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tangare [24]
3 years ago
7

A 30° incline permanently sits on a 1.1 meter high table. Starting from rest a ball rolls off the incline with a velocity of 2m/

s.
a. Calculate the length of the incline
b. calculate the horizontal and vertical components of the velocity at the end of the incline.​
Physics
1 answer:
exis [7]3 years ago
3 0
It would be a good game for you but if I get a pic I don’t want you can you come to my crib I just
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Interference is an example of which aspect of electromagnetic radiation
Phantasy [73]

Answer:

We experience interference while listening to the radio. A radio station works by sending and receiving radio waves. Since the radio waves are being interfered with other waves which must have a wave nature.

The interference is the net result of two individual waves. It can be constructive or destructive interference and is the property of waves and not particles.

This interference is an example of electromagnetic radiation. Thus we experience wave behavior of electromagnetic radiation in our daily communications.

7 0
3 years ago
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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(
Natasha2012 [34]

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

5 0
3 years ago
6^5/2^3= what is the answer please​
kogti [31]

Answer:

972

Explanation:

6^5/2^3

=7776/8

=972

6 0
3 years ago
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Determine the net work a hiker must do on a 3.35-kg backpack to carry it up a
Sphinxa [80]

Answer:

410.4J

Explanation:

Step one:

given

mass= 3.35kg

weight= 3.35*9.81= 32.86N

h=12.49m

Required

The net work done

Step two:

the work done is given  as

WD= force* distance

WD= 32.86*12.49

WD= 410.4J

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