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zloy xaker [14]

3 years ago

6

How many values must a 1-D motion problem give you in order to solve it?

1 answer:

MrRa [10]3 years ago

3 0

Answer:

I think it is B: 3

Explanation:

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g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin

DedPeter [7]

Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as $mgh=0.5mv^{2}$ since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then $v=\sqrt {2gh}$ and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that $v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s$

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3 years ago

What type of kinetic energy does flowing water have?

mina [271]

Answer:

none it's Hydroelectric energy

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3 years ago

What is the Nearest Galaxy from the Milky Way

Vsevolod [243]

Answer:

the andromeda galaxy

Explanation:

hope it helps , pls mark me as brainliest

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3 years ago

Read 2 more answers

A moose runs through the woods and covers 825 m North in 118 s. What is the average velocity of the moose?

weqwewe [10]

Total displacement=825m
Total Time=118s

Average Velocity=Total Displacement/Total Time

$\\ \sf\longmapsto Average\: Velocity=\dfrac{825}{118}$

$\\ \sf\longmapsto Average\:Velocity\approx 8m/s$

4 0

3 years ago

a pool ball leaves a 0.60 meter high table with an initial high table with an initial horizontal velocity of 2.4m/s. predict the

timurjin [86]

Ball leaves the table of height 0.60 m

Initial speed of the ball is in horizontal direction which is 2.4 m/s

Now here we can say that time taken by the ball to hit the ground will be given by kinematics equation in vertical direction

$\Delta y = v_y*t + \frac{1}{2}at^2$

here in y direction we can say

$\Delta y = 0.60 m$

$v_y = 0$

$a = 9.8 m/s^2$ due to gravity

now from above equation we have

$0.60 = 0 + \frac{1}{2}*9.8*t^2$

$t = 0.35 s$

Now in the same time the distance that ball move in horizontal direction is given as

$d = v_x* t$

given that

$v_x = 2.4 m/s$

$d = 2.4 * 0.35$

$d = 0.84 m$

So it will move a total distance of 0.84 m where it will land

5 0

3 years ago