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Ira Lisetskai [31]

3 years ago

13

Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 30 mph. at the same tim

e, another car is 1/2 mile east of the intersection, driving east (away from the intersection) at an unknown speed. the officer's radar gun indicates 15 mph when pointed at the other car (that is, the straight-line distance between the officer and the other car is increasing at a rate of 15 mph). what is the speed of the other car?

2 answers:

IgorLugansk [536]3 years ago

6 0

Answer:

$v = 51.2 mph$

Explanation:

Velocity of police car

$v_p = 30 mph$ towards North

velocity of the car with respect to police car

$v_{cp} = 15 mph$

Speed of the other car is given with respect to police car so let say the speed of other car is "v"

Now the component of this speed along the line joining two cars is

$v_1 = v cos45$

similarly speed of police car along the line joining two cars is given as

$v_2 = v sin45$

now we can find the relative speed

$v_{12} = v_1 - v_2$

$15 mph = vcos45 - 30cos45$

$v = 51.2 mph$

aleksandr82 [10.1K]3 years ago

3 0

In triangle ABC , using Pythagorean theorem

BC = sqrt(AB² + AC²)

r = sqrt(y² + x²) eq-1

taking derivative both side relative to "t"

dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))

dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))

dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)

15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)

v₂ = 51.2 m/s

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When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

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4 years ago

Explain how all three of Newton’s Laws of Motion apply to weightlifting.

Lubov Fominskaja [6]

Answer:

Newton's first law, an object will stay at rest only if there are no external forces acting on the object or if the net external force acting on the object is zero. Since the barbell is at rest (not moving), the net external force acting on it must be zero

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4 years ago

Which is younger, the rock layers or the fault?

jeka94

<h2>FAULT</h2>

The principle of cross-cutting relationships states that a fault or intrusion is younger than the rocks that it cuts. The fault labeled 'E' cuts through all three sedimentary rock layers (A, B, and C) and also cuts through the intrusion (D). So the fault must be the youngest formation that is seen.

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6 0

3 years ago

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A current of 0.96 amps flows through a copper wire 0.44 mm in diameter when connected to a potential difference of 15 v. how lon

FinnZ [79.3K]

By using Ohm's law, we can calculate the resistance of the wire. Ohm's law states that:
$V=IR$
where V is the potential difference across the conductor, I is the current and R the resistance. Rearranging the equation, we get
$R= \frac{V}{I}= \frac{15 V}{0.96 A}=15.6 \Omega$

Now we can use the following equation to calculate the length of the wire:
$R= \frac{\rho L}{A}$ (1)
where
$\rho$ is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area
In this problem, we have a wire of copper, with resistivity $\rho=1.68 \cdot 10^{-8} \Omega m$ . The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{0.44 mm}{2}=0.22 mm=0.22 \cdot 10^{-3} m$
And the cross-sectional area is
$A=\pi r^2=\pi (0.22 \cdot 10^{-3}m)^2=1.52 \cdot 10^{-7} m^2$

So now we can rearrange eq.(1) to calculate the length of the wire:
$L= \frac{RA}{\rho}= \frac{(15.6 \Omega)(1.52 \cdot 10^{-7} m^2)}{1.68 \cdot 10^{-8} \Omega m}=141.1 m$

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3 years ago

A specimen of aluminum having a rectangular cross section 9.5 mm × 12.9 mm (0.3740 in. × 0.5079 in.) is pulled in tension with 3

ryzh [129]

Answer:

The resultant strain in the aluminum specimen is $4.14 \times {10^{ - 3}}$

Explanation:

Given that,

Dimension of specimen of aluminium, 9.5 mm × 12.9 mm

Area of cross section of aluminium specimen,

$A=9.5\times 12.9=123.84\times 10^{-6}\ mm^2A=122.55\times 10^{-6}\ m^2$

Tension acting on object, T = 35000 N

The elastic modulus for aluminum is, $E=69\ GPa=69\times 10^9\ Pa$

The stress acting on material is proportional to the strain. Its formula is given by :

$\epsilon=\dfrac{\sigma}{E}$

$\sigma$ is the stress

$\epsilon=\dfrac{F}{EA}\\\epsilon=\dfrac{35000}{69\times 10^9\times 122.5\times 10^{-6}}\\\epsilon=4.14\times 10^{-3}$

Thus, The resultant strain in the aluminum specimen is $4.14 \times {10^{ - 3}}$

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4 years ago

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