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3 years ago

5

Tom Cruise jumped from one building to other building while filming the roof chase scene in Mission: Impossible - Fallout. He di

d not land on the roof of the other building safely and broke his ankle. If we assume the heights of the two building are the same and the distance between the buildings is 5 m, what is the minimum speed to land on the roof of the other building? Assume the jumping angle is 15 degrees and the air friction is negligible. (15 points) Vo =? O = 15° D = 5 m

1 answer:

OLga [1]3 years ago

3 0

Answer:

$v_0=9.9\ m.s^{-1}$

Explanation:

Given:

angle of launch of projectile from horizontal, $\theta=15^{\circ}$

range of projectile, $R=5\ m$

<u>We have formula for the range of projectile:</u>

$R=\frac{v_0^2\times sin\ 2\theta}{g}$

putting the respective values

$5=\frac{v_0^2\times sin\ 30^{\circ}}{9.8}$

$v_0=9.9\ m.s^{-1}$ is the velocity with which Tom should jump to land on the other roof.

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1. The distance between a trough and a crest on a transverse wave is 12.0cm. If cycles of the wave pass a fixed point in one sec

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Answer:

0.12m/s

Explanation:

v=λf

Given that, λ = 12cm = 0.12m

T = 1second

(A period T is the time required for one complete cycle of vibration to pass a given point)

frequency 'f' is unknown but we can get frequency from f = 1/T = 1/1 = 1Hz

therefore, v= 0.12 × 1 = 0.12m/s

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2 years ago

What region of the electromagnetic spectrum contains wavelengths the size of an atom? (10^-10 meters) What frequency does this w

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If a photon has a wavelength of about 10^-10 meters, it would be certainly in the X rays region and it would have a frequency of 2.99792458x10^18 Hz, that you can easily calculate using the relation that states that the product of the wavelength and the frequency of an electromagnetic wave aka light is always equal to 299,792,458 m/s which is the exact value of the speed of light!

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4 years ago

A bowling ball weighing 71.2 N (16.0 lb) is attached to the ceiling by a 3.80-m rope. The ball is pulled to one side and release

Elina [12.6K]

Answer:

a) For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

$a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2$

b) For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

$T= mg +ma$

In order to find the mass we know that $W=mg$ and solving for m we got:

$m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg$

And replacing into the tension we got:

$T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N$

Explanation:

For this case we have the following data given:

$W= 71.2 N$ represent the weigth for the object

$R=3.8 m$ the length of the rope or the radius

$v= 4.2 m/s$ represent the velocity of the bowling ball

Part a

For this case we want to find the acceleration of the bowling ball, and we now that the only acceleration for this case would be the centrifugal acceleration becuase since the pendulum is in phase with the equilibrium point the tangential acceleration is 0, and if we find the centrifugal acceleration we got:

$a= \frac{v^2}{R}= \frac{(4.2 m/s)^2}{3.8 m}=4.64 m/s^2$

Part b

For this case the tendsion is an opposite force against the weight and the centrifugal force acting in the ball so then we can find the tension like this:

$T= mg +ma$

In order to find the mass we know that $W=mg$ and solving for m we got:

$m =\frac{W}{g}=\frac{71.2 N}{9.8 m/s^2}=7.265 kg$

And replacing into the tension we got:

$T= m(g+a) =7.265 kg *(9.8 m/s^2 +4.64 m/s^2)=104.907 N$

7 0

3 years ago

On a cello, the string with the largest linear density (1.56 *10-2 kg/m) is the C string. This string produces afundamental freq

Mrrafil [7]

1) Wavelength of the wave: 1.6 m

2) Speed of the wave: 104.6 m/s

3) Tension in the string: 170.7 N

Explanation:

1)

For the standing waves on a string, the wavelength of the wave is related to the length of the string by

$\lambda = 2 L$

where

$\lambda$ is the wavelength

L is the length of the string

For the string in this problem.

L = 0.8 m is its length (I assume there is a mistake in the text, since 0.08 m is not a realistic value for the length of the string)

Therefore, the wavelength of the wave on the string is

$\lambda=2(0.8)=1.6 m$

2)

The speed of a wave is calculated through the wave equation:

$v=f\lambda$

where

f is the frequency

$\lambda$ is the wavelength

For the standing wave on this string, the fundamental frequency is

$f=65.4 Hz$

while the wavelength is

$\lambda=1.6 m$

Therefore, the speed of the wave is

$v=(65.4)(1.6)=104.6 m/s$

3)

The speed of the wave is related to the tension in the string by

$v=\sqrt{\frac{T}{\mu}}$

where

v is the speed

T is the tension

$\mu$ is the linear density of the string

For this string,

v = 104.6 m/s

$\mu=1.56\cdot 10^{-2} kg/m$

Therefore, the tension in the string is

$T=\mu v^2 = (1.56\cdot 10^{-2})(104.6)^2=170.7 N$

Learn more about waves:

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3 years ago

The uniform beam is supported by two rods AB and CD that have cross-sectional areas of 10 mm 2 and 15 mm, 2 respectively. Determ

Ilya [14]

Answer:

2 N / m

Explanation:

It is necessary to determine the maximum w so that the normal stress in AB and CD rods does not exceed the allowable stress.

The cross-section of respective rods are:

A_ab = 10 mm^2

A_cd = 15 mm^2

Step 1 :The maximum load is determined from the condition that the normal stress (б)

is not higher than (б_allow).

б_ab = F_ab / A_ab =< б_allow

б_cd = F_cd / A_cd =< б_allow

Step 2 : Compute the static size

We apply the Equilibrium conditions:

1) Sum of Moments (A) = 0

-w * 6 * 0.5 * 6 * 0.75 + 6*F_cd = 0

F_cd = (13.5 / 6) * w = 2.25*w KN

2) Sum of Forces (y-direction) = 0

F_cd + F_ab - 0.5*6*w = 0

F_ab = 3*w - 2.25*w = 0.75*w KN

Step 3 : Compute Load w

1) Sector AB :

F_ab / A_ab =< б_allow

0.75*w / 10 * 10^-6 =< 300KPa

0.75*w =< 3

w =< 4 N / m

2) Sector CD :

F_cd / A_cd =< б_allow

2.25*w / 15 * 10^-6 =< 300KPa

2.25*w =< 4.5

w =< 2 N / m

Answer: w = 2 N / m

6 0

4 years ago