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iren2701 [21]
2 years ago
7

10. A flashlight runs on a battery. If the light is left on the battery runs out and the flashlight

Physics
1 answer:
schepotkina [342]2 years ago
3 0
Well I believe the claim is if you use all its energy fr a long period of time it does so the claim is if the light is left on the battery runs out and flashlight stops working
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Which of the following is the best example of a good hypothesis?
Musya8 [376]

Answer:

Which of the following is the best example of a good hypothesis?

B. A cheetah can run faster than a tiger This Is A good Hypothesis Because You Can Test this With a Experiment

xXxAnimexXx

Have a great day!

4 0
3 years ago
Read 2 more answers
An incompressible fluid (water) is flowing through a pipe of diameter 20 cm with
sergey [27]

Answer:

115 kPa

Explanation:

Use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Assuming no elevation change, h₁ = h₂.

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Plugging in values:

(582,000 Pa) + ½ (1000 kg/m³) (1.28 m/s)² = P + ½ (1000 kg/m³) (30.6 m/s)²

P = 115,000 Pa

P = 115 kPa

3 0
3 years ago
Driving Zone 7 corresponds with
stepan [7]
Your driving zone refers to the areas of space around your car, it refers to all the area around your car as far as your eyes can see. 
Each car has seven zones numbered from 1 to 7. Driving zone 7 corresponds with THE SPACE YOUR VEHICLE IS OCCUPYING. The other zones are as follows:
zone 1 = area directly infront of your car
zone 2 = your left lane
zone 3 = your right lane
zone 4 = left rear of your car
zone 5 = right rear of your car 
zone 6 = area directly behind your car.
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3 0
2 years ago
The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an
9966 [12]

Answer:

-4.941*10^8J.

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

\Delta U = \Delta_{perogee}-\Delta_{Apogee}

Where,

U= \frac{-GmM_e}{r}

Replacing

\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}

\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})

Our values are given by,

m = 85.5Kg

M_e = 5.97*10^{24}Kg

r_A = 7330Km

r_p = 6610Km

G = 6.67*10^{-11}Nm^2/Kg^2

Replacing at the equation,

\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})

\Delta U = -4.941*10^8J

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was -4.941*10^8J.

4 0
3 years ago
A clown at a birthday party has brought along a helium cylinder, with which he intends to fill balloons. When full, each balloon
Furkat [3]

Answer:

11

Explanation:

According to Boyle's law:

P\times V=constant

Thus,

P_{cylinder}\times V_{cylinder}=n\times P_{balloon}\times V_{balloon}

Where, n is the number of the balloons

From the question, it is given that:

For balloon:

P = 1.2\times 10^5 Pa

V = 0.040 m³

For cylinder:

P = 1.80\times 10^7 Pa

V = 0.0031 m³

So,

1.80\times 10^7\times 0.0031=n\times 1.2\times 10^5\times 0.040

n = 11.625

<u>So, Maximum number of balloons = 11</u>

7 0
3 years ago
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