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ElenaW [278]
3 years ago
14

A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this b

y launching the ball at the relatively low angle of 33.1 ° above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.
Physics
1 answer:
pychu [463]3 years ago
7 0

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle=33.1^{\circ}

Let u be the initial speed of ball

Range=\frac{u^2\sin 2\theta }{g}

167=\frac{u^2\sin (66.2)}{9.8}

u^2=1788.71

u=\sqrt{1788.71}

u=42.29 m/s

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An alert driver can apply the brakes fully in about 0.5 seconds. How far would the car travel if it
dybincka [34]

Answer:

The car would travel after applying brakes is, d = 14.53 m

Explanation:

Given that,

The time taken to apply brakes fully is, t = 0.5 s

The velocity of the car, v = 29.06 m/s

The distance traveled by the car in 0.5 s, d = ?

The relation between the velocity, displacement, and time is given by the formula                

                                d = v x t    m

Substituting the values in the above equation,

                                  d = 29.06 m/s x 0.5 s

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Therefore, the car would travel after applying brakes is, d = 14.53 m

8 0
4 years ago
A diffraction grating with 750 slits per mm is illuminated by light which gives a first-order diffraction angle of 34.0°. What i
lesya [120]
<h2>Answer: 745.59 nm</h2>

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

Where:

d is the width of the slit

\lambda is the wavelength of the light  

n is an integer different from zero

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda (2)

We know:

\theta_{1}=34\°

In addition we are told the diffraction grating has 750 slits per mm, this means:

d=\frac{1mm}{750}

Solving (2) with the known values we will find \lambda:

\lambda=(\frac{1mm}{750})sin(34\°) (3)

\lambda=0.00074559mm (4)

Knowing 1mm=10^{6}nm:

\lambda=745.59nm  >>>This is the wavelength of the light, wich corresponds to red.

6 0
4 years ago
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