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stiv31 [10]
3 years ago
14

What type of plate boundary exists today along the himalayas?

Physics
1 answer:
lakkis [162]3 years ago
3 0
Continental convergent boundaries
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Does sound travels through your skull to your ear when you speak ? If yes then how ?
Maksim231197 [3]
Sound travels through waves, more specifically, through vibrations. They do not go from skull to ear, but they can go from ear to brain, or skull to brain. Ear to brain is simply vibrations traveling from outer ear, to inner ear, to the brain. Skull to brain, otherwise known as "bone conduction", has the vibrations hitting the skull, then to the temproal bone, then to the inner ear where the brain picks it up.
5 0
3 years ago
What did Blaise Pascal discover, and how did he impact Physics?
yan [13]

Answer:

Well it really depends because he found out about a lot of things in a lot of scientific fields actually. He is known for contributing Pascal's triangle and probability theory. He also invented an early digital calculator and a roulette machine. In the field of physics, Blaise contributed to the study of atmospheric pressure by discovering that vacuums are real and exist in the real world.

6 0
3 years ago
If the bat emits a sound at a frequency of 80.2 kHzkHz and hears it reflected at a frequency of 83.5 kHzkHz while traveling at a
solong [7]

To solve this problem we will apply the concepts related to the Doppler Effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other.

Mathematically it can be described as

v' = v (\frac{f' ( v - v_{bat}) - f ( v + v_{bat})}{f'(v-v_{bat}) + f ( v + v_{bat})})

Where,

v = Speed of sound

f' = Reflected frequency

f = Emitted Frequency

v_{bat} = Speed of the object

Assuming that the speed of the medium is the speed of sound of 343m / s, we will then have that:

v' = 343 (\frac{(83.5) (343 -3.9) - 80.2(343+3.9)}{83.5(343-3.9) + 80.2(343+4.3)})

v' = 3.0134m/s

8 0
4 years ago
The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob
Crazy boy [7]

Answer:

104.3 cm  or 179.7

Explanation:

First find time that it takes for the object to hit the ground

\sqrt{(2H)/g}  ->   \sqrt{(2 x 179)/ 9.8} = 6.04s\\*

Then find xf of projectile xf= 5.9(6.04) = 37.7\\\\

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

hope that helps

4 0
3 years ago
Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time accordin
Vera_Pavlovna [14]

Answer:

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

Explanation:

This is an angular kinematic exercise the equation for the angular position

the particle A

       θ = θ₀ + ω₀ t + ½ α t²

They say for the particle B

     w₀B = ½ w₀

     αB = 2 α

In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial

      t´ = t - t_1

l

et's write the equation of particle B

      θ = θ₀ + w₀B t´ + ½ αB t´2

replace

     θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²

     θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

4 0
3 years ago
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