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rosijanka [135]
3 years ago
13

How many joules are expended by a 70 kg man climbing up 6m of stairway? How much work is done if the climbing takes place on the

surface of the moon? (Assume that the acceleration due to gravity on the moon’s surface is 1.6 m/s2
Physics
1 answer:
GuDViN [60]3 years ago
7 0

Answer:

On the earth surface ,W= 4200 J

On the moon surface ,W= 672 J

Explanation:

Given that

mass of the man ,m = 70 kg

Height of the stairway ,h= 6 m

On the earth surface :

We know that acceleration due to gravity at earth ,g= 10 m/s²

The work done by man is given as

W= m g h

Now by putting the values in the above equation we get

W= 70 x 10 x 6 J

W=4200 J

On the moon  surface :

The gravity on the moon ,g= 1.6 m/s²

The work done W

W= 70 x 1.6 x 6 J

W=672 J

On the earth surface ,W= 4200 J

On the moon surface ,W= 672 J

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lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
3 years ago
Who has the greater velocity, an astronaut who has just completed an orbit of the Earth or you when you have just traveled from
Alik [6]

Answer:

the answer is

Explanation:

constant acceleration

because when the object's velocity is changing then the object is accelerating or decelerating

as acceleration describe changing of velocity so the answer is constant acceleration

Acceleration is defined as the rate of change of velocity.

Acceleration = (Change in velocity) / time taken

Acceleration = (Final velocity - initial velocity) / time

As the object velocity changes by the same amount in each second, it means the acceleration is constant.

Hope I can help u

4 0
2 years ago
A motorboat traveling on a straight course slows
never [62]

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

  • 65 km/h → 18.0556 m/s
  • 35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

  • v² = v₀² + 2aΔx

Substitute the known values into the equation.

  • (9.72222)² = (18.0556)² + 2a(45)
  • 94.52156173 = 326.0046914 + 90a
  • -231.4831296 = 90a
  • a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

3 0
3 years ago
Betty is sitting on of her surfboard out in the ocean. She is waiting for the perfect wave to come along so she can ride it in t
givi [52]

Crest is the part of the wave which does this.

A sound wave is the sample of disturbance resulting from the movement of strength visiting through a medium, including air, water or every other liquid or stable remember as it propagates far from the supply of the sound.

The sound waves are generated by a sound source, such as the vibrating diaphragm of a stereo speaker. The sound source creates vibrations in the surrounding medium. because the supply continues to vibrate the medium, the vibrations propagate far from the supply at the rate of sound, hence forming the sound wave.

A sound wave is not a transverse wave with crests and troughs, however alternatively a longitudinal wave with compressions and rarefactions.

Learn more about wave here:- brainly.com/question/1199084

#SPJ1

7 0
10 months ago
Kiran drove from City A to City B, a distance of 228 mi. She increased her speed by 12 mi/h for the 400-mi trip from City B to C
Degger [83]

Answer:

From city A to city B her speed was 38 mi/h

Explanation:

The traveled distance can be calculated using this equation:

From city A to city B

228 mi = v · t₁

Where:

v = velocity

t₁ = time it took Kiran to travel the 228 mi from city A to city B

From city B to city C

400 mi = (v + 12 mi/h) · t₂

We also know that the entire trip took 14 h, then:

t₁ + t₂ = 14 h

So, we have a system of three equations with three unknwons:

228 mi = v · t₁

400 mi = (v + 12 mi/h) · t₂

t₁ + t₂ = 14 h

Let´s solve the third equation for t₁:

t₁ = 14 h - t₂

Now let´s replace t₁ in the first equation and solve it for t₂

228 mi = v · t₁

228 mi = v · (14 h - t₂)

228 mi/v - 14 h =  - t₂

t₂ = 14 h - 228 mi/v

Now let´s replace t₂ in the second equation:

400 mi = (v + 12 mi/h) · t₂

400 mi = (v + 12 mi/h) · (14 h - 228 mi/v)

400 mi = 14 h · v - 228 mi + 168 mi - 2736 mi²/(v · h)

400 mi = 14 h · v - 60 mi - 2736 mi²/(v · h)

460 mi = 14 h · v - 2736 mi²/(v · h)

Multiplicate by v both sides of the equation:

460 mi · v = 14 h · v² - 2736 mi²/h

0 = 14 h · v² - 460 mi · v - 2737 mi²/h

Solving the quadratic equation:

v = 38 mi/h

(The other solution of the equation is negative, and therefore discarded)

From city A to city B her speed was 38 mi/h

4 0
3 years ago
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