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PolarNik [594]
2 years ago
11

The lowest point in Death Valley is 75.5 m below the sea level. The summit of nearby Mt. Whitney has an elevation of 4360 m. Wha

t is the change in potential energy of an energetic 63 kg hiker who makes it from the floor of Death Valley to the top of Mt. Whitney?
Physics
1 answer:
sineoko [7]2 years ago
8 0

Answer:2738.477 kJ

Explanation:

Given

Lowest Point of death valley is 75.5 m below sea level

Height of Mt. Whitney is 4360 m

mass of hiker =63 kg

Change in potential energy=m\times g\times (h_2-h_1)

Considering sea level a datum

\delta PE=63\times 9.8\times (4360-(-75.5))=2738477.7 J

\delta PE=2738.477 kJ

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A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur
emmasim [6.3K]

• Net vertical force on the block:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

(<em>n</em> = magnitude of normal force, <em>w</em> = weight)

<em>n</em> = <em>w</em> = <em>m g</em>

(<em>m</em> = mass, <em>g</em> = 9.8 m/s²)

<em>n</em> = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ <em>F</em> = -<em>f</em> = <em>m a</em>

(<em>f</em> = mag. of friction, <em>a</em> = acceleration)

We have <em>f</em> = <em>µ</em> <em>n</em> = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) <em>a</em>

<em>a</em> = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆<em>x</em> in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆<em>x</em>

∆<em>x</em> = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

8 0
2 years ago
a pillow , a textbook and a paper airplane are dropped from the top of a tall building at the same time. consider what you have
MAVERICK [17]

A textbook would hit the ground first


Factors:

-Textbook weighs most

-Pillow is flat and fluffy not very aerodynamic) also is very light

-Paper airplane will glide to the ground do to its wings and will hit the ground last

3 0
3 years ago
Read 2 more answers
A small boat sailed straight north out of a harbor in strong east wind (blowing from west to east). After sailing for 120 minute
Amiraneli [1.4K]

it can be said that  the speed of the east wind is

v=0.3608m/s

From the question we are told

A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).

After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,

  • what is the speed of the east wind?.

<h3> the speed of the east wind</h3>

Generally the equation for the distance  is mathematically given as

BA=3000sin60

BA=2598.07m

Therefore

the speed of the east wind

V_w=\frac{BA}{120*60}\\\\V_w=\frac{2598.07}{120*60}

v=0.3608

For more information on this visit

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2 years ago
Mars has two moons. What does the orbital speed of these moons depend on?
Radda [10]

Answer:

their masses and their distances from Mars

Explanation:

Hope this helped

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8 0
3 years ago
A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to
Len [333]

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

v = 10.07968...

5 0
3 years ago
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