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Mariana [72]
2 years ago
6

A sound is first produced by making something . The sound then travels through a to reach the ears, which are the parts of the b

ody that allow for sounds to be heard.
Physics
2 answers:
r-ruslan [8.4K]2 years ago
8 0
A sound is first produced by making something<span> vibrate</span><span>. The sound then travels through a </span><span> medium</span><span> to reach the ears, which are the parts of the body that allow for sounds to be heard.

Vibrate and Medium are the correct answers
</span>
steposvetlana [31]2 years ago
5 0

Vibrate

Medium

are both correct on edu.

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A car drives at a constant speed around a banked circular track with a diameter of 136 m . The motion of the car can be describe
galina1969 [7]

Answer:

speed = 44.9m/s

x = 35.5 m,  y = 58.0m

Explanation:

A car on a circular track with constant angular velocity ω can be described by the equation of position r:

\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}

The velocity v is given by:

\overrightarrow {v(t)} = \overrightarrow{\frac{dr}{dt}}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}

The acceleration a:

\overrightarrow {a(t)} = \overrightarrow{\frac{dv}{dt}}= -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}

From the given values we get two equations:

-\omega^2 Rsin(\omega t)=-15.4\\-\omega^2 Rcos(\omega t)=-25.4

We also know:

\overrightarrow {a(t)} = -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}=-\omega^2\overrightarrow{r(t)}

The magnitude of the acceleration a is:

a=\sqrt{(-15.4)^2+(-25.4)^2}=29.7

The magnitude of position r is:

r=R=68m

Plugging in to the equation for a(t):

\overrightarrow {a(t)} =-\omega^2\overrightarrow{r(t)}

and solving for ω:

|\omega|=0.66

Now solve for time t:

\frac{sin(0.66t)}{cos(0.66t)}=tan(0.66t)=\frac{15.4}{25.4}\\t=0.83

Using the calculated values to compute v(t):

\overrightarrow {v(t)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}\\\overrightarrow {v(t)}=44.88cos(0.55)\hai{i}-44.88sin(0.55)\hat{j}\\\overrightarrow {v(t)}=38.3\hat{i}-23.5\hat{j}

The speed of the car is:

\sqrt{38.3^2 + (-23.5)^2} = 44.9

The position r:

\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}\\\overrightarrow {r(t)} = 68sin(0.55)\hat{i} + 68cos(0.55)\hat{j}\\\overrightarrow {r(t)} = 35.5{i} + 58.0\hat{j}

5 0
2 years ago
Read 2 more answers
The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f
zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

                 584 = Ha + Lg

an also this mean that torques acting on the body is balanced

         So,   0.410 Ha  = 0.840 Lg

    Making Lg the subject of formula in the equation above we

   Lg = 0.4881 Ha

 Considering the first equation and replacing Lg with this recent equation we have

                      584 = Ha + 0.4881 Ha

          Therefore Ha = 392.44 N

This value obtained is  for both hands for each hand we divide by 2

Therefore we have for each hand = 392.44/2 =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

             Lg = 0.4881 ×  392.44

                  = 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

                  191.22/2 = 95.8 N.

8 0
2 years ago
A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0
1 year ago
Julius competes in the hammer throw event. the hammer has a mass of 7.26 kg and is 1.215 m long. what is the centripetal force o
TEA [102]
Centripetal force <span>a force that acts on a body moving in a circular path and is directed toward the center around which the body is moving. It is calculated by the expression:

F = mv^2/r

where m is the mass, v is the velocity and r is the radius.

F = 7.26(31.95)^2 / (1.215) = 6100 N</span>
8 0
2 years ago
Read 2 more answers
Two climbers are on a mountain. Simon, of mass m, is sitting on a snow covered slope that makes an angle θ with the horizontal.
elena-14-01-66 [18.8K]

Answer:

Explanation:

It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.

Let in equilibrium , tension in rope be T

For balancing Joe

T = M g

For balancing Simon

friction + T = mgsinθ

μmgcosθ+T = mgsinθ

μmgcosθ+Mg = mgsinθ

M = (msinθ - μmcosθ)

M = m(sinθ - μcosθ)

5 0
1 year ago
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