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NNADVOKAT [17]
3 years ago
8

HELP!!!

Physics
1 answer:
aniked [119]3 years ago
4 0

Answer:

p = m .v momentum = mass • velocity. [kg • m/s] [kg] [m/s]. Kinetic Energy. KE = 12 • m • v ... 1. A 1500 kg car traveling at 15 m/s to the south collides with a 4500 kg truck that is ... What is the final velocity of the two-vehicle mass? ... m/s. What is the velocity of the joined cars after the collision? ... 5) = (1.5x104+1.5x604) VELVE.

Explanation:

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Explanation:

6 0
3 years ago
Read 2 more answers
A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
3 years ago
Two 8.0 Ω lightbulbs are connected in a 12 V parallel circuit. What is the power of both glowing bulbs?
kati45 [8]

Answer:

96w

Explanation:

p=Iv..where v=12 and I=8.0

8 0
3 years ago
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