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Usimov [2.4K]
3 years ago
10

John is riding the bus to the next samuell football game. The bus is traveling to the right at 20 m/s and Jon is walking towards

the back (left) of the bus at 5 m/s. What is Jon’s relative velocity to a stationary observer?
Physics
1 answer:
Elenna [48]3 years ago
4 0

Answer:

Vrel_jon's = 15 [m/s] to the right

Explanation:

Relative velocity is defined as the relative motion between two bodies, taking into account the directions of motion.

Relative velocity is defined as the relative motion between two bodies, taking into account the directions of motion. The relative velocity is defined as the algebraic sum of the velocities, if the movements are opposite the vectors are subtracted, as will be done below.

Vrel = 20 - 5 = 15 [m/s]

A person watching Jon sees him moving to the right at a speed of 15 [m/s]

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
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Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

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3 years ago
) what is kinetic energy, and how does it differ from potential energy?
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Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen
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Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

\lambda = 436.1 \ nm

or,

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or,

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Distance between the 1st and 2nd dark fringes,

(y_2-y_1) = 6\times 10^{-3} \ m

As we know,

⇒ \frac{d}{L} (y_2-y_1) = \lambda

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By substituting the values, we get

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       =\frac{0.31\times 6\times 10^3}{436.1}

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       =4.26 \ m

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