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vaieri [72.5K]
3 years ago
12

A cabbie is trying to stop when he notices a fare is whistling them over. The

Physics
1 answer:
liberstina [14]3 years ago
7 0
  • K.E=18750J
  • Mass=m=2100kg
  • Velocity=v

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2

\\ \sf\longmapsto 18750=1050v^2

\\ \sf\longmapsto v^2=\dfrac{18750}{1050}

\\ \sf\longmapsto v^2=17.85m^2

\\ \sf\longmapsto v=\sqrt{17.85}

\\ \sf\longmapsto v=4.1m/s

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Analyze the data to identify the mathematical relationship between the
koban [17]

Answer:

C.) Wave B has about 2.25 times more energy than wave A

Explanation:

5 0
2 years ago
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th
Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0
3 years ago
In a ________________ wave, such as a sound wave, the particles in the medium vibrate in the same direction that the wave travel
Varvara68 [4.7K]
The answer would be A, transverse.
6 0
3 years ago
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by
Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

v^{2}  = u^{2} + 2as\\u = 0 m/s\\v^{2}  =  2 * 0.731 * 5\\v^{2}  = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s

Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

7 0
3 years ago
Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out
blondinia [14]

Answer:

\theta = 20.98 degree

Explanation:

As we know that the speed of the sound is given as

v = 332 + 0.6 t

now at t = 273 k = 0 degree

v = 332 m/s

so we have

a sin\theta = N\lambda

a sin\theta = N(\frac{v_1}{f})

now when temperature is changed to 313 K we have

t = 313 - 273 = 40 degree

now we have

v = 332 + (0.6)(40)

v_2 = 356 m/s

a sin\theta' = N(\frac{v_2}{f})

now from two equations we have

\frac{sin19.5}{sin\theta} = \frac{332}{356}

so we have

sin\theta = 0.358

\theta = 20.98 degree

7 0
3 years ago
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