Henry sets his keys on a wooden table.

A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m alo

HACTEHA [7]

Answer:

$B=2.91\ \mu T$

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

$B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}$

Put all the values,

$B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T$

So, the required magnetic field is equal to $2.91\ \mu T$ .

3 0

3 years ago

Is it daylight all over the world at the same time then nighttime all over the world at the same time?explain.

Afina-wow [57]

It is neither. on one half of the world at a time it is daytime(because that is the side facing the sun, therefore reciving light from the sun) and on the other side on earth it would be nighttime because that side is facing away from the sun. please review this on google if you are unsure.

5 0

3 years ago

a 0.199 kg snowball moving west makes an inelastic collision with a 2.89 kg box moving 0.523 m/s west. afterward,they move west

kogti [31]

Answer:

The initial velocity of the snowball was 22.21 m/s

Explanation:

Since the collision is inelastic, only momentum is conserved. And since the snowball and the box move together after the collision, they have the same final velocity.

Let $m_1$ be the mass of the ball, and $v_1$ be its initial velocity; let $m_2$ be the mass of the box, and $v_2$ be its velocity; let $v_f$ be the final velocity after the collision, then according to the law of conservation of momentum:

$m_1v_1+m_2v_2=v_f(m_1+m_2)$ .

From this we solve for $v_1$ , the initial velocity of the snowball:

$\boxed{v_1=\frac{v_f(m_1+m_2)-m_2v_2}{m_1}}$

now we plug in the numerical values $m_1=0.199\:kg$ , $m_2=2.89\:kg$ , $v_2=0.523\:m/s$ , and $v_f=1.92\:m/s$ to get:

$v_1=\frac{1.92*(0.199+2.89)-2.89*0.523}{0.199}$

$\boxed{v_1=22.21\:m/s}$

The initial velocity of the snowball is 22.21 m/s.

<em>P.S: we did not take vectors into account because everything is moving in one direction—towards the west.</em>

4 0

3 years ago

honor physics The velocity of a 200 kg object is changed from 5 m/s to 25 m/s in 50 seconds by an applied constant force. a. Wha

Ksivusya [100]

Answer:

(a) 4000 kgm/s.

(b) 80 N

Explanation:

(a) Change in momentum: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is Ns or kgm/s

Mathematically, Change in momentum is expressed as

ΔM = mΔv ..................................... Equation 1

Where ΔM = change in momentum, m = mass of the object, Δv = change in velocity = v₂ - v₁

Given: m = 200 kg, Δv = v₂ - v₁ = 25-5 = 20 m/s.

Substituting into equation 1

ΔM = 200(20)

ΔM = 4000 kgm/s.

Hence the change in momentum = 4000 kgm/s

(b)

Force: This can be defined as the ratio of the change in momentum of a body to the time required for the change.

F = ΔM/t.............................. Equation 2

Where F = force, ΔM = change in momentum, t = time.

Given: ΔM = 4000 kgm/s, t = 50 second.

Substituting into equation 2

F = 4000/50

F = 80 N.

Hence the force = 80 N

7 0

3 years ago

A 5.6 V battery is connected in series with a 41 mH inductor, a 100 Ω resistor, and an open switch. At what time after the switc

Vinil7 [7]

The current in a direct current resistor inductor circuit is given by:

i(t) = (-ℰ/R)e^{-Rt/L} + ℰ/R

Where i(t) is the current, t is time, ℰ is the battery's terminal voltage, R is the resistor's resistance, and L is the inductor's inductance.

Given values:

ℰ = 5.6V

R = 100Ω

L = 4.1×10⁻²H

Plug in the values to get i(t):

i(t) = -0.056e^{-2440t} + 0.056

We want to calculate when the current is 0.012A, i.e. find a time t when i(t) = 0.012A. So let us set i(t) equal to 0.012 and solve for t:

-0.056e^{-2440t} + 0.056 = 0.012

0.056e^{-2440t} = 0.044

e^{-2440t} = 0.786

-2440t = ㏑(0.786)

t = -㏑(0.786)/2440

t = 9.87×10⁻⁵s

3 0

4 years ago