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3 years ago

What is the area under the curve for the histogram below?

Physics

2 answers:

emmasim [6.3K]3 years ago

8 0

Answer:

it's 194

Explanation:

you have to add up every value of people through each hour of the day.

olga2289 [7]3 years ago

5 0

Well i honestly wouldn’t know

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A light bulb can be all of the following except

Dennis_Churaev [7]

Answer: A light bulb can be all of the following except option C (a consumer product if it is used to light the office of the board of directors.)

Explanation:

Products are classified as being BUSINESS or CONSUMER products according to the buyer's intended use of the product.

-Consumer products: these are sold goods that are used for personal, family, or household use. The intention of the buyer is for the products to satisfy his personal needs and desires. Example of some of the consumer products include: toothpaste, eatables and clothes.

Business products: products that are not for personal use but for the manufacturing of other goods are called business products.

Therefore a bulb is not serving as a personal use when used to light the office of the board of directors rather it's serving as a business product .

3 0

3 years ago

Density of water is 1000kg/m3. Find out the amount of mass of 5m3 water

Licemer1 [7]

Answer:

5000 m³

Explanation:

density = mass / volume

mass = density × volume

mass = 1000 × 5

mass = 5000 ( kg )

4 0

3 years ago

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,

Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B $=\ 1.0\ Kg.$

Let mass of ball $A$ and $B\ is\ m$

Final velocity of ball $A\ is\ v_1$

Final velocity of ball $B\ is\ v_2$

initial velocity of ball $A\ is\ u_1$

Initial velocity of ball $B\ is\ u_2$

Momentum after collision $=mv_1+mv_2$

Momentum before collision $= mu_1+mu_2$

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, $mu_1+mu_2=mv_1+mv_2$

Plugging each trial in this equation we get,

First Trial

$mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3$

momentum before collision $\neq$ moment after collision

Second Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1$

moment before collision $=$ moment after collision

Third Trial

$mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1$

momentum before collision $\neq$ moment after collision

Fourth Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0$

momentum before collision $\neq$ moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0

3 years ago

Read 2 more answers

If the beam carries 1015 electrons per second and is accelerated by a 350 kV source, find the current and power in the beam.

Y_Kistochka [10]

To solve this problem we will apply the concept of current defined as the electron charge flow by the number of electrons per second. That is,

I = q*N

Here q is Flow of electric charge in one second and N the number of electron flow per second.

A the same time the power is described as the applied voltage for the current.

P = VI

We know the charge of electron, $q = 1.602 * 10^{-19}$ Coulombs, then the current is

$I = (1.602*10^{-19})(10^{15})$

$I = 0.1602 mA$

And the power in the Beam is

$P = VI$

$P = (350*10^3)(0.1602)$

$P = 0.05607 Watts$

3 0

3 years ago

Which type of wave has particles moving in an elliptical or circular motion?

mote1985 [20]

$\huge \rm༆ Answer ༄$

The Correct choice is ~

$\sf \boxed{ \sf surface \: \: wave}$

_________________________________

$\\ \\ \\ \\ \\ \\$

$꧁ \:\large \cal{Kaul}\: ꧂$

6 0

2 years ago

Read 2 more answers