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3 years ago

How many mL of beakers u need in separating salt, sand, iron filings, and beads

Physics

1 answer:

grin007 [14]3 years ago

3 0

Explanation:

a 50-mL beaker with the same label.

hope it helps

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how much gravitational potential energy do you give a 70 kg person when you lift him up 3 m in the air?

SCORPION-xisa [38]

Given gravitational potential energy when he's lifted is 2058 J.

Kinetic energy is transferred to the person.

Amount of kinetic energy the person has is -2058 J

velocity of person = 7.67 m/s².

<h3>Explanation:</h3>

Given:

Weight of person = 70 kg

Lifted height = 3 m

1. Gravitational potential energy of a lifted person is equal to the work done.

$PE_g=W=m\times g\times h\\Acceleration due to gravity = g = 9.8 \ m/s^2 \\PE_g= m = m\times g\times h= 70\times 9.8 \times 3 = 2058\ kg.m/s^2 = 2058\ J$

Gravitational potential energy is equal to 2058 Joules.

2. The Gravitational potential energy is converted into kinetic energy. Kinetic energy is being transferred to the person.

3. Kinetic energy gained = Potential energy lost = $-PE_g = -2058\ kg.m/s^2$

Kinetic energy gained by the person = (-2058 kg.m/s²)

4. Velocity = ?

Kinetic energy magnitude= $\frac{1}{2} m\times v^2 = m\times g \times h$

Solving for v, we get

$v=\sqrt{2gh} =\sqrt{2\times 9.8 \times 3} = \sqrt{58.8} = 7.67 m/s^2$

The person will be going at a speed of 7.67 m/s².

4 0

3 years ago

What is the specific heat of a material that has a mass of 5 grams and increases 125ºC when it receives 150J of heat?

rjkz [21]

Answer:

0.24 ? I hope that was the answer you were looking for.

Explanation:

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What is normal for a spring that obeys hook's law ?

Bezzdna [24]

A spring that obeys Hooke's law has a spring force constant of 272 N/m. This spring is then stretched by 28.6 cm

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A race car has a centripetal acceleration of 15.625 m/s2 as it goes around a curve. If the curve is a circle with radius 40 m, w

myrzilka [38]

The centripetal acceleration is given by
$a_c = \frac{v^2}{r}$
where v is the tangential speed and r the radius of the circular orbit.

For the car in this problem, $a_c = 15.625 m/s^2$ and r=40 m, so we can re-arrange the previous equation to find the velocity of the car:
$v= \sqrt{a_c r}= \sqrt{(15.625 m/s^2)(40 m)}=25 m/s$

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3 years ago

Bagaimana cara untuk melatih kemampuan melempar tangkap yang baik dalam bola basket

anyanavicka [17]

Answer:

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