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3 years ago

What is potential energy? Energy released Energy stored Moving Energy

Physics

2 answers:

oksian1 [2.3K]3 years ago

8 0

Energy stored- it’s inside

Sergeu [11.5K]3 years ago

4 0

Potential energy is stored :)

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a 1500 kg car traveling at 15 m/s to the south collides with a 4500 kg truck that is intially at rest at a spotlight. The car an

harkovskaia [24]

Answer:

3.75 m/s south

Explanation:

Momentum before collision = momentum after collision

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Since the car and truck stick together, v₁ = v₂.

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

Given m₁ = 1500 kg, u₁ = -15 m/s, m₂ = 4500 kg, and u₂ = 0 m/s:

(1500 kg) (-15 m/s) + (4500 kg) (0 m/s) = (1500 kg + 4500 kg) v

-22500 kg m/s = 6000 kg v

v = -3.75 m/s

The final velocity is 3.75 m/s to the south.

4 0

3 years ago

Read 2 more answers

A person pushes a 10 kg box from rest and accelerates it to a speed of 4 m/s with a constant force. If the box is pushed for a t

lutik1710 [3]

The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration a is such that

4 m/s = a (2.5 s) → a = (4 m/s) / (2.5 s) = 1.6 m/s²

Then the force applied to the box has a magnitude F such that

F = (10 kg) (1.6 m/s²) = 16 N

7 0

3 years ago

A 15-watt bulb is connected to a circuit that has a total of 60. Ω of resistance. How many electrons are passing through that bu

Mariulka [41]

Answer:

3.2075*10^16

Explanation:

Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.

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2 years ago

Click the links to open the resources below. These resources will help you complete the assignment. Once you have created your f

Paul [167]

Answer:

Explanation:

lesgse in no

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3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

The current I is any motion of charge from one region to another, so this is given by:

 $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$

Finally, for the drift velocity magnitude vd, we find:

 $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

3 years ago