Ask question

Ask question

All categories

3 years ago

7

a 4m long straight wir that carries acurrent of 0.5A is placed perpendicular to a uniform magnetic field. if the size of magneti

c deflection force is 0.4N, then what is the strength of magbetic field?

1 answer:

PolarNik [594]3 years ago

5 0

Answer:

B=0.2T

Explanation:

given required solution

l=4m B=? <em>F</em><em>=</em><em>BIL</em>

i=0.5A B=F/IL

F=0.4N B=0.4N/0.5A*4m

B=0.4/2=0.2T

You might be interested in

Lithium is more active than aluminium<br> A.True<br> B.false

WITCHER [35]

A:True because Lithium is the lighest solid and metal and the third lightest element.

5 0

3 years ago

Read 2 more answers

A device that uses electricity and magnetism to create motion is called a (motor magnet generator) . In a reverse process, a dev

Tems11 [23]

the answers are a. and c.

I hope I help you out!

6 0

3 years ago

Read 2 more answers

A tennis ball of mass 0.060 kg travels horizontally at a speed of 25m/s. The ball hits a tennis

stepladder [879]

It would be A because a is perfect

7 0

3 years ago

Read 2 more answers

What is the acceleration of a 10 kg mass pushed by a 5 N force?

yanalaym [24]

$g-\ gravitational \ acceleration \\ g= \frac{G}{m} = \frac{5N}{10kg}= \\ g=0,5 \frac{N}{kg}$

8 0

3 years ago

Read 2 more answers

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0

3 years ago