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3 years ago

What is the difference between a permanent magnet and a temporary magnet

2 answers:

7 0

A permanent magnet does not loose its magnetism easily. It can only be lost by permanent destruction say by fire. Example of this is magnetism from a bar magnet. This magnetism is not induced by say electric current.

A temporary magnet looses its magnetism easily. The magnetism is gained or loosed for the action of carrying out a particular task. This magnetism can be achieved by the gain or loss by passing of electric current over say an iron core.

lara31 [8.8K]3 years ago

4 0

Pernament magnet is one that retains its magnetic properties for a long period of time. Temporary megnet only maintains its magnetism while in a magnetic field produced by permanent magnet or electric current.

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Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an up

grin007 [14]

Answer:

The velocity is $v_2= 0.45 \ m/s$

Explanation:

From the question we are told that

The initial speed of the hot water is $v_1 = 0.85 \ m/s$

The pressure from the heater $P_1 = 450 \ KPa = 450 *10^{3} \ Pa$

The height of the hot water before flowing is $h_1 = 0 \ m$

The height of bathtub above the heater is $h_2 = 3.70 \ m$

The pressure in the pipe is $P_2 = 414 KPa = 414 *10^{3} \ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

Apply Bernoulli equation

$P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2$

Substituting values

$(450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )$

=> $v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}$

=> $v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}$

=> $v_2= 0.45 \ m/s$

4 0

3 years ago

Can someone please help me on thisss

Ivahew [28]

Answer 2 i did this before

4 0

3 years ago

Please help<br> Right answers please<br> Will mark brainliest

balandron [24]

Answer:

a. 45 N. / b. 0.08 m/s^2. / c. 102 N

F = ma

F = 15(3)

F = 45 newtons

F/m = a

20/250 = a

0.08 m/s^2 = a

R = ma

R =1.5(68)

102 N

3 0

3 years ago

a train travles at a speed of 30m/s. the train starts at an initial position of 1000 meters and travels for 30 seconds. what is

ycow [4]

Answer:

1900 meters

Explanation:

30m/s x 30 second = 900 meters

+ 1000 meters starting position

= 1900meters

5 0

3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

3 years ago