deflection.
This deflection is due to zero current.
Due to magnetic field , current in coil changes.
Mathematically, $\alpha =-\frac{d\beta }{dt}$
The presence of conducting loop is not necessary for having electric field.

$e= -\frac{d\beta }{dt}$

$e=-L\frac{d\beta }{dt}$

$i= i0 \left \{ {{y=0.63} \atop {x=t}} \right.$

i=0.63 io

To know more about voltmeter-

<u>brainly.com/question/1511135</u>

#SPJ4

8 0

1 year ago