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algol [13]
3 years ago
5

What is one way to store a lot of potential energy in the field between two charged objects?

Physics
1 answer:
Alekssandra [29.7K]3 years ago
3 0

Answer:

Explanation:

Two charged particles will vary in energy as we alter the distance between them.

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The formula used to find force is F=m*v.<br> true or false
zepelin [54]

It's true IF ' m ' stands for mass and ' v ' stands for acceleration. Otherwise it's false.

4 0
2 years ago
Which equation describes the line containing the points (-2, 3) and (1, 2)​
snow_lady [41]

Answer:

y =  \frac{ - 1}{3}x +  \frac{7}{3}

Explanation:

\frac{y - 3}{2 - 3}  =  \frac{x + 2}{1 + 2}  \\  \ - y + 3 =  \frac{x + 2}{3}  \\  y =   \frac{ - 1}{3}x  +  \frac{7}{3}

8 0
2 years ago
A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo
m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

   E_{mf} = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

7 0
3 years ago
Which of the following have the same density. number 26
marusya05 [52]

1 cubic cm is the same as 1 mL, so the answer would be C.

8 0
2 years ago
A jeweler working with a heated 47 g gold ring must lower the ring's temperature to make it safe to handle. If the ring is initi
Gelneren [198K]

Mass of gold m₁ = 47 g

Initial temperature of gold T₁ = 99 C

Specific heat of gold C₁ = 0.129 J/gC

final temperature T₂ = 38 C

Heat needed by the gold to cool down

Q =m₁ * C₁* ( T₁ - T₂)

Q = (47)(0.129)(99-38)

Q = 369.843 J

This heat will be given by the water

we need to find out mass of water m₂

and initial temperature of water is T₃ = 25 C

Specific heat of water C₂ = 4.184 J/gC

Q = m₂*C₂*(T₂ - T₃)

369.843 = m₂(4.184)(38-25)

m₂ = 6.8 g

6 0
3 years ago
Read 2 more answers
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