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3 years ago

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The arrival time of an elevator in a 12-story dormitory is equally likely at any time range during the next 5.5 minutes. (a) Cal

culate the expected arrival time. (Round your answer to 2 decimal place.) (b) What is the probability that an elevator arrives in less than 2.6 minutes

1 answer:

Bumek [7]3 years ago

6 0

Answer:

(a). 2.8 minutes.

(b). 0.4732

Explanation:

Without mincing words, let's dive straight into the solution to the question.

So, for the part (a), the expected arrival time can be calculated as given below.

The distribution falls between the ranges of 0(lower boundary) to 5.5minutes(upper boundary).

Therefore, the expected time = (0 + 5.5)÷ 2 = 2.75 minutes = 2.8 minutes(to 2 decimal places).

(b). The probability that an elevator arrives in less than 2.6 minutes can be calculated as given below;

Recall: We have that the upper boundary = 0 and the lower boundary = 5.5 minutes for the distribution. Also, the upper limit is equal to 2.6 minutes and the lower limits = 0 minutes.

Therefore, 1/ (5.5 - 0) = 1/5.5 = 0.182.

Therefore, the probability that an elevator arrives in less than 2.6 minutes = 0.182 ( 2.6 - 0).

The probability that an elevator arrives in less than 2.6 minutes = ( 0.182 × 2.6).

The probability that an elevator arrives in less than 2.6 minutes = 0.4732.

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The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This sp

Delvig [45]

Answer:

Explanation:

Here is the full question and answer,

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.

Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

Answer

A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute $9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g and $60^\circ$ for $\theta$ in above equation.

$t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\$

Spitted water will travel $0.80{\rm{ m}}$ horizontally.

Displacement of water in this time period is

$x = vt\cos \theta$

Substitute $\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}$ for $t\rm 60^\circ[tex] for [tex]\theta$ and $0.80{\rm{ m}}$ for x in above equation.

$\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.

Calculate the height of insect above the surface.

Vertical component of the velocity is,

${v_v} = v\sin \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

Horizontal component of the velocity is,

${v_h} = v\cos \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

When horizontal $({0.60\;{\rm{m}}}$ distance away from the fish.

The time of flight for distance (d) is ,

$t = \frac{d}{{{v_h}}}$

Substitute $0.60\;{\rm{m}}$ for d and $1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_h}$ in equation $t = \frac{d}{{{v_h}}}$

$\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\$

Distance of the insect above the surface is,

$s = {v_v}t + \frac{1}{2}g{t^2}$

Substitute $2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_v}$ and $0.3987{\rm{ s}}$ for t and $- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g in above equation.

$\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\$

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3 years ago

Hello, I need help. Due right now

ryzh [129]

Answer: See below

Explanation:

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$m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2}=\left(m_{1}+m_{2}\right) \vec{v} \\1383(-11.2 {\math})+1732(31.3 {\math}) \\=(1383+1732) \vec{v} \\-15489.6 {\math}+54211.6 {\math}=3115 \vec{v} \\ \vec{v}=(17.4 {\math}-4.972 {\math}) \ \mathrm{m/s}$

$\text { So magnitude } \vec{|v|} =\sqrt{(17.4)^{2}+(-4.972)^{2}} \\ =\sqrt{327.605} \\ =18.099 \mathrm \ {m/s} \\$

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3 years ago

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ad-work [718]

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4 years ago

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.

snow_lady [41]

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is $v = 3.79 *10^{5} \ m/s$

Explanation:

From the question we are told that

The magnitude of the electric field is $E = 144 \ kV /m = 144*10^{3} \ V/m$

The magnetic field is $B = 0.38 \ T$

The force due to the electric field is mathematically represented as

$F_e = E * q$

and

The force due to the magnetic field is mathematically represented as

$F_b = q * v * B * sin(\theta )$

Now given that it is perpendicular , $\theta = 90$

=> $F_b = q * v * B * sin(90)$

=> $F_b = q * v * B$

Now given that it is not deflected it means that

$F_ e = F_b$

=> $q * E = q * v * B$

=> $v = \frac{E}{B }$

substituting values

$v = \frac{ 144 *10^{3}}{0.38 }$

$v = 3.79 *10^{5} \ m/s$

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