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mamaluj [8]
3 years ago
5

checking your understanding: which of the following would pose the mist immediate danger to your hearing???? a) a singing trio t

hst couldn't stay in tune,,,b) a supersonic jet flying over your house,,,,c) sitting in thr third row of a loud rock concert...,,,d) operatinf a vacuum cleaner for 30 minutes a day....which are the correctly answer...
Physics
1 answer:
professor190 [17]3 years ago
7 0
I believe the answer is C.

Concerts are blaring music which is extremely unhealthy for your ears. Not to mention is a rock concert where loud sounds and screaming are common. Then on top of that you are sitting in the third row. This would be an extreme hazard to your ears and probably the worst out of these 4 choices.

Thanks for using brainly :)

I hope this helped. Have a great night!
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Serggg [28]

Answer:

Part a)

t = 16.8 s

d = 100.8 m

Part b)

v_f = 2.86 m/s

Explanation:

Part a)

Constant speed by which the student will run is given as

v = 5 m/s

now after some time if student is going to overtake the position of bus

so here the final positions will be same

so we have

x_{bus} = x_{student}

0 + \frac{1}{2}at^2 + d = v_{student} t

\frac{1}{2}(0.170)t^2 + 60 = 5 t

0.085 t^2 - 5t + 60 = 0

so it is

t = 16.8 s

So student will run the total distance

d = vt

d = (6)(16.8)

d = 100.8 m

Part b)

Speed of bus when student reach the bus is given as

v_f = v_i + at

v_f = 0 + (0.170)(16.8)

v_f = 2.86 m/s

4 0
3 years ago
A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc
Nikitich [7]

Answer:

x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

Explanation:

From the law of conservation of energy

Energy lost  by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x

x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}

x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

The required distance from A to B is x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

5 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0
3 years ago
A small electric motor produces a force of 5 N that moves a remote-control car 5 m every second. How much power does the motor p
kati45 [8]

Answer:

Explanation:

Given:

Force, f = 5 N

Velocity, v = 5 m/s

Power, p = energy/time

Energy = mass × acceleration × distance

Poer, p = force × velocity

= 5 × 5

= 25 W.

Note 1 watt = 0.00134 horsepower

But 25 watt,

0.00134 hp/1 watt × 25 watt

= 0.0335 hp.

4 0
3 years ago
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Answer:

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