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3 years ago

12

WHAT TYPE OF MOON IS THIS WILL MARK AS BRAINLIEST :)

2 answers:

Liula [17]3 years ago

5 0

It is a waning Crescent moon

weqwewe [10]3 years ago

4 0

The answer is new moon. Hope this helps

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Technician A says test lights are great for quick tests on non-computerized circuits. Technician B says you can use a test light

Tpy6a [65]

Answer:

that technician A is right

Explanation:

The test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits, these two values are high and can light the bulb. In digital circuits the current is very small in the order of milliamps, so there is not enough power to turn on these lights.

From the above it is seen that technician A is right

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4 years ago

Melanie watched the path a baseball followed after a pitcher threw it. She noticed that the ball traveled horizontally away from

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4 years ago

Gunther needs a 75% concentrated solution. When making his solution at room temperature, he could only make a solution that was

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3 years ago

Read 2 more answers

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

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4 years ago

Lucy boards a crowded bus as it sits in the station. There are no available seats, so Lucy stands in the center aisle next to a

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