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3 years ago

batman recharges his bat taser with 2500 joules of energy in 15 minutes. how much power does it draw from the mains

1 answer:

6 0

The power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s):
P(W) = E(J) / t(s)

2.78 Watts

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a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric fiel

Lesechka [4]

Answer:

-1748*10^N/C

Explanation:

See attached file

8 0

3 years ago

Two cars collide at an intersection. Car A , with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, i

devlian [24]

Answer:

The answer of the part (a) is v2 = 7.09 m/s

and the answer of the part (b) is vA1 = 5.25 m/s

Explanation:

Explanation of the both parts of answer is in the following attachments

6 0

3 years ago

If 1.8 amp flows through a 40 ohm resistor, the voltage across the resistor is :

GenaCL600 [577]

Answer:

72 volts.

Explanation:

To solve this, we have to use the Ohm's law.

The ohm's law tells us that the voltage drop of a resistor is directly proportional to the current applied to the conductor.

$V=I*R$

in this case the current is 1.8 amps and the resistor is 40 ohm

$V=1.8*40$

$V=72 volts$ .

5 0

3 years ago

The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur

Nimfa-mama [501]

The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

P₂ is the pressure in the hose

ρ is the density of the fluid

h₁ is the initial height

h₂ is the final height

v₁ is the initial velocity of the fluid

v₂ is the final velocity of the fluid and

g is the acceleration due to gravity

Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

= (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

= (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

= (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

= 5.92 × 10⁵ Pa

= 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

<u>brainly.com/question/9506577</u>

#SPJ4

6 0

2 years ago

Imagine a place in the cosmos far from all gravitational and frictional influences. Suppose that you visit that place (just supp

xz_007 [3.2K]

The rock will continue to travel in a straight line with a constant velocity for ever... The reason is, once it leaves your hand there is no force acting on the rock, so it will just continue to move in a natural motion which is constant velocity.

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3 years ago