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alina1380 [7]
3 years ago
5

Many hard-of-hearing people like sound compressed because they remain sensitive to ________ sounds. prolonged high-pitched frequ

ent unpredictable loud
Physics
1 answer:
pychu [463]3 years ago
7 0

Answer:

loud

Explanation:

Noise exposure results in a decrease in the sensitivity of auditory sensory cells thereby affecting the hearing ability of the concerned.

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Random kinetic energy possessed by objects in a material at finite temperature. An object that feels hot has a lot of this.
gregori [183]
Internal energy or thermal energy.
3 0
3 years ago
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A cricket can travel approximately 8 m/s. How many meters could a cricket travel in 75<br> s?
nika2105 [10]

Answer:

Your answer will be 600meters

7 0
2 years ago
A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff
Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0
1 year ago
Estimated speed of the vehicle
SSSSS [86.1K]

Really, Gundy ? ! ?

The formula for the car's speed is given and discussed in the box.  The formula is

v = √(2·g·μ·d)

Then they <em>tell</em> you that μ is 0.750 , and then they <em>tell</em> you that d = 52.9 m .  Also, everybody knows that 'g' is gravity = 9.8 m/s² .

They also tell us that the mass of the car is 1,000 kg, and they tell us that it took 3.8 seconds to skid to a stop.  But we already <em>have</em> all the numbers in the formula <em>without</em> knowing the car's mass or how long it took to stop.  The police don't need to weigh the car, and nobody was there to measure how long the car took to stop.  All they need is the length of the skid mark, which they can measure, and they'll know how fast the guy was going when he hit the brakes !

Now, can you take the numbers and plug them into the formula ? ! ?

v = √(2·g·μ·d)

v = √( 2 · 9.8 m/s² · 0.75 · 52.9 m)

v = √( 777.63 m²/s²)

v = 27.886 m/s

Rounded to 3 digits, that's  <em>27.9 m/s </em>.

That's about 62.4 mile/hour .



3 0
3 years ago
A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane
Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, \phi=8.6\times 10^{-3}\ T/m^2

The magnetic flux linked through the loop is :

\phi=B.A

\phi=BA\ cos\theta

Here, \theta=0

B=\dfrac{\phi}{A}

or

B=\dfrac{\phi}{\pi r^2}

B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0
3 years ago
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