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3 years ago

8

A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of t

he tuning fork?

1 answer:

Ierofanga [76]3 years ago

4 0

<u>We are given:</u>

The tuning fork vibrates at 15660 oscillations per minute

<u>Period of one back-and forth movement:</u>

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations <em>(1 minute = 60 seconds)</em>

<em>dividing the values</em>

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

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A 63-kg hiker is climbing the 828-m-tall Burj Khalifa in Dubai. If the efficiency of converting the energy content of the bars i

konstantin123 [22]

Answer:

$T_f=5854.76$ °C

Explanation:

Given:

mass of hiker, m= 63 kg

height to be climbed, h= 828 m

energy produced by an energy bar, $E= 1.10\times 10^6 J$

heat capacity of the hiker, $c=75.3 J.mol^{-1}.K^{-1}= 4.184 J.kg^{-1}.K^{-1}$

initial body temperature of hiker, $T_i=36.6 \degree C$

<em>The efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body.</em>

We find the energy required for climbing 828 m height:

W=m.g.h

$W=63\times 9.8\times 828$

W= 511207.2 J

∵Hike eats 2 energy bars= $2\times 1.1\times 10^{6} J$

Energy produced $= 2.2\times 10^{6} J$

Now, according to her efficiency:

Total energy required for producing the work of W= 511207.2 J which is required to climb the given height will be (say, E):

$25\% of E= 511207.2$

$\Rightarrow E= 511207.2\times \frac{100}{25}$

$E=2044828.8 J$

&

Amount of total energy (E) converted into heat(Q) is:

$Q=2044828.8-511207.2\\Q=1533621.6J$

As we know that:

heat, $Q=m.c. (T_f-T_i)$ .................(1)

where:

$T_f$ is the final temperature

Putting respective values in the eq. (1)

$1533621.6= 63\times 4.184\times (T_f-36.6)$

$(T_f-36.6)\approx 5818.16$

$T_f\approx 5854.76$ °C

4 0

3 years ago

In the sum A→+B→=C→, vector A→ has a magnitude of 13.6 m and is angled 40.2° counterclockwise from the +x direction, and vector

il63 [147K]

Answer:

$|B|=27.00425726m$

$\alpha =210.3781372$ °

Explanation:

Let's use the component method of vector addition:

$A_x=13.6cos(40.2)=10.38762599\\A_y=13.6sin(40.2)=8.778224553\\Cx=13.8cos(20.7+180)=-12.90912763\\Cy=13.8sin(20.7+180)=-4.877952844$

Now, we know:

$C_x=A_x+B_x\\\\C_y=A_y+b_y$

So:

$B_x=C_x-A_x=-23.29675362\\B_y=C_y-A_y=-13.6561774$

Now lets calculate the magnitude of the vector B:

$|B|=\sqrt{(B_x)^{2} +(B_y)^{2} }=27.00425726m$

Finally its angle is given by:

$\alpha =(arctan(\frac{B_y}{B_x}))+180=30.37813438+180=210.3781344$ °

Keep in mind that I added 180 to the angles of C and B to find the real angles measured from the + x axis counter-clock wise.

8 0

3 years ago

a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy(KE) when it jumps PLEASE HELP ME

Kay [80]

Answer:

0.91 J

Explanation:

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the lemming in this problem, when he jumps, we have:

m = 0.0780 kg is the mass

v = 4.84 m/s is the speed

Substituting into the equation, we find:

$K=\frac{1}{2}(0.0780)(4.84)^2=0.91 J$

8 0

3 years ago

2. Consider the scenario below. If the surface is frictionless, how much force does the 2 kg box exert on

vekshin1

Answer:

5 kg box

Explanation:

6 0

3 years ago

A motorboat traveling from one shore to the other at a rate of 5m/s east encounters a current flowing at a rate of 3.5m/s north

nikdorinn [45]

Answer:

Resultant velocity will be equal to 6.10 m/sec

Explanation:

We have given a motorbike is traveling with 5 m/sec in east

And a current is flowing at a rate of 3.5 m /sec in north

We know that east and north is perpendicular to each other

So resultant velocity will be vector sum of both velocity

So resultant velocity $v=\sqrt{5^2+3.5^2}=6.10m/sec$

So resultant velocity will be equal to 6.10 m/sec

6 0

4 years ago