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vodomira [7]
3 years ago
6

Is a Joule the same as a Kelvin?

Physics
2 answers:
earnstyle [38]3 years ago
7 0
No,because a Kevin is 7.242971666663E+22 times Smaller than a Joule.
Harlamova29_29 [7]3 years ago
3 0

Answer:

no

Explanation:

joule is the si unit of work and kelvin is temperature

plz make me brainliest

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A stone on ground is zero energy​
NNADVOKAT [17]

Answer:

A stone on the ground does not have zero energy…there is an internal potential in every object. Aldo is not in action or in any mechanical motion it is being acted upon by gravity and also molecular forces and energy.

<em>Hope</em><em> </em><em>this</em><em> helps</em><em> </em><em>!</em>

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3 years ago
What did Wundt's pendulum experiment demonstrate?
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C. <span>People need time to shift between two different stimuli.</span>
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A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
What kind of acceleration occurs when an object speeds up?
Fantom [35]

Answer:

Postive Acceleration

Explanation:

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2 years ago
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A VHF television station assigned to channel 4 transmits its signal using radio waves with a frequency of 66.0 MHz. Calculate th
soldi70 [24.7K]

Answer:

The wavelength is 4.55 m.

Explanation:

Given data

  • Frequency (ν): 66.0 MHz = 66.0 × 10⁶ Hz = 66.0 × 10⁶ s⁻¹
  • Speed of light (c): 3.00 × 10⁸ m/s
  • Wavelength (λ): To be found

We can determine the wavelength of the radio waves using the following expression.

c = λ × ν

λ = c / ν

λ = (3.00 × 10⁸ m/s)/66.0 × 10⁶ s⁻¹

λ = 4.55 m

6 0
2 years ago
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