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Citrus2011 [14]
2 years ago
5

What is the area of the bottom of a tank 30.0 cm long and 15.0 cm wide?​

Physics
2 answers:
REY [17]2 years ago
6 0

Answer:

450cm^2

Explanation:

Length (l) = 30cm

Width (w)= 15cm

Area = l × w = 30cm × 15cm = 450cm^2

Igoryamba2 years ago
5 0

Answer:

perimeter of the bottom of the tank is 450 cm

Explanation:

If you want to get area you need the Length, width,and height.

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When a carousel is in motion, the movement of the carousel horse can BEST be described as
lara [203]
Kinetic energy because the horse is in motion
7 0
3 years ago
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C

4 0
3 years ago
DESCRIBE THE REQUIREMENTS OF AN INTERNET CONNECTION?<br>please tell me the answer​
AlekseyPX

Answer: The basic requirements for connecting to the Internet are a computer device, a working Internet line, and the right modem for that Internet line. In addition, software programs such as Internet browsers, email clients, Usenet clients, and other special applications are needed in order to access the Internet.

Explanation: brainleist pls :)

4 0
3 years ago
A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net
Schach [20]

Explanation:

Let us first calculate  long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

negative sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0
2 years ago
A reconnaissance plane flies 605 km away from
kolezko [41]

Answer:

                      v_{avg}  = 355 m/s  

Explanation:

Distance = 605 km

Initial speed = v_{i} = 284 m/s

Final velocity = v_{f} = 426 m/s

Average speed = ?

There is two method two find average speed. In first method, using 3rd equation of motion, we find acceleration.

                        2as = v_{f}^{2}+v_{i}^{2}

Then using first equation of motion, we find time

                        v_{f} = v_{i}+at

Then using the formula of average velocity, we find average velocity

                         v_{avg}=\frac{total-distance}{total-time}

Second method is very simple

                                  v_{avg}=\frac{v_{f}+v_{i} }{2}

                                   v_{avg}=\frac{426+284}{2}

                                   v_{avg}  = 355 m/s      

8 0
3 years ago
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