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frutty [35]
3 years ago
14

Casey remembers that molecular motion increases as temperature increases. Identify what effect increased molecular motion.

Physics
2 answers:
adelina 88 [10]3 years ago
8 0
It's probably something that makes it hot
Zanzabum3 years ago
3 0
Since Casey remembered that molecular motion increases as TEMPERATURE increases, I would have to say the effect is some form of heat source
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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
what is the gravitational potential energy of the rock if its weight is 15 newtons its mass is 1.53 grams and it is 30 meters ab
Leviafan [203]

Answer:

16. An object has a gravitational potential energy 41,772.5 Jof and has a mass of 1550 kg. How high is it above the ground? Plz help

5 0
3 years ago
An antelope moving with constant acceleration covers the distance 72.0 m between two points in time 7.10 s . Its speed as it pas
Studentka2010 [4]

Answer:

a. 4.28 m/s

b. 1.65 m/s²

Explanation:

a.

Speed: This can be defined as the ratio of distance to time. The S.I unit of speed is m/s.

From newton's equation of motion,

S = (v+u)t/2 ............................... Equation 1

Where S = distance, u = initial speed, v = final speed, t = time.

Given: S = 72 m, v = 16.0 m/s, t = 7.10 s.

Substitute into equation 1

72 = (16+u)7.10/2

making u the subject of the equation,

7.1(16+u) = 72×2

113.6+7.1u = 144

7.1u = 144-113.6

7.1u = 30.4

u = 30.4/7.1

u = 4.28 m/s

Thus the speed at the first point = 4.28 m/s

b.

Acceleration: This can be defined as the rate of change of velocity. The S.I unit of acceleration is m/s²

a = (v-u)/t............................ Equation 2

Where a = acceleration.

Given: v = 16.0 m/s, u = 4.28 m/s, t = 7.10 s

Substitute into equation 2,

a = (16-4.28)/7.1

a = 11.72/7.10

a = 1.65 m/s²

Thus the acceleration = 1.65 m/s²

3 0
3 years ago
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