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3 years ago

15

What would a permanetly magnetic substance be called

2 answers:

evablogger [386]3 years ago

7 0

Could also be called ferromagnetic !

Leya [2.2K]3 years ago

5 0

Answer:

A Permanent magnet or a static magnet

Explanation:

A permanent magnet is an object made from a material that is magnetized and creates its own persistent magnetic field. Materials that can be magnetized, which are also the ones that are strongly attracted to a magnet, are called ferromagnetic (or ferrimagnetic).

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Which unit is used to measure resistance?<br><br> a. volts<br> b. charges<br> c. amperes<br> d. ohms

Ilia_Sergeevich [38]

Choose last

d ohms

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3 years ago

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A truck weighs 500 N on Earth. What's the mass of the truck?

Jobisdone [24]

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A parachutist's fall to Earth is determined by two opposing forces. A gravitational force of 605 N acts on the parachutist. Afte

Doss [256]

Answer:15.8 m/s

Explanation:

Given

Gravitational Force $F_g=605 N$

Air resistance $F_d=689 N$

mass of Parachutist $=\frac{605}{9.8}=61.73 kg$

Velocity after $3 s$

$v=u+at$

$v=0+9.8\times 3$

$v=29.4 m/s$

Now Parachutist opens the parachute

Net for on Parachutist

$605-689=-84$

$84 N$ drag force

therefore deceleration $\frac{84}{61.73}=1.36 m/s^2$

velocity after 10 s is

$v_1=v+at$

$v_1=29.4-1.36\times 10$

$v_1=15.8 m/s$

7 0

4 years ago

Using the superposition method, calculate the current through R5 in Figure 8-71

Natasha2012 [34]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

8 0

3 years ago

Feliz [49]

40 seconds I’m pretty sure sorry if I’m wrong

6 0

3 years ago

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