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2 years ago

5

If you run 12 km at an average speed of 4 km/h, how much time does it take?

1 answer:

Shtirlitz [24]2 years ago

4 0

Answer:

3 hours, I think

Explanation:

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A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele

Fantom [35]

The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, $\Delta t$ :
$I= \frac{Q}{\Delta t}$
First we need to find the net charge flowing at a certain point of the wire in one second, $\Delta t=1.0 s$ . Using I=0.92 A and re-arranging the previous equation, we find
$Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C$

Now we know that each electron carries a charge of $e=1.6 \cdot 10^{-19} C$ , so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
$N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}$

3 0

2 years ago

A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is

AleksandrR [38]

(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

- its weight, equal to $W=mg$ (m=mass, g=acceleration of gravity), downward

- the buoyant force, equal to $B=m_w g$ ( $m_w$ =mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

$W'=W-B$

which can be rewritten as

$m'g = mg-m_w g$

where m' is the apparent mass of the rock. Using:

m = 540 g

m' = 342 g

we find the mass of water displaced

$m_w = m-m'=540 g-342 g=198 g$

(b) $1.98\cdot 10^{-4} m^3$

If the rock is completely submerged, the volume of the rock corresponds to the volume of water displaced.

The volume of water displaced is given by

$V_w = \frac{m_w}{\rho_w}$

where

$m_w = 198 g = 0.198 kg$ is the mass of the water displaced

$\rho_w = 1000 kg/m^3$ is the density of the water

Substituting,

$V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3$

And so this is also the volume of the rock.

(c) $2727 kg/m^3$

The average density of the rock is given by

$\rho = \frac{m}{V}$

where

m = 540 g = 0.540 kg is the mass of the rock

$V=1.98\cdot 10^{-4} m^3$ is its volume

Substituting into the equation, we find

$\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3$

3 0

3 years ago

How to find moment of inertia of hemisphere

gtnhenbr [62]

<span>It’s 2/5 MR^2 where M is mass and R is the radius of the bas</span>

6 0

3 years ago

When a piano tuner strikes both the A on the piano and a 440 Hz tuning fork, he hears a beat every 2 seconds. The frequency of t

AysviL [449]

Answer:

The frequency is $f = 439.5 \ Hz$

Explanation:

From the question we are told that

The frequency of the tuning fork is $f_t = 440 \ Hz$

The beat period is $T = 2 \ s$

Generally the beat frequency is mathematically represented as

$f_b = \frac{1}{T}$

$f_b = \frac{1}{2}$

$f_b = 0.5 \ Hz$

The beat frequency is also represented mathematically as

$f_b = f_t - f$

Where $f$ is the frequency of the piano

So

$f = 440 - 0.5$

$f = 439.5 \ Hz$

3 0

2 years ago

Answer meeeeeeeeeeeeeee

konstantin123 [22]

Answer:

option A is correct because air friction is greater than gravity

Explanation:

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5 0

2 years ago

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