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3 years ago

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In a shuffleboard game, the puck slides a total of 12 m before coming to rest. If the coefficient ofkinetic friction between the

puck and the horizontal board is 0.28, what was the initial speed of thepuck?

1 answer:

Artist 52 [7]3 years ago

4 0

Answer:

puck decelerates due to the kinetic frictional force μk mg

Explanation:

given data

total distance = 12 m

coefficient of kinetic friction = 0.28

solution

we will apply equation of motion that is

v² - u² = 2 × a × s ................1

we know acceleration will be

a = $\frac{-u^2}{2\times S}$

Then we have

Force = mass × acceleration .................2

m × $\frac{-u^2}{2\times S}$ = -μk mg

The puck decelerates due to the kinetic frictional force μk mg

and frictional force is negative as it opposes the motion.

so we get initial velocity of the puck which is strike.

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A small office experiences frequent power outages. Users at the location reports that after rebooting their Windows workstations

Katyanochek1 [597]

Answer:

Explanation:

Usually office's networks works using the TCP/IP protocol, this mean that each workstation has an assigned IP. This IP are the directions gave to each computer on the network,

This IP are usually asigned by a DHCP Server. Servers every time a new device connect to the Network, assigns to it an IP, this way every Computer will know how to send or request information from that Computer.

In the TCP/IP protocol is not posibble for two computer to have the same IP. This causes what it call IP Conflict, IP Conflict make imposible the comunications between those devices and the Network.

On Power outages all the Computer are turn off, because of this they disconnect from the Network and need to requests IP one more time to gain comunication,

We this in mind, we can predict that frequent power outages can cause for all the Computer in the Network to request new IPs very frequently this will cause IP Conflict in several units, disconnecting them from the network and making them no longer able to find files on the network. To fix this, what can be done it to establish Static IP to every workstation, this way we they request a new IP, the DHCP Serve will assign them the same IP form every session.

This way, the IP Conficlt can end. Another option is to make the DHCP to assign new IP to every unit one more time, eliminating the machines with same IP.

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3 years ago

What effect does Earth’s spherical shape have on the amount of sunlight that reaches the equator?

vaieri [72.5K]

A

Explanation:

The earth is spherical. So it's middle part is bulgjng outside. So more sunlight will be incident on the latitude near the equator. This will heat the air and it will rise up. This will cause high pressure difference and polar disturbances.

I hope this satisfies you. I hope u will follow me and make this the brainliest answer.

5 0

3 years ago

A force of 350n Acts on an object of mass 17.5 kg what acceleration does it produce

UNO [17]

Answer:

20 m/s/s

Explanation:

F=ma, 350=17.5 * a, a=20 m/s/s

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3 years ago

The distance between the earth and sun is 1.5 x 108 kilometers and the speed of light is 3.00 x 108 meters per second. Calculate

butalik [34]

Answer:

time = 8.3333 minutes.

Explanation:

distance between earth and sun = 1.5 * $10^{8}km$

speed of light = 3* $10^{8}m/s$

convert the distance unit from km to m so we can have uniform units.

distance between earth and sun = 1.5 * $10^{8}*1000$ m

distance between earth and sun = 1.5 * $10^{11}$ m

speed = distance /time

time = distance / speed

time = $\frac{1.5*10^{11} }{3*10^{8} }$

$= 0.5*10^{3}$

time =500 sec

time = 500/60 minutes

time = 8.3333 minutes.

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3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

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3 years ago