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3 years ago

How do scientists interpret data?

Physics

1 answer:

Vladimir79 [104]3 years ago

6 0

Answer:

Explanation:

Scientist use many techniques to analyze and interpret data. To interpret data they make tables and graphs of data which are two of the most useful techniques in data analysis.

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An electromagnet is a device in which moving electric charges (current) in a coil of wire create a magnet. What’s one advantage

docker41 [41]

Electromagnets have certain advantages over permanent magnets. One of the major advantage is, the amount of electric current can easily be controlled, in turn, magnetic field can be controlled. Out of all the options, option D would be the best answer.

In short, Your Answer would be option D) Electromagnets can easily be turned on and off.

Hope this helps!

5 0

3 years ago

Given two vectors A=4i^+3j^ and vector<br> B=5i^-2j^.find the magnitude of each vector

Flura [38]

Answer:

<em>Magnitude of A=5</em>

<em>Magnitude of B=5.39</em>

Explanation:

<u>The magnitude of Vectors in Rectangular Form</u>

Given a vector v in its rectangular form:

$\mathbf{v}=x\hat i+y\hat j$

The magnitude of v is:

$\mid\mid\mathbf{v}\mid \mid=\sqrt{x^2+y^2}$

We are given the vectors

$\mathbf{A}=4\hat i+3\hat j$

$\mathbf{B}=5\hat i-2\hat j$

Their magnitudes are:

$\mid\mid\mathbf{A}\mid \mid=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}$

$\mid\mid\mathbf{A}\mid \mid=5$

$\mid\mid\mathbf{B}\mid \mid=\sqrt{5^2+(-2)^2}=\sqrt{25+4}=\sqrt{29}$

$\mid\mid\mathbf{B}\mid \mid=\sqrt{29}=5.39$

4 0

3 years ago

which statement is true of a wave that's propagating along the pavement and girders of a suspension bridge?

Blizzard [7]

<u>Question:</u>

Which statement is true of a wave that’s propagating along the pavement and girders of a suspension bridge?

A. The wave is mechanical, with particles vibrating in a direction that is parallel to that of the wave, forming compressions and rarefactions.

B. The wave is electromagnetic, with particles vibrating in a direction that is parallel to that of the wave, forming compressions and rarefactions.

C. The wave is mechanical, with particles vibrating in a direction that is perpendicular to that of the wave, forming crests and troughs.

D. The wave is electromagnetic, with particles vibrating in a direction that is perpendicular to that of the wave, forming crests and troughs.

E. The wave is mechanical, with particles vibrating in a direction that is perpendicular to that of the wave, forming compressions and rarefactions.

Answer:

The statement “The wave is mechanical, with particles vibrating in a direction that is perpendicular to that of the wave forming compressions and rarefactions” is true of a wave that’s propagating along the pavement and girders of a suspension bridge.

<u>Explanation: </u>

As it is stated about the wave propagates in the given question, the first thing to confirm is the wave should be a mechanical wave as electromagnetic wave will not penetrate the suspension bridge.

Then as both the pavements and girders are rigid structures present in suspension bridge, the wave can propagate in them which are forming compressions and rarefactions.

The transverse waves cannot pass through them because of the polarization property of transverse waves. Also as they propagate in compression and rarefactions mode, the particles will tend to vibrate only in perpendicular direction to that of the wave propagation. So, the last statement is only true.

3 0

3 years ago

Read 2 more answers

u are holding the axle of a bicycle wheel with radius30 cm and mass 1.05 kg. You get the wheel spinning at arate of 77 rpm and t

DedPeter [7]

Answer:

-5.9 rad/s^{2}

Explanation:

radius (r) = 30 cm = 0.3 m

mass (m) = 1.05 kg

initial speed (u) = 77 rpm

final speed (v) = 0 rpm

time (t) = 1.37 s

angular acceleration = $\frac{(final speed-initial speed)rad/s}{time}$

therefore

initial speed (U) = 77 rpm = 77 x (2π/60) = 8.06 rad/s

final speed (v) = 0 rpm = 0 rad/s

angular acceleration = $\frac{0-8.06}{1.37}$ = -5.9 rad/s^{2}

5 0

3 years ago

Read 2 more answers

In the figure below (Figure 1), the upper ball is released from rest, collides with the stationary lower ball, and sticks to it.

Georgia [21]

(a) The frequency of the motion after the collision is 0.71 Hz.

(b) The maximum angular displacement of the motion after the collision is 16.3⁰.

<h3>Speed of the 2.2 kg ball when it collides with 2.7 kg ball</h3>

The speed of the 2.2 kg ball which was initially 10 cm higher that 2.7 kg ball is calculated as follows;

K.E = P.E

¹/₂mv² = mgh

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 0.1)

v = 1.4 m/s

<h3>Final speed of both balls after collision</h3>

The final speed of both balls after the collision is determined from the principle of conservation of linear momentum.

Pi = Pf

m₁v₁ + m₂v₂ = vf(m₁ + m₂)

2.2(1.4) + 2.7(0) = vf(2.2 + 2.7)

3.08 = 4.9vf

vf = 3.08/4.9

vf = 0.63 m/s

<h3>Maximum displacement of the balls after the collision</h3>

P.E = K.E

$mgh_f = \frac{1}{2} mv_f^2\\\\h_f = \frac{v_f^2}{2g} \\\\h_f = \frac{(0.63)^2}{2(9.8)} \\\\h_f = 0.02 \ m$

<h3>Maximum angular displacement</h3>

The maximum angular displacement of the balls after the collision is calculated as follows;

$cos \theta = \frac{L - h_f}{L} \\\\cos\theta = \frac{0.5 - 0.02}{0.5} \\\\cos\theta = 0.96\\\\\theta = cos^{-1}(0.96)\\\\\theta = 16.3 \ ^0$

<h3>Frequency of the motion</h3>

$f = \frac{1}{2\pi} \sqrt{\frac{g}{L} } \\\\f = \frac{1}{2\pi } \sqrt{\frac{9.8}{0.5} } \\\\f = 0.71 \ Hz$

Learn more about maximum angular displacement here: brainly.com/question/13665036

4 0

2 years ago