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hammer [34]
2 years ago
14

A toboggan and rider have a total mass of 100 kg and travel down along the (smooth) slope defined by the equation y=0.2x2. At th

e instant x=8 m, the toboggan's speed is 4 m/s. At this point, determine the rate of increase in speed and the normal force which the toboggan exerts on the slope. Neglect the size of the toboggan and rider for the calculation
Physics
1 answer:
creativ13 [48]2 years ago
7 0

Answer:

9.35 m/s^2

n = 292.24

Explanation:

y = 0.2×(x^2) , the equation that desribes the slope is parabolic, then;

the tangent line at a point can be found by differentiating the equation:

dy/dx = 0.4×x

then, at x = 8 m:

dy/dx = 0.4×8 = 3.2

that represents the slope of the tangent line

then to find the incline:

tan(∅) = 3.2

      ∅ = 72.65°

at the point, the toboggan is sitting at 72.65° on the incline.

since the surface is smooth, no friction.

the only force playing is the parallel component of gravity, then by Newton's second law:

F = m×a

a = F/m

  = m×g×sin(∅)/m

  = g×sin(∅)

  = (9.8)×(sin(72.65°))

  = 9.35 m/s^2

then the normal force is given by:

n = m×g×cos(∅) = (100)×(9.8)×(cos(72.65))

n = 292.24

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Aneli [31]

Answer: 0.392 m/s

Explanation:

The Doppler shift equation is:

f'=\frac{V+V_{o}}{V-V_{s}} f

Where:

f=8(10)^{4} Hz is the actual frequency of the sound wave

f'=8.002(10)^{4} Hz is the "observed" frequency

V=1570 m/s is the speed of sound

V_{o}=0 m/s is the velocity of the observer, which is stationary

V_{s} is the velocity of the source, which are the red blood cells

Isolating V_{s}:

V_{s}=\frac{V(f'-f)}{f'}

V_{s}=\frac{1570 m/s(8.002(10)^{4} Hz-8(10)^{4} Hz)}{8.002(10)^{4} Hz}

Finally:

V_{s}=0.392 m/s

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2 years ago
Sound waves with a constant frequency of 250 hertz are traveling through air at stp. what is the wavelength of the sound waves
Deffense [45]

Answer:

wave length is 1.2m

Explanation:

since formula of wave length is v/f

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f(frequency 250hertz)

then wave length is 300÷250 which give 1.2m

5 0
1 year ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

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vova2212 [387]
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SVETLANKA909090 [29]

Answer:

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