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zzz [600]
2 years ago
12

A physics student with a stopwatch drops a rock into a very deep well, and measures the time between when he drops the rocks and

when he hears the sound of the rock hitting the water below. If the speed of sound is 343 m/s, and the student measures a time of 6.20 s, how deep is the well?
Physics
1 answer:
tankabanditka [31]2 years ago
6 0

Answer:

h= 161.06 m

Explanation:

Given that

Speed of the sound ,C= 343 m/s

Total time ,t= 6.2 s

lets take the depth of the well = h

The time taken by stone before striking the water = t₁

we know that

h=\dfrac{1}{2}gt_1^2

t_1=\sqrt{\dfrac{2h}{g}}

The time taken by sound =t₂

t_2=\dfrac{h}{343}

The total time

t = t₁+ t₂

6.2 = \sqrt{\dfrac{2h}{g}}+\dfrac{h}{343}

6.2 = \sqrt{\dfrac{2h}{9.81}}+\dfrac{h}{343}

Now by solving the above equation we get

h= 161.06 m

Therefore the depth of the well will be 161.06 m.

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Answer:

Explanation:

This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is

V_y=Vsin\theta which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):

V_y=12.2sin(81.5) and get

V_y=12.1 m

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2 years ago
Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a
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To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

\omega = 1.25 rad/s \rightarrow The angular speed

\alpha = 0.745 rad/s2 \rightarrow The angular acceleration

r = 4.65 m \rightarrow The distance

The relation between the linear velocity and angular velocity is

v = r\omega

Where,

r = Radius

\omega = Angular velocity

At the same time we have that the centripetal acceleration is

a_c = \frac{v^2}{r}

a_c = \frac{(r\omega)^2}{r}

a_c = \frac{r^2\omega^2}{r}

a_c = r \omega^2

a_c = (4.65 )(1.25 rad/s)^2

a_c = 7.265625 m/s^2

Now the tangential acceleration is given as,

a_t = \alpha r

Here,

\alpha = Angular acceleration

r = Radius

\alpha = (0.745)(4.65)

\alpha = 3.46425 m/s^2

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

|a| = \sqrt{a_c^2+a_t^2}

|a| = \sqrt{(7.265625)^2+(3.46425)^2}

|a| = 8.049 m/s^2 \approx 8.05 m/s2

Therefore the correct answer is C.

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2 years ago
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser
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Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

V_L=V_oe^{-\frac{Rt}{L}}

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s

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A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir
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Answer:

The distance and height of the object  is 6 m and 2 m.

The image is virtual and upright.

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Given that,

Focal length = 0.25 m

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\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}

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Put the value into the formula

m=-\dfrac{0.24}{-6}

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Using formula of magnification

m=\dfrac{h'}{h}

h=\dfrac{0.080}{0.04}

h=2\ m

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Hence, The distance and height of the object  is 6 m and 2 m.

The image is virtual and upright.

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