What is a key phrase for remembering the order of the planets?

likoan [24]

Answer :

C) Atom

Actual answer should be element but depends on the way you interpret the question and the options given to answer it

Explanation :

Atom is the most basic unit of any substance and is what molecules are made of.

Hope it helps if it does let me know by thanking

5 0

3 years ago

A .5kg bird is perched on its nest so that it has 50J of potential energy. how far is it off the of the ground?

pshichka [43]

It is 10.20 m from the ground.

<u>Explanation:</u>

<u>Given:</u>

m = 0.5 kg

PE = 50 J

We know that the Potential energy is calculated by the formula:

$P. E = m \times g \times h$

where m is the is mass in kg; g is acceleration due to gravity which is 9.8 m/s and h is height in meters.

PE is the Potential Energy.

Potential Energy is the amount of energy stored when an object is stationary.

Here, if we substitute the values in the formula, we get

$P. E = m \times g \times h$

50 = 0.5 × 9.8 × h

50 = 4.9 × h

$h = \frac {50} {4.9}$

h = 10.20 m

3 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago

A car starts moving from the position of rest with uniform acceleration of 8m/s^calculate the distance travelled by it during 10

ad-work [718]

We calculate the coordinates at t₁ = 9 min and t₂ = 10 min, since the 10th minute is between t₁ and t₂.

As it leaves from rest, it means that the initial speed is zero

t₁=9 min=540 s

t₂=10 min=600 s

x₁=at₁²/2=8*540²/2=4*291600=1166400 m

x₂=at₂²/2=8*600²/2=4*360000=1440000 m

Δx=x₂-x₁=1440000-1166400=273600 m represents the distance traveled by the car in the 10th minute of travel

8 0

3 years ago

Please help ASAP. How is heat related to work?

earnstyle [38]

Energy is the ability to do work so I would say that thermal or heat energy is a type of work. Don’t know if this will work but that’s what I would put.

7 0

3 years ago