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3 years ago

How many regions does the deltoid muscle have

Physics

2 answers:

pentagon [3]3 years ago

6 0

There are 3 regions in the deltoid muscle. :D

Lapatulllka [165]3 years ago

5 0

There are 3 regions!!!

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I don’t understand how to do this

anzhelika [568]

Explanation:

A) Use Hooke's law to find the spring constant.

F = kx

40 N = k (0.4 m)

k = 100 N/m

B) Period of a spring-mass system is:

T = 2π √(m / k)

T = 2π √(2.6 kg / 100 N/m)

T = 1 s

Frequency is the inverse of period.

f = 1 / T

f = 1 Hz

5 0

3 years ago

An ac generator with Em = 223 V and operating at 399 Hz causes oscillations in a series RLC circuit having R = 222 Ω, L = 147 mH

Doss [256]

Answer:

Xc= 17.267 Ω, Z= 415.5 Ω, I= 0.537 A

Explanation:

Em = 223 V

f= 300 Hz, R = 222 Ω, L = 147 mH, C = 23.1 μF

Capacitive reactance = Xc=?

Xc= $\frac{1}{2\pi fC}$

Xc=1/2pi *399*23.1*10^-6

Xc= 17.267 Ω

b).

Z= $\sqrt{ R^2 + (Xl - Xc)^2}$

Xl= 2π * f * L

Xl= 2π * 399 * 147 * $10^{-3}$

Xl= 368.5 Ω

Z= $\sqrt{ R^2 + (Xl - Xc)^2}$ = $\sqrt{222^2 + (368.5 - 17.267)^2}$

Z= 415.5 Ω

c).

Current:

I= V / Z= Em / Z

I= 223/415.5

I= 0.537 A

3 0

3 years ago

A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

Mathematical expression for the Newton's second law of motion is given as:

$F=\frac{dp}{dt}$ ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

$p=m.v$

Now, equation (1) becomes:

$F=\frac{d(m.v)}{dt}$

∵mass is constant at speeds v << c (speed of light)

$\therefore F=m.\frac{dv}{dt}$

and, $\frac{dv}{dt} =a$

where: a = acceleration

$\Rightarrow F=m.a$

also

$F\propto \frac{1}{dt}$

so, more the time, lesser the force.

& Impulse:

$I=F.dt$

$I=m.a.dt$

$I=m.\frac{dv}{dt}.dt$

$I=m.dv=dp$

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0

3 years ago

Passengers on a carnival ride move at constant speed in a horizontal circle of radius 5.0 m, making a complete circle in 4.0 s.

Nataliya [291]

Answer:

bonita sisisisiisisisisiisissiissiisiiss

7 0

2 years ago

Sound wave cannot propagate through vacuum.Why? Plz give reason in short....

Olenka [21]

Answer:

Explanation:

If a perfect vacuum existed in any volume, then no sound would be able to propagate through it, because a sound wave is a pressure wave, and there would be identically zero pressure. Of course, we could get into speculations about “dark energy” or “vacuum energy” supporting pressure waves, but let’s not go there.

4 0

3 years ago