Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.40 kg mass to the

_______is a process changing un magnetized <br> material to magnet

Tcecarenko [31]

Answer:

magnetization

Explanation:

by single stocking , double stocking and electricity

6 0

3 years ago

98 POINTS FOR ANYONE THAT DOES THIS NOW!!

Varvara68 [4.7K]

In simpler terms, a proton and neutron weigh 1 amu (atomic mass unit) each.

The nucleus has 15 protons and 18 neutrons. Since a proton's and a neutron's weight is only 1 amu, we can simply add the number of protons and neutrons to find the total mass of the nucleus:

$15 + 18 = \boxed{33}$

The nucleus' mass is 33 amu.

8 0

3 years ago

Read 2 more answers

A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil

krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

Emf induced in the coil represented by formula

∈ = $-N\frac{d\phi}{dt}$ ...................(1)

Where:

. $\phi = BAcos\theta$ { B is magnetic field }

{A is cross-sectional area}

. $N =$ No. of turns in coil.

. $\frac{d\phi}{dt} =$ Rate change of induced Emf.

Here,

Considering the case :-

$N1 = 2N$ & $\frac{d\phi1}{dt} = 2\frac{d\phi}{dt}$

Putting these value in the equation (1) and finding the new emf induced (∈1)

∈1 = $-N1\times\frac{d\phi1}{dt}$

∈1 = $-2N\times2\frac{d\phi}{dt}$

∈1 = $4 [-N\times\frac{d\phi}{dt}]$

∈1 = 4∈ ...............{from Equation (1)}

Hence,

The Resultant Induced Emf in coil is 4∈.

8 0

4 years ago

A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the

kozerog [31]

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

$\theta=tan^{-1}(\frac{v_y}{v_x})$ (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

$v_x=v_o$

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

$v_y^2=v_{oy}^2+2gy$ (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

$v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}$

Finally, you use the equation (1) to find the angle:

$\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°$

The angle below the horizontal is 12.60°

7 0

3 years ago

A ball is droped from a height of 16m how much time will pass before the ball hits the ground

sergey [27]

Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

$h= \frac{1}{2}g{t}^{2}$

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

Also, let take

$g = 9.8m{s}^{-2}$