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Valentin [98]
3 years ago
6

Can anyone please help me on this question (DUE TOMORROW) (13 POINTS GO AWAY)

Physics
1 answer:
julia-pushkina [17]3 years ago
8 0

4.) Quantitative: notice the root word 'quantity', which is a certain percentage of something.

a.) Quantitative, since it represents a certain percentage. The weight before the camp could also be written as an improper fraction.

b.) This is also quantitative because it can be written as a ratio(2.99:500), which can also be written as a fraction(\frac{2.99}{500} or percentage.

c.) This is qualitative. Joan has a higher percentage than Emile.

d.) Qualitative. It judges the quality of something instead of amount. To make it quantitative requires a fraction of percentage. In this case, maybe you could add how he can run y miles every x seconds.

5.)

a.)This depends on whether or not the air is moving. If the classroom's AC is on, or the window is open, then the correct unit to use is CFM (cubic feet per minute). If not, the common unit is milliliters, but must also include temperature and air pressure as well as the volume of the room.

b.)  Simply calculate the volume of the can using inches cubed, since volume is the measure of how much space an object takes up.

c.) The unit would be \frac{kg}{ft^3}. If \frac{mass}{volume} = density, then simply multiply your density by your mass.

d.) I personally would use centimeters. Inches would work too, but centimeters would be much more accurate.

e.) Use the same formula for c.), which means kilograms per cubic foot.

Hope this helped.

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Light is traveling through the different media shown. In which medium does light travel fastest?
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Light travelling in a vacuum is the fastest thing in the universe. The speed would be 2.99x10^8 m/s. The answer to this question is 'vacuum', where light can travel the fastest. I hope this helps you. You're welcome!
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A light-year is the distance light travels in one year (at speed = 2.998 × 108 m/s). (a) how many meters are there in 11.0 light
larisa [96]
<span>The answers are as follows:

(a) how many meters are there in 11.0 light-years?

11.0 light years ( 365 days / 1 year ) ( 24 h / 1 day ) ( 60 min / 1 h ) ( 60 s / 1 min ) ( 2.998x10^8 m/s ) = 1.04x10^17 m

(b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years?

1.04x10^17 m ( 1 au / </span>1.50 × 10^8 km <span>) ( 1 km / 1000 m) = 693329.472 au

(c) what is the speed of light in au/h? au/h

</span>2.998 × 10^8 m/s ( 1 au / 1.50 × 10^8 km ) ( 1 km / 1000 m) ( 3600 s / 1 h ) = 7.1952 au/h

8 0
2 years ago
Why is warm up important?
atroni [7]
Because it stretches and makes your muscles ready for the activity
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3 years ago
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A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the c
Semmy [17]

First establish the summation of the forces acting int the ladder

Forces in the x direction Fx = 0 = force of friction (Ff) – normal force in the wall(n2)

Forces in the y direction Fy =0 = normal force in floor (n1) – (12*9.81) –( 60*9.81)

So n1 = 706.32 N

Since Ff = un1 = 0.28*706.32 = 197,77 N = n2

Torque balance along the bottom of the ladder = 0 = n2(4 m) – (12*9.81*2.5 m) – (60*9.81 *x m)

X = 0.844 m

5/ 3 = h/ 0.844

H = 1.4 m can the 60 kg person climb berfore the ladder will slip

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*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co
sp2606 [1]

Answer: The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

I = \sqrt{\frac{P}{R} }

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }

Then, simplify the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} }

Rationalize the denominator:

I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}

Simplify the numerator by finding its factors:

I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}

The current must be equal to \frac{\sqrt{33} }{6} amps, or ~0.9574 amps.

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