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3 years ago

T or F: In situations of inhalation poisoning, you should call for help before you attempt a rescue.

Physics

1 answer:

siniylev [52]3 years ago

5 0

Answer:

True

Explanation:

When attempting to rescue an individual who is suffering from inhalation poisoning you could possibly do more harm than good. It is advised to call professionals and experts within that field to prevent further damage.

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Work output of a large machine in a factory is 89,000 joules, and it’s input is 102,000 joules. Work output of a similar machine

mojhsa [17]

(89000/102000)×100
=87.25%

(92000/104000)×100
=88.46%

efficiency is (output/input)×100
if u get confused which way input and output should go, remember the smaller value is always output and it's above in the fraction, then only it's possible to get a efficiency lower than 100.

7 0

3 years ago

Read 2 more answers

Write HCl+MgO---> MgCl 2 + H 2 O as a balanced chemical equation

Effectus [21]

2HCl + MgO ---> MgCl2 + H2O

3 0

2 years ago

Point charge q1 of 30 nC is separated by 50 cm from point charge q2 of -45 nC. As shown in the diagram, point a is located 30 cm

Angelina_Jolie [31]

Answer:

E1 = 2996.667N/C E2 = 11237.5N/C

Explanation:

E1 = kQ1/r^2

=8.99 x 10^9 x 30 x 10^-9/(30x10^-2)^2

= 2996.667N/C

E2 = kQ2/r^2

= 8.99 x 10^9 x 50 x 10^-9/(20x10^-2)^2

= 11237.5N/C

The direction are towards the point a

6 0

3 years ago

Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$