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adelina 88 [10]
3 years ago
14

A 500 kg car traveling at 20 m/s rear ends another car of 600 kg at rest. The collision is great enough that the two cars stick

together after they collide. How fast will both cars be going after the collision?
Physics
1 answer:
aksik [14]3 years ago
4 0

Answer: its 50

Explanation:

im waffling does anybody have syrup

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Please help with this question !!!!!
lesantik [10]

Answer:

te puedo ayudar pero corazón y estrella

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7 0
3 years ago
Pierre is a 375 kg great white with an average speed of 3 m/s. When Pierre spots a seal, he increases his velocity to 7 m/s. Aft
tangare [24]

Answer:

1890J

Explanation:

375+45 = 420kg (total mass)

kinetic energy = 1/2 × mass × velocity²

1/2 × 420 × 3² = 1890J

5 0
3 years ago
An object stays at rest until what happens to it?
bixtya [17]
D. an outside or unbalanced force acts upon the object.
4 0
3 years ago
A 2 kg object has a specific heat capacity of 1,700 J/(kg \cdot⋅oC)
Nutka1998 [239]

The amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

  • The amount of heat absorbed by an object can be calculated by using the following expression:

  • Q = m.c.∆T

Where;

  1. Q = amount of heat absorbed or released (J)
  2. m = mass of object
  3. c = specific heat capacity (J/g°C)
  4. ∆T = change in temperature (°C)

  • According to this question, 2 kg object has a specific heat capacity of 1,700J/kg°C and was raised from a temperature of 15 Celsius to 25 Celsius. The heat absorbed is calculated as follows:

  • Q = 2 × 1700 × {25 - 15}

  • Q = 3400 × 10

  • Q = 34000J

  • Therefore, the amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.

Learn more about how to calculate heat absorbed at: brainly.com/question/11194034?referrer=searchResults

8 0
2 years ago
*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude
kvv77 [185]

Answer:

Approximately 3.86\; {\rm N} (given that the magnitude of this charge is -7.35\; {\rm \mu C}.)

Explanation:

If a charge of magnitude q is placed in an electric field of magnitude E, the magnitude of the electrostatic force on that charge would be F = E\, q.

The magnitude of this charge is q = 7.35\; {\rm \mu C}. Apply the unit conversion 1\; {\rm \mu C} = 10^{-6}\; {\rm C}:

\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}.

An electric field of magnitude E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}} would exert on this charge a force with a magnitude of:

\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0
2 years ago
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