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3 years ago

PLEASE HELP!

Physics

1 answer:

Kobotan [32]3 years ago

3 0

Answer:

Explanation:

Cause a monster truck don

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Calculate the current through the ammeter (look at photo)

AleksAgata [21]

Answer:

6 amps

Explanation:by Kirchhoff's loop rule the current at any point in the loop must be equal or charge would be building up. The current at the ammeter is equally to the total current through the sun of the paths in parallel which it is in series with

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2 years ago

If 4000 C of charge pass through a section of wire in 50 seconds, what would the current be?

9966 [12]

Answer:

Explanation:

Given that,

Charge through a wire is

Q = 4000C

Time for charge to pass through the wire is

t = 50seconds

Then,

Current through the wire is the rate of charge passing through the wire

Q = it

Where

Q is charge.

I is current

t is time taken

Therefore,

I = Q / t

I = 4000 / 50

I = 80 Amps

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4 years ago

E4.Suppose Galileo’s pulse rate was 75 beats per minute. How many beats per second is this? What is the time in seconds between

Umnica [9.8K]

Answer

given,

Galileo's pulse rate = 75 beats per minutes

Beats per second

1 minutes = 60 s

pulse rate = $\dfrac{75}{60}$

= 1.25 beats/s

times in second

$t = \dfrac{1}{1.25}$

t = 0.8 s/pulse

distance of the object

initial velocity = 0 m/s

$s = ut + \dfrac{1}{2}gt^2$

$s = \dfrac{1}{2}\times 9.8\times 0.8^2$

s = 3.14 m

distance in feet

1 m = 3.28 ft

3.14 m = 3.14 x 3.28 ft

= 10.3 ft

distance in feet is equal to 10.3 ft.

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4 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

3 years ago

Draw a stationary wave and show the position of node and antinode

Aneli [31]

I hope that the attachment helps you..

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3 years ago