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Aleks04 [339]
3 years ago
6

When lights are turned off, the electrical circuit is _________, because _________. A. closed, because it stops electricity from

flowing through the circuit.
B. open, which stops electricity from flowing completely through the circuit.
C. tangled, it is in a mess. D. broken, and does not let electricity flow through.
Physics
2 answers:
Eva8 [605]3 years ago
3 0

Answer:

I would say B makes the most sense here

8_murik_8 [283]3 years ago
3 0
It is A because everything does not make sense, b, when ever it’s answer for because, it does not work out, a closed it so it will stop it from flowing from the circuit, when it opens electricity flows through.
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A bicyclist starts at point P and travels around a triangular path that takes her through points Q and R before returning to the
REY [17]

Answer:

Displacement by cyclist is zero.

Explanation:

In the given question bicyclist is travelling in a rectangular track having P , Q and R edges.

The bicyclist starts from P and travel through Q and R and returned to P again.

We need to find its displacement.

We know displacement  of a body is its difference between its initial position to final position.

Here in the given question the bicyclist returns to P again.

Therefore, total displacement by bicyclist is zero.

Hence, this is the required solution.

4 0
4 years ago
If the frequency of the 13C signal of TMS is 201.16 MHz, the two 13C signals of acetic acid at 179.0 and 20.0 ppm are separated
Lelu [443]

The difference in frequency of the two signals is 1.33 \times 10^{10} \ kHz.

The given parameters;

  • <em>frequency of the 13 C signal = 201.16 MHz</em>

The energy of the 13 C signal located at 20 ppm is calculated as follows;

E = hf\\\\E_1 = h \frac{c}{\lambda} \\\\E_1 =  \frac{(6.626 \times 10^{-34})\times 3\times 10^8}{20 \times 10^{-6}} \\\\E_1 = 9.94 \times 10^{-21} \ J

The energy of the 13 C signal located at 179 ppm is calculated as follows;

E_2 = \frac{hc}{\lambda} \\\\E_2 = \frac{(6.626\times 10^{-34})\times (3\times 10^{8})}{179 \times 10^{-6} } \\\\E_2 = 1.11 \times 10^{-21} \ J

The difference in frequency of the two signals is calculated as follows;

E_1- E_2 = hf_1 - hf_2\\\\E_1 - E_2 = h(f_1 - f_2)\\\\f_1 - f_2 = \frac{E_1 - E_2 }{h} \\\\f_1 - f_2 = \frac{(9.94\times 10^{-21}) - (1.11 \times 10^{-21})}{6.626\times 10^{-34}} \\\\f_1 - f_2 = 1.33 \times 10^{13} \ Hz\\\\f_1 - f_2 = 1.33\times 10^{10} \ kHz

Thus, the difference in frequency of the two signals is 1.33 \times 10^{10} \ kHz.

Learn more here:brainly.com/question/14016376

4 0
3 years ago
A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform abov
Rina8888 [55]

Answer:

20.96 m/s

Explanation:

Using the equations of motion

y = uᵧt + gt²/2

Since the puck slides off horizontally,

uᵧ = vertical component of the initial velocity of the puck = 0 m/s

y = vertical height of the platform = 2 m

g = 9.8 m/s²

t = time of flight of the puck = ?

2 = (0)(t) + 9.8 t²/2

4.9t² = 2

t = 0.639 s

For the horizontal component of the motion

x = uₓt + gt²/2

x = horizontal distance covered by the puck

uₓ = horizontal component of the initial velocity = 20 m/s

g = 0 m/s² as there's no acceleration component in the x-direction

t = 0.639 s

x = (20 × 0.639) + (0 × 0.639²/2) = 12.78 m

For the final velocity, we'll calculate the horizontal and vertical components

vₓ² = uₓ² + 2gx

g = 0 m/s²

vₓ = uₓ = 20 m/s

Vertical component

vᵧ² = uᵧ² + 2gy

vᵧ² = 0 + 2×9.8×2

vᵧ = 6.26 m/s

vₓ = 20 m/s, vᵧ = 6.26 m/s

Magnitude of the velocity = √(20² + 6.26²) = 20.96 m/s

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3 years ago
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KengaRu [80]

The list of choices you provided with your question
is utterly devoid of any such examples.

6 0
4 years ago
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What is the difference between 1D motion and 2D motion?<br> I need explanation, pleases.
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Explanation:

1D motion is motion in only one direction.  To put it simply, it's motion along a line.  For example, a train on a straight track has 1D motion.

2D motion is motion in two directions.  A good example of this is a projectile launched at an angle.  It has both vertical motion and horizontal motion.

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