What is the key topic to consider when researching a potential employer

What is the mass of a 2 kg object on the Earth and on the moon?

zubka84 [21]

Answer:

Same

Explanation:

Mass is the quantity of matter in a certain object.

WHEREVER you take a 2kg object, the mass will remain 2kg. All that changes is the Weight ..Weight the force which the centre of a Planet uses to pull everything towards itself.

On earth, it is 9.81 whereas on the Moon it is 1.6

7 0

3 years ago

Peter went swimming at a lake with his friends. They arrived early in the morning. Peter walked on the sand around the lake with

lapo4ka [179]

The sun was shining on the sand heating it up causing ot to become hot

4 0

3 years ago

You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is

katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

$K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}$

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

$\dfrac{1}{2} \times 166 \times v^2 = 17457.0912$

$v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5$

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

$\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2) }}$

Which gives;

$\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66) } \approx 8.1$

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

$Time = \dfrac{Distance}{Velocity}$

$\mathrm{The \ time \ the\ Lone \ Ranger \ and \ Tonto \ have, \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s$

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

5 0

3 years ago

If you press as hard as you can against a brick wall until you begin to sweating and breathing hard did you do any work on the w

Reika [66]

No, as long as nothing moves you aren't doing work, since W=F.s. You have no s, so no W

4 0

3 years ago

A canoe is designed to have very little drag when it moves along its length. Riley, mass 62 kg, sits in a 21 kg canoe in the mid

vovangra [49]

(1) Riley velocity relative to the water after she jumps is 1.7 m/s.

(2) The velocity of the canoe relative to the water is 5.02 m/s.

The given parameters;

mass of Riley, m₁ = 62 kg
mass of canoe, m₂ = 21 kg
velocity of the Riley after jumping, v₁ = 1.7 m/s

(1) Since the lake is still, her velocity relative to the water after she jumps is 1.7 m/s.

(2) Apply the principle of conservation of linear momentum;

m₁v₁ = m₂v₂

62 x 1.7 = 21v₂

105.4 = 21v₂

$v_2 = \frac{105.4}{21} \\\\v_2 = 5.02 \ m/s$

Thus, after her leap, the velocity of the canoe relative to the water is 5.02 m/s.

Learn more here:brainly.com/question/12803552

7 0

3 years ago