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Vlad1618 [11]
4 years ago
5

A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a consta

nt force (in contrast to the constant power in Part A). If such a sports car went from zero to 31.0 mph in time 1.50 s , how long would it take to go from zero to 62.0 mph
Physics
1 answer:
Artist 52 [7]4 years ago
8 0

Answer:

3 seconds

Explanation:

To solve this problem, we have to find the acceleration of the car.

Acceleration is given as:

a = (v - u) / t

Where v = final velocity

u = initial velocity

t = time taken.

From the question,

u = 0 m/s

v = 31 mph = 13.86 m/s

t = 1.5 seconds

Acceleration, a, will be:

a (13.86 - 0) / 1.5

a = 13.86 / 1.5

a = 9.24 m/s²

We are told that this kind of car operates with constant force. This means it operates with constant acceleration (since force and acceleration are directly proportional)

Therefore, when the final velocity of the car is 62 mph (27.72 m/s), time taken is:

a = (v - u) / t

=> t = (v - u) / a

t = (27.72 - 0) / 9.24

t = 3 seconds

It will take the car 5 seconds to go from zero to 62 mph.

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Match the description to the property.
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Light of wavelength 660 nm passes through two narrow slits 0.65 mm apart. The screen is 2.55 m away. A second source of unknown
stiv31 [10]

Answer:

Explanation:

Distance of 2 nd order fringe  x₁ = 2 λ D/d [ λ is wave length , D is distance of screen , d is slit distance.

x₁ = 2 x 660 x10⁻⁹x 2.55/.65 x 10⁻³.= 5.17846 x 10⁻³ m.

Distance of fringe for 2nd radiation = 5.178 x 10⁻³ -1.17 x 10⁻³ = 4.008 x10⁻³

4.008 x 10⁻³ = 2 x λ x 2.55/.65 x 10⁻³

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A star's transverse velocity depends on which two factors?
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3 years ago
Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acti
Tems11 [23]

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

\dfrac{dF_{net}}{df}=0

We need to calculate the unknown force

Using formula of net force

\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}

Put the value into the formula

\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}

\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}

The magnitude of net force,

F_{net}=\sqrt{F_{x}^2+F_{y}^2}

F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}

F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}

F_{net}=\sqrt{F^2+4899.78+36.232F}

On differentiating w.r.to F

(\dfrac{dF_{net}}{dF})^2=2F+36.232

0=2F+36.232

F=-\dfrac{36.232}{2}

F=-18.116\ lb

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.

6 0
3 years ago
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