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2 years ago

12

A bowling ball with a mass of 7.0 kg, generates 56.0 kgᐧm/s units of momentum. What

1 answer:

Misha Larkins [42]2 years ago

7 0

Answer:

8 m/s

Explanation:

$p = mv \\ 56 = (7.0)(v) \\ v = 8.0 \: m {s}^{ - 1}$

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Plants release water vapor into the atmosphere through the process called _____.

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The process is called transpiration <span />

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The fastest possible conduction velocity of action potentials is observed in ________.

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Thick, myelinated neurons

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3 years ago

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

$\frac{dx}{dt} = 4ft/s$

We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

$\frac{x}{6} tan\theta$

$x = 6tan\theta$

Differentiating each side with respect to t, we get,

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}$

$\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}$

Replacing the value of the velocity

$\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2$

$\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta$

The value of $cos \theta$ could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

$cos\theta = \frac{6}{10}$

$\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2$

$\frac{d\theta}{dt} = \frac{24}{25}$

Search light is rotating at a rate of 0.96rad/s

4 0

2 years ago

A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is

lys-0071 [83]

Answer:

$\omega_f=571.42\ rpm$

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, $a=1.5\ m/s^2$

Displacement, d = 1.3 m

The angular acceleration is given by :

$\alpha =\dfrac{a}{r}$

$\alpha =\dfrac{1.5}{0.033}$

$\alpha =45.46\ rad/s^2$

The angular displacement is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{1.3}{0.033}$

$\theta=39.39\ rad$

Using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

Here, $\omega_i=0$

$\omega_f=\sqrt{2\alpha \theta}$

$\omega_f=\sqrt{2\times 45.46\times 39.39}$

$\omega_f=59.84\ rad/s$

Since, 1 rad/s = 9.54 rpm

So,

$\omega_f=571.42\ rpm$

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0

3 years ago

A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun

cluponka [151]

Answer:Highest frequency =618.89Hz

Lowest frequency=582.22Hz

Explanation:

The linear velocity of a sound generator is related to angular velocity and is given as

Vs = rω where

r = the radius of circular path = 1.0 m

ω is the angular velocity of the sound generator. = 100 rpm

1 rev/min = 0.10472 rad/s

100rpm =10.472 rad/ s

Vs = rω

= 1m x 10.472rad/ s= 10.472m/s

A) Highest frequency heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,

f max = (v/ v-vs) fs

Where , v is the speed of the sound in air at 20 degrees celcius =

343 metres per second

vs is the linear velocity of the sound generator=10.472m/s

fs is the frequency of the sound generator= 600 Hz

f max = (343/ 343 - 10.472) x 600

=343/332.528) x600

=618.89Hz

B) Lowest frequency heard by a student in the classroom = Minimum frequency

f min = (v/ v+vs) fs

(343/ 343 + 10.472) x 600

=343/353.472) x 600

=582.22hz

5 0

3 years ago