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3 years ago

What is the mass of 3.70 L of water? Remember 1000 L = 1 m^3 (Unit=Kg)

Physics

1 answer:

xxMikexx [17]3 years ago

8 0

Answer:

3.7kg

Explanation:

The following data were obtained from the question:

Volume = 3.7L

Mass =?

Next, we shall convert 3.7L to m³.

This is illustrated below:

1000L = 1m³

Therefore, 3.7L = 3.7/1000 = 0.0037m³

Now, we can obtain the mass of the water as shown below:

Density of water = 1000kg/m³

Volume of water = 0.0037m³

Mass of water =..?

Density = Mass /volume

1000kg/m³ = Mass /0.0037m³

Cross multiply

Mass = 1000Kg/m³ × 0.0037m³

Mass = 3.7Kg

Therefore, the mass of the water is 3.7Kg.

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A power plant produces 1000 MW to suply a city 40Km away.Current flows from the power plant on a single wire of resistance0.050

Westkost [7]

Answer:

The current in wire resistance 2Ω

a). 8696 A

b). fraction power 15.1% a 115kV

Explanation:

Resistance

$R=0.05$ Ω/Km*40km

R=2Ω

P=1000 MW

a).

$P=V*I\\I=\frac{P}{V}=\frac{1000x10^{6}W}{115x10^{3}k } =8696.65A$

Using law ohm

b).

$V=I*R\\I=\frac{V}{R}$

$P=I*I*R\\P=I^{2} *R\\P=8696.65^{2}*2\\P=151.228 x10^{6} W$

$e=\frac{151.228x10^{6} }{1000x10^{6} }*100= 15.12$ %

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3 years ago

33 POINTS How can I get the temperature? SOUND SPEED 340 m/s = 331 m/s + (0,6xTemperature)

inessss [21]

(340-331)/0.6 = temp

9/0.6

90/6

30/2

15 degrees

6 0

3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

3 years ago

A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?

jok3333 [9.3K]

Answer:

$S_a_v_e_r_a_g_e=48km/h$

Explanation:

Ok, the average speed can be calculate with the next equation:

$S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}$ (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

$Total\hspace{3}distance=2*d$

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

$Total\hspace{3}time=t1+t2$

From basic physics we know:

$t=\frac{d}{S1}$

so:

$t1=\frac{d}{S1}$

$t2=\frac{d}{S2}$

Using the previous information in equation (1)

$S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }$

Factoring:

$S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}$ (2)

Finally, replacing the data in (2)

$S_a_v_e_r_a_g_e=\frac{2*40*60}{60+40} =48km/h$

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3 years ago

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 118 km

ruslelena [56]

Answer:

x component 60.85 m

y component 101.031 m

Explanation:

We have given distance r = 118 km

Angle which makes from ground = 58.9°

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(b) Y component of distance is given by $r_Y=rcos\Theta =118\times sin58.9=118\times 0.8562=101.0316m$

These are the x and y component of position vector

6 0

3 years ago