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2 years ago

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Ten students stand in a circle and are told to make a transverse wave. What best describes the motion of the students? Each stud

ent bumps the shoulder of his or her neighbor. At the exact time, all students take a step to the right. The students skip clockwise, going in a circular motion. One at a time, each student lifts his or her hands up and then down.

2 answers:

ELEN [110]2 years ago

6 0

Answer:

The correct option here is D.

<em>~Hope this helps! :)</em>

Bond [772]2 years ago

3 0

Lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students. Option D is correct.

<h3>What is a Transverse wave?</h3>

The wave in which the oscillation of particles is is perpendicular to the direction of energy transfer.

The students can make a transverse wave by raising their hands up and then down, one student at a time.

The raised hand represents the oscillation of particles while the sequence of the raising hand represents the direction of energy transfer.

Therefore, lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students.

Learn more about Transverse waves:

brainly.com/question/3813804

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A baseball has a mass of 0.15 kg and radius 3.7 cm. In a baseball game, a pitcher throws the ball with a substantial spin so tha

mylen [45]

Answer:

Rotational kinetic energy = 0.099 J

Translational kinetic energy = 200 J

The moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$ .

Explanation:

Rotational kinetic energy is given by

$\text{RKE} = \frac{1}{2}I\omega^2$

where <em>I</em> is the moment of inertia and <em>ω</em> is the angular speed.

For a solid sphere,

$I = \frac{2}{5}mr^2$

where <em>m</em> is its mass and <em>r</em> is its radius.

From the question,

<em>ω</em> = 49 rad/s

<em>m</em> = 0.15 kg

<em>r</em> = 3.7 cm = 0.037 m

$\text{RKE} = \frac{1}{2}\times \frac{2}{5} mr^2\omega^2 = \frac{1}{5} mr^2\omega^2$

$\text{RKE} = \frac{1}{5} (0.15\ \text{kg})(0.037\ \text{m})^2(49\ \text{rad/s})^2 = 0.099\text{ J}$

Translational kinetic energy is given by

$\text{TKE} = \frac{1}{2} mv^2$

where <em>v</em> is the linear speed.

$\text{TKE} = \frac{1}{2} (0.15\ \text{kg})(52\ \text{m/s})^2 = 200\text{ J}$

5 0

2 years ago

Read 2 more answers

Two long parallel wires carry currents of 20 a and 5.0 a in opposite directions. the wires are separated by 0.20 m. what is the

Alex777 [14]

The two wires carry current in opposite directions: this means that if we see them from above, the magnetic field generated by one wire is clock-wise, while the magnetic field generated by the other wire is anti-clockwise. Therefore, if we take a point midway between the two wires, the resultant magnetic field at this point is just the sum of the two magnetic fields, since they act in the same direction.

Therefore, we should calculate the magnetic field generated by each wire and then calculate their sum. We are located at a distance r=0.10 m from each wire.

The magnetic field generated by wire 1 is:
$B_1= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(20 A)}{2 \pi (0.10 m)}= 4 \cdot 10^{-5}T$

The magnetic field generated by wire 2 is:
$B_2= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(5.0 A)}{2 \pi (0.10 m)}= 1 \cdot 10^{-5}T$

And so, the resultant magnetic field at the point midway between the two wires is
$B=B_1 + B_2 = 4 \cdot 10^{-5} T + 1 \cdot 10^{-5}T=5 \cdot 10^{-5} T$

8 0

3 years ago

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 ms from an initial spe

dezoksy [38]

Answer:

the magnitude of the average contact force exerted on the leg is 3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F $_{hand$

using equation for impulse in momentum

F $_{hand$ × t = m( v - v₀ )

we substitute

F $_{hand$ × 0.00265 = 1.75( 0 - 5.25 )

F $_{hand$ × 0.00265 = 1.75( - 5.25 )

F $_{hand$ × 0.00265 = -9.1875

F $_{hand$ = -9.1875 / 0.00265

F $_{hand$ = -3466.98 N

Next we determine force on the leg F $_{leg$

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F $_{leg$ = - F $_{hand$

we substitute

F $_{leg$ = - ( -3466.98 N )

F $_{leg$ = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N

5 0

2 years ago

A right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 lb/ft3

tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ $ft^{3}$ to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = $\pi r^{2}$ = $\pi 4^{2}$ =16 $\pi$ $ft^{2}$

Now,

Work done required = $\int_{0}^{8} 52\times 16\pi (21 - y)dy$

= $832\pi \int_{0}^{8} (21 - y)dy$

= 832 $\pi($ $[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\$ )

= 113152 $\pi$ = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0

3 years ago

For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103

Alona [7]

Answer:

a) P = 44850 N

b) $\delta l =0.254\ mm$

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

$\sigma=\frac{\textup{Load}}{\textup{Area}}$

on substituting the values, we get

$345\times10^6=\frac{\textup{Load}}{0.00013}$

or

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation ( $\delta l$ ) due to an axial load is given as:

$\delta l =\frac{PL}{AE}$

on substituting the values, we get

$\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}$

or

$\delta l =0.254\ mm$

3 0

2 years ago