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REY [17]
2 years ago
7

Energy is the ability to do work. All of the following describe work being done EXCEPT:

Physics
1 answer:
lbvjy [14]2 years ago
4 0

Answer:

A. blowing out a candles flame

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please answer 1-3 and use a separate piece of paper thank you i've been trying those problems for 3 day.... ikr Sad
11Alexandr11 [23.1K]
\frac{total distance}{ total time}

\frac{72m}{36s} = 2m/s

Same thing for the rest 
 2) 5km/h
 3)360km/h

4 0
3 years ago
An object accelerating may be changing in what two ways
Oksana_A [137]
Velocity and direction
8 0
3 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
A 12 oz can of soda is left in a car on a hot day. In the morning the soda temperature was 60oF with a gauge pressure of 40 psi.
Neko [114]
In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:

P/T = Constant

Then
\frac{ P_{1} }{ T_{1} } = \frac{ P_{2} }{ T_{2} }

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)

Substituting;
P_{2} = \frac{ P_{1}  T_{2} }{ T_{1} } = \frac{40*305}{289}  =42.21 psi
3 0
3 years ago
Julian and Joshua each maintain a constant speed as they run laps around a 400-meter track. In the time it takes Julian to compl
Marysya12 [62]

Answer:

the time Joshua travels 1 mile is 12.5 min

Explanation:

Let's start by finding the distance traveled on each lap,

Let's reduce everything to the SI system

    R = 400 m

    d = 1 mile (1609 m / 1 mile) = 1609 m

    L = 2 pi R

    L = 2 pi 400

    L = 2513 m

Let us form a rule of proportions if 2 turns of Julian is 3 turns Joshua, for 1 turn of Joshua how many turns Julian took

    lap Julian = 2/3 turn Joshua

Let's calculate what distance is the same for both of them since they are on the same track

    1 lap = 2513 m

    d. Julian = 2/3 2513 m

    d Julian = 1675 m distance Joshua

Let us form the last rule of three or proportions if 1609 m you travel in 12 min how long it takes to travel 1675 m

    t Julian = 1675/1609 12

    t = 12.5 s

Since this is the distance Joshua travels, this is the time Joshua travels 1 mile

5 0
3 years ago
Read 2 more answers
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