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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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What is the pressure drop due to the bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

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Answer:

$\Delta P=1581357.92\ Pa$

Explanation:

Given:

diameter of hose pipe, $D=0.09\ m$

diameter of nozzle, $d=0.03\ m$

volume flow rate, $\dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}$

<u>Now, flow velocity in hose:</u>

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<u>Now, flow velocity in nozzle:</u>

$v_n=\frac{\dot V}{\pi.d^2\div 4}$

$v_n=\frac{0.04\times 4}{\pi\times 0.03^2}$

$v_n=56.5884\ m.s^{-1}$

We know the Bernoulli's equation:

$\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}+Z_2$

when the two points are at same height then the eq. becomes

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8 0

3 years ago

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Nana76 [90]

Answer:

Part a: The potential at point 3 and point 6 are V and 0 respectively.

Part b: The charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c: The net charge is 6CV.

Part d: The equivalent capacitance is 6C

Explanation:

As the question is not given here ,the complete question is found online and is attached herewith.

Part a:

V1 = V, V3 = V1 = V

and, V6 = V1-V = 0

The potential at point 3 and point 6 are V and 0 respectively

Part b

Q1 =CV = Q,

Q2 = 2C *V = 2Q

Q3 = 3 C*V = 3Q

So the charges Q1, Q2 and Q3 are CV, 2CV and 3CV respectively.

Part c

Total charge of the system,

Q_net =Q1+Q2+Q3= (1+2+3) CV = 6 CV

So the net charge is 6CV.

Part d:

Equivalent Capacitance = Net charge / Voltage

Eq. C = 6CV/V = 6C

So the equivalent capacitance is 6C

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3 years ago

The cylindrical tub of a dryer in a laundromat rotates counterclockwise about a horizontal axis at 41.5 rev/min as it dries the

frozen [14]

Answer:

$\theta = 49.81^0$

Explanation:

Given that:

$\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m$

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

where;

$v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77$

Again:

$\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0$

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3 years ago

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mestny [16]

Answer:

a. Δx = 2.59 cm

Explanation:

mb = 0.454 kg , mp = 5.9 x 10 ⁻² kg , vp = 8.97 m / s , k = 21.0 N / m

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mb * (0) + mp * vp = ( mb + mp ) * vf

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Solve to Δx '

Δx = √ ( mp² * vp² ) / ( k * ( mp + mb )

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Δx = 0.02599 m ⇒ 2.59 cm

8 0

3 years ago