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3 years ago

15

What is the weight of a 5.0-kilogram object at the surface of Earth?

1 answer:

Liono4ka [1.6K]3 years ago

6 0

Answer:

49 N (d)

Explanation:

w= mg = 5 kg * 9.8 m/s^2 = 49 N

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A molecule that speed up a reaction

Diano4ka-milaya [45]

Answer:

I think it's a catalyst.

Explanation:

Definition: a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change

6 0

3 years ago

Read 2 more answers

You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.

Alchen [17]

Answer:

$h = 5.505\,m$

Explanation:

Let assume that ball begins its movement at a height of zero. The height can be found by energy approach, to be precise, by Principle of Energy Conservation:

$\frac{1}{2}\cdot (0.145\,kg)\cdot (12\,\frac{m}{s} )^{2}= \frac{1}{2}\cdot (0.145\,kg)\cdot (6\,\frac{m}{s} )^{2}+(0.145\,kg)\cdot (9,81\,\frac{m}{s^{2}} )\cdot h$

$h = 5.505\,m$

4 0

3 years ago

Edgar is walking 0.5 M/S toward the back of a train that is traveling forward at 6.0 M/S west. What is Edgars velocity relative

iragen [17]

Answer:

Just in 1 dimension

Vpg = Vps + Vsg

Where

Vpg= relative to the ground

Vps= person relative to the ship

Vsg= ship relative to the ground

Vpg= Vps+Vsg ;

Vpg= 0.5 + 6

Vpg= 6,5 m/s

8 0

4 years ago

What makes the subject of star formation so difficult and complex?

sdas [7]

Stars live to long for just one person to see it die, usually it out lives all of us

5 0

3 years ago

An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

LenaWriter [7]

Answer:

$t=6.4534 s$

Explanation:

This is an exercise where you need to use the concepts of <em>free fall objects</em>

Our <u>knowable variables</u> are initial high, initial velocity and the acceleration due to gravity:

$y_{0}=75m$

$v_{oy} =20m/s$

$g=9.8 m/s^{2}$

At the end of the motion, the <u><em>rock hits the ground</em></u> making the final high y=0m

$y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}$

If we <em>evaluate the equation</em>:

$0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$

This is a classic form of <u><em>Quadratic Formula</em></u>, we can solve it using:

$t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}$

$a=-4.9\\b=20\\c=75$

$t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s$

$t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s$

Since the <u><em>time can not be negative</em></u>, the <em>reasonable answer</em> is

$t=6.4534s$

8 0

3 years ago