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FrozenT [24]
3 years ago
5

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 28.0 m/s. With what maximum sp

eed can it go around a curve having a radius of 73.0 m?
Physics
1 answer:
Dima020 [189]3 years ago
5 0

Given Information:  

Radius = r₁ = 150 m

Radius = r₂ = 73 m

Maximum speed = v₁ = 28.0 m/s

Required Information:  

Maximum speed = v₂ = ?

Answer:

Maximum speed v₂ = 19.53 m/s

Explanation:

The maximum acceleration of the truck is

a = v²₁/r₁

a = 28²/150

a = 5.226 m/s²

Now if we change the radius to r₂ = 73 m

Then the maximum speed of the truck is

v₂ = √a*r₂

v₂ = √5.226*73

v₂ = 19.53 m/s

So the speed of the truck is decreased when the radius of curve is decreased.

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natta225 [31]

Answer:

around 23 miles per hour

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3 years ago
A turtle ambles leisurely, as turtles tend to do, when it moves from a location with position vector 1,=1.91 m and 1,=−2.73 m in
Elena-2011 [213]

Answer:

Components: 0.0057, -0.0068. Magnitude: 0.0089 m/s

Explanation:

The displacement in the x-direction is:

d_x = 3.65-1.91=1.74 m

While the displacement in the y-direction is:

d_y = -4.79 -(-2.73)=-2.06 m

The time taken is t = 304 s.

So the components of the average velocity are:

v_x = \frac{d_x}{t}=\frac{1.74}{304}=0.0057 m/s

v_y = \frac{d_y}{t}=\frac{-2.06}{304}=-0.0068 m/s

And the magnitude of the average velocity is

v=\sqrt{v_x^2+v_y^2}=\sqrt{(0.0057)^2+(-0.0068)^2}=0.0089 m/s

8 0
3 years ago
The energy of an electromagnetic wave is related to its amplitude. <br> a. True<br> b. False
barxatty [35]
The answer to your question is True
6 0
3 years ago
Read 2 more answers
Question 8
viktelen [127]

Answer: D(t) = 8.e^{-0.4t}.cos(\frac{\pi }{6}.t )

Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:

y = a.sin(\omega.t) or y = a.cos(\omega.t)

where:

|a| is initil displacement

\frac{2.\pi}{\omega} is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

y=a.e^{-ct}.cos(\omega.t) or y=a.e^{-ct}.sin(\omega.t)

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

y=a.e^{-ct}.cos(\omega.t)

period = \frac{2.\pi}{\omega}

12 = \frac{2.\pi}{\omega}

ω = \frac{\pi}{6}

Replacing values:

D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)

The equation of displacement, D(t), of a spring with damping factor is D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t).

3 0
3 years ago
A suspension bridge with weight uniformly distributed along its length has twin towers that extend 65 meters above the road surf
Fynjy0 [20]

Answer:

16.25 m

Explanation:

we know that the equation pf parabola

y=kx^2

from bellow figure the coordinate of parabola is (600,65) that is y=600 and x=65

putting the the value of y and x in the equation of parabola

65=k600^2

k=0.0001805

now the equation is

y=0.0001805x^2

we have to find the value of y at x=300m

so y=0.0001805\times 300^2

y=16.25 m

4 0
3 years ago
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