Pls help will mark brain

How large a current would a very long, straight wire have to carry so that the magnetic field 2.20 cm from the wire is equal to

choli [55]

Biot-Savart Law - Magnetic field around a straight conductor carrying a current I.

B = μ_0 I/2πr

where:
B is strength of magnetic field in T.
μ_0 is permeability of free space = 1.25664 x 10^ -6 T-m/A.
I is current in A.
r is distance from conductor in m.

<span>So I = 2πrB /μ_0</span>

6 0

3 years ago

An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce

ankoles [38]

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

$v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}$

From them we can get $v^2=v_0^2+2ad$ .

We have:

$v-v_0=at\\t=\frac{v-v_0}{a}$

And substitute:

$x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}$

We multiply both sides by 2a, and continue:

$2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2$

Being d the displacement $x-x_0$ , we have $v^2=v_0^2+2ad$

For our exercise, we will write this as:

$d=\frac{v^2-v_0^2}{2a}$

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

$d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m$

For the time we use:

$t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s$

6 0

3 years ago

We measure the loudness of sound in decibels.<br> a. True<br> b. False

Tatiana [17]

Answer: The correct answer is True.

Explanation:

Loudness of sound is referred to how soft or loud a sound is for the listener.

This term is measured in a unit known as decibels referred to as dB.

This unit is used to measure the relative intensity of sounds on a scale from zero to 100 dB.

More the value of decibels, it will be uncomfortable for a person to hear that sound.

So Yes, the loudness of sound is measured in decibels.

7 0

3 years ago

Complete the ray diagram and label incident ray, refracted ray, angle of incidence, and angle of refraction

Novay_Z [31]

Answer:

Solution

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Verified by Toppr

(a) The labelled diagram is shown.

(b) The refractive index of diamond is 2.42. Refractive index of diamond is the ratio of the speed of light in air to the speed of light in diamond.i.e.,

μ=

Speedoflightindiamond

Speedoflightinair

and, the ratio of these velocities is 2.42. i.e., This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air. In other words, the speed of light in diamond is

1/2.42

times the speed of light in vacuum.

Explanation:

a) Draw and label the diagram given :

(i) Incident ray

(ii) Refracted ray

(iii) Emergent ray

(iv) Angle of reflection

(v) Angle of deviation

(v) Angle of emergence

(b) The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?

4 0

3 years ago

What pressing problem did the invention of the cotton gin solve?

AlexFokin [52]

Are you answering a question or asking? You have already seemed to get the answer, A is the correct answer :I

4 0

3 years ago