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2 years ago

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A Honda Civic and an 18 wheeler approach a right angle intersection and then collide. After the collision, they become interlock

ed. If their mass ratios were 1: 4 and their respective speeds as they approached were both 13 m/s, find the magnitude and direction of the final velocity of the wreck. (Please do not worry no one was harmed in making of this question) A. 16.3 m/s at 79° B. 10.7 m/s at 79° C. 12.5 m/s at 59° D. 15.7 m/s at 59°

1 answer:

Goryan [66]2 years ago

8 0

Answer:

The concept of conservation of momentum is applied in the particular case of collisions.

The general equation ig given by,

$M_1V_1 + M_2V_2 = (M_1+M_2) * V_f,$

Where,

$M_2 = 4 M_1$

The crash occurs at an intersection so we must separate the two speeds by their respective vector: x, y.

In the case of the X axis, we have that the body $M_2$ has a speed = 0, this because it is not the direction in which it travels, therefore

$M_1* 13 = (M_1+M_2) * V_{fx} \\M_1*13 = (M_1+4M_1)*V_{fx}\\M_1*13=5M_1*V_{fx}\\Vx = \frac{13}{5}m/s$

The same analysis must be given for the particular case in the Y direction, where the mass body $M_1$ does not act with its velocity here, therefore:

$M_2* 13 = (M_1+M_2) * V_{fy},\\4*M_1* 13 = 5Ma * V_{fy} ,\\V_{fy} = \frac{52}{5}m/s ,$

We have the two components of a velocity vector given by $V_f = \frac{13}{5}\hat{i} + \frac{52}{5}\hat{j}$

Get the magnitude,

$V_f = \sqrt{(\frac{13}{5})^2+(\frac{52}{5})^2}$

$V_f = 10.72 m/s$

With a direction given by

$Tan^{-1} \frac{4}{1} = 75.96 \°$

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A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an

Whitepunk [10]

Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

F = Force exerted by empty rack = Weight of Empty Rack = 40 lb

A = Base Area = 452.4 in²

Therefore,

P₁ = 40 lb/452.4 in²

P₁ = 0.088 psi

Now, we calculate the pressure exerted by the rack along with the coat.

P₂ = F/A

where,

P₂ = Pressure exerted by rack filled with coats= ?

F = Force exerted by filled rack = Weight of Filled Rack = 65 lb

A = Base Area = 452.4 in²

Therefore,

P₂ = 65 lb/452.4 in²

P₂ = 0.144 psi

Now, the difference between both pressures is:

ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

<u>ΔP = 0.056 psi</u>

8 0

2 years ago

A girl weighing 200 newtons hangs from three pulley systems.

a_sh-v [17]

Most likely the answer is b

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2 years ago

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A straight wire carries a current of 3 A which is in the plane of this page, pointed toward the top of the page. A particle of c

Ilia_Sergeevich [38]

Answer:

The magnitude of the magnetic force exerted on the moving charge by the current in the wire is 2.18 x $10^{-8}$ N

The direction of the magnetic force exerted on the moving charge by the current in the wire is radially inward

Explanation:

given information:

current, I = 3 A

$q_{0}$ = +6.5 x $10^{-6}$ C

r = 0.05 m

v = 280 m/s

and direction of the magnetic force exerted on the moving charge by the current in the wire, we can use the following formula:

F = qvB sin θ

where

F = magnetic force (N)

q = electric charge (C)

v = velocity (m/s)

θ = the angle between the velocity and magnetic field

to find B we use

B = μ $_{0}$ I/2πr

μ $_{0}$ = 4π x $10^{-7}$ or 1.26 x $10^{-6}$ N/ $A^{2}$ , thus

B = 4π x $10^{-7}$ x 3 / 2π(0.05)

= 1.2 x $10^{-5}$ T

Now, we can calculate the magnitude force

F = qvB sin θ

θ = 90°, because the speed and magnetic are perpendicular

F = 6.5 x $10^{-6}$ x 280 x 1.2 x $10^{-5}$ sin 90°

= 2.18 x $10^{-8}$ N

Using the hand law, the magnetic direction is radially inward

8 0

2 years ago

If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee

Alexus [3.1K]

Answer:

457.81 Hz

Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz

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2 years ago

A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C

Sveta_85 [38]

Answer:

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190 $\frac{J}{kg k}$

The heat of vaporization is = 2256 $\frac{KJ}{kg}$

Mass = 0.030 kg

Energy needed to heat water to boiling point = $m c ( T_{2} - T_{1} )$

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy ( $E_{1}$ ) = 10.08 KJ

Energy needed to convert liquid water into vapor at the boiling point

$E_{2}$ = 0.030 × 2256 = 67.68 KJ

Thus the total energy needed E = $E_{1}$ + $E_{2}$

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

9 0

2 years ago