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3 years ago

12

What is the RICE regimen for injuries?

1 answer:

amid [387]3 years ago

7 0

Answer:

<h2>R.I.C.E. stands for rest, ice, compression, and elevation</h2>

Explanation:

I hope this helps

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a truck is moving around a circular curve @ a uniform velocity of 13m/s. if the centifrical force on truck is 3,300N with truck

VMariaS [17]

<u>Answer</u>

81.94 m

<u>Explanation</u>

The centripetal force of an object moving in a circular path is given by:

F = mv²/r Where m is the mass of the object, v is the constant velocity and r is the radius of the curve.

F = mv²/r

3,300 = (1600×13²)/r

3,300 = 270,400/r

r = 270,400/3,300

= 81.94 m

8 0

3 years ago

A magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that a

scoundrel [369]

To solve this problem we will apply the concepts related to magnetic flux and induced voltage. This last expression understood as the variation of the magnetic flux over time and, in turn, the magnetic flux expressed as the variation of the magnetic field in a certain area.

Magnetic flux through the circular coil is given as

$\Phi_C = B(\pi r^2)$

The induced voltage is the change of the magnetic flux across the time, then

$\epsilon_{emf,C} = \frac{B(\pi r^2)}{t}$

At the same time the magnetic flux through the square coil would be given as,

$\Phi_S = B(r^2)$

And the induced voltage EMF will be

$\epsilon_{emf,s} = \frac{B(r^2)}{t}$

Equating both expression we have

$\epsilon_{emf,s} = \frac{\epsilon_{emf,C}tr^2}{\pi r^2t}$

$\epsilon_{emf,s} = \frac{0.74V}{\pi}$

$\epsilon_{emf,s} = 0.23355V$

Therefore the emf induced in the square coil is 0.23355V

7 0

3 years ago

An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases

andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0

3 years ago

Which feature of a longitudinal wave corresponds to a trough in a transverse wave?

Rama09 [41]

It’s the crest, the crest is the top part of the wave and the trough is the bottom so they correspond

3 0

3 years ago

Read 2 more answers

En una práctica un beisbolista lanza verticalmente la bola con una velocidad de 12m/s en dirección ascendente. ¿Cuál será la alt

fgiga [73]

Answer:

Thus, the maximum height is 7.35 m.

Explanation:

initial velocity, u = 12 m/s

acceleration due to gravity, g = 9.8 m/s^2

Velocity at maximum height, v = 0 m/s

Let the maximum height is h.

Use third equation of motion

$v^{2}=u^{2}-2 g h\\0 = 12\times 12 - 2 \times 9.8\times h\\h =7.35 m$

8 0

3 years ago