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ELEN [110]
3 years ago
6

Explain what an electric current could do if applied to a pickle!!..... I’m waiting........

Physics
1 answer:
Airida [17]3 years ago
7 0
If you apply an electric current to a pickle, the salt water will act as a conductor and will cause the pickle to glow in the dark!!!

-yw!! <3
You might be interested in
1. What is a conversion factor?
timurjin [86]

Answer:

arithmetical multiplier for converting a quantity expressed in one set of units into an equivalent expressed in another.

Explanation:

5 0
2 years ago
M
cupoosta [38]

Answer:

<h3>The answer is 1.92 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 2.5 g

Volume = 1.3 cm³

We have

density =  \frac{2.5}{1.3}  \\  = 1.923076...

We have the final answer as

<h3>1.92 g/cm³</h3>

Hope this helps you

5 0
3 years ago
An astronaut and his space suit have a combined mass of 157 kg. The
alexgriva [62]

Answer:

v₃ = 9.62[m/s]

Explanation:

To solve this type of problem we must use the principle of conservation of linear momentum, which tells us that the momentum is equal to the product of mass by velocity.

We must analyze the moment when the astronaut launches the toolkit, the before and after. In order to return to the ship, the astronaut must launch the toolkit in the opposite direction to the movement.

Let's take the leftward movement as negative, which is when the astronaut moves away from the ship, and rightward as positive, which is when he approaches the ship.

In this way, we can construct the following equation.

-(m_{1}+m_{2})*v_{1}=(m_{1}*v_{2})-(m_{2}*v_{3})

where:

m₁ = mass of the astronaut = 157 [kg]

m₂ = mass of the toolkit = 5 [kg]

v₁ = velocity combined of the astronaut and the toolkit before throwing the toolkit = 0.2 [m/s]

v₂ = velocity for returning back to the ship after throwing the toolkit [m/s]

v₃ = velocity at which the toolkit should be thrown [m/s]

Now replacing:

-(157+5)*0.2=(157*0.1)-(5*v_{3})\\(5*v_{3})= 15.7+32.4\\v_{3}=9.62[m/s]

6 0
3 years ago
PLEASE HELP A student climbs to the top of a press box where the cameras are. He wonders how many meters he is off of the ground
Feliz [49]

The height of the rail on top of the press box where the ball was dropped from is 11.025 m.

The given parameters:

  • Time of motion of the ball, t = 1.5 s
  • Let the height of the rail = h

<h3>Maximum height of fall;</h3>
  • The maximum height through which the ball was dropped from is calculated by applying second equation of motion;

h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = \frac{1}{2} (9.8) (1.5)^2\\\\h = 11.025 \ m

Thus, the height of the rail on top of the press box where the ball was dropped from is 11.025 m.

Learn more about height of projectiles here: brainly.com/question/10008919

3 0
2 years ago
A 1250 kg car starts are rest then speeds up to 15 m/s. What is the impulse of the car?
Masteriza [31]

Answer:

B)18750

Explanation:

it because

1250 \times 15 = 18750

7 0
3 years ago
Read 2 more answers
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