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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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If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens woul
Brilliant_brown [7]

Answer:

3×10^7 m/s or 0.10c (e)

Explanation: If the actual value of the speed of light were to be put into consideration.

Given that the speed of light is c = 3.0×10^8m/s

The alien spaceship is approaching at the rate of 10% of the speed of light.

10% of 3.0×10^8m/s

10/100 × 3.0×10^8m/s

0.1 ×3.0×10^8m/s

3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c

7 0
1 year ago
What best describes myotibrils
krok68 [10]
Myofibrils are composed of long proteins such as actin, myosin, and titin, and other proteins that hold them together. These proteins are organized into thin filaments and thick filaments, which repeat along the length of the myofibril in sections called sarcomeres. Muscles contract by sliding the thin (actin) and thick (myosin) filaments along each other.
8 0
1 year ago
At what distance above earth would a satellite have a period of 125 min?
Nezavi [6.7K]
Rw^2 = GmM/r^2 
<span> Leads to 
</span><span> w^2 r^3 = GM 
</span><span> (2pi /T) ^2 r^3 = GM 
</span><span> 4pi^2 r^3 = GM T^2 
</span><span> r^3 = GM T^2 / 4pi^2 
</span><span> Work out r^3 then r. 
</span> T = 125 min = 125(60) = 7500 s 
<span> R = 6.38E6 m 
</span><span> m = 5.97E24 kg 
</span><span> G = 6.673E-11 
</span> r=<span> 8279791.78</span><span> m
 Since r = radius R of Earth + height above urface,h 
</span><span> h = r - R = </span><span> 8279791.78 - </span>6.38E6 = <span> <span>1899791.78 m
 h=</span></span><span> <span>1899.79178 Km</span></span>
5 0
1 year ago
Read 2 more answers
A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
Elan Coil [88]

Answer:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.  </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.  </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.  </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we use our previous value for the work:

W_{ff}=-F_f*(hy/sin\theta)   Solving for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

Vf^2=Vo^2+2*a*d

Solving for a:

a=-3.844m/s^2

Now, by dynamics:

|F_f|=|m*a|

|F_f|=207.58N

8 0
1 year ago
A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.
Anna007 [38]

Answer:

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

So here as per force equation in vertical direction we know that

N - mg = ma

so normal force on the block is given as

N = mg + ma

now the magnitude of kinetic friction on the block is given as

F_f = \mu N

F_f = \mu (mg + ma)

now when bag is sliding on the floor then net deceleration of the block due to friction is given as

a = - \frac{F_f}{m}

a = -\mu_k(g + a)

now we know that bag hits the opposite wall at L distance away in time t

so we have

d = v t + \frac{1}{2}at^2

L = vt - \frac{1}{2}(\mu_k)(g + a) t^2

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

8 0
1 year ago
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