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3 years ago

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On a map, a measurement of 1.0 cm represents a velocity of 100 m/s. If you were to draw a vector on the map that represents 80 m

/s, how long would the vector be?
A. 8.0 cm B. 1.25 cm C. 1.25 mm D. 8.0 mm

1 answer:

zimovet [89]3 years ago

5 0

Answer:

80 $\frac{m}{s}$ = 8 millimeter

Hence, option D is correct

Explanation:

1 centimeter on map represents 100 $\frac{m}{s}$

10 millimeter on map represents 100 $\frac{m}{s}$ .

So, 1 millimeter on map represents 10 $\frac{m}{s}$ .

10 $\frac{m}{s}$ = 1 millimeter

Thus,

80 $\frac{m}{s}$ = 8 millimeter

Hence, option D is correct

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At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?

notka56 [123]

The speed of car is 100.8km/h

$KE = \frac{1}{2} m(v \: truck ) {}^{2}$

$= 0.5 \times 18777 \times \frac{26}{3.6} \times \frac{26}{3.6}$

$= 489708.79j$

$= \frac{1}{2} \times 1248 \times( v \: car) {}^{2}$

$= 624v {}^{2}$

$so \: v {}^{2} = \frac{489708.7}{624}$

$= 784$

$v = \sqrt{784}$

$v = 28m/s$

v car= 28×3.6

=100.8km/h

Hence, the speed of the car is 100.8km/h

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2 years ago

a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde

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The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

$The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal$

The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

$Volume \ of \ a \ cone = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot h$

$Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}$

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

$\dfrac{dV}{dr} = \dfrac{d}{dr} \left(\dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0$

3·r²·s² - 4·r⁴ = 0

3·r²·s² = 4·r⁴

3·s² = 4·r²

$\underline{\left \right. The \ radius, \, r =\sqrt{ \dfrac{3}{4}} \cdot s}$

$\underline{The \ height, \, h =\sqrt{s^2 - \dfrac{3}{4}\cdot s^2} = \dfrac{s}{2}}}$

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3 years ago

The y-position of a damped oscillator as a function of time is shown in the figure.

NISA [10]

(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

<h3>at time, t = 0, y = 3.5</h3>

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

<h3>at time, t = 1 cm, y = - 3cm</h3>

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) = e^(-bx)

ln[-0.857 / cos(ω)] = -bx

ln[-0.857 / cos(ω)] / b = - x ---- (1)

<h3>at time, t = 2 cm, y = - 2cm</h3>

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx

ln[-0.57 / cos(2ω)] /2b = - x ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) = cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω - cosω - 6 = 0

let cosω = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω = arc cos(-0.92)

ω = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

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This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag

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Answer:

Explanation:

Given

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Vera_Pavlovna [14]

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