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AysviL [449]
3 years ago
11

Find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the

zero of potential energy at sea level.
Physics
1 answer:
Agata [3.3K]3 years ago
7 0

Answer:

P = 1470980 J

Explanation:

We have,

Mass of the hiker is 79 kg

It is required to find the potential energy associated with a 79-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level.

It is given by :

P=mgh\\\\P=79\times 9.8\times 1900\\\\P=1470980\ J

So, the potential energy of 1470980 J is associated with a hiker.

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If a vehicle accelerating at 2.7 m/s2what is it's velocity at 20 meters 12.63m/s,1.03m/s,10.39m/s,6.39m/s
netineya [11]

Answer:

The final velocity of the vehicle is 10.39 m/s.

Explanation:

Given;

acceleration of the vehicle, a = 2.7 m/s²

distance moved by the vehicle, d = 20 m

The final velocity of the vehicle is calculated using the following kinematic equation;

v² = u² + 2ah

v² = 0 + 2 x 2.7 x 20

v² = 108

v = √108

v = 10.39 m/s

Therefore, the final velocity of the vehicle is 10.39 m/s.

5 0
3 years ago
In which way are the cathode rays deflected in the x-plates of the Cathode Ray Oscilloscope​
MrRissso [65]

Answer:

The cathode ray is deflected vertically to the fluorescent screen

Explanation:

.

4 0
2 years ago
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A race car starting from rest accelerates uniformly at a rate of 3.90 m/s^2 what is the cars speed after it has traveled 200 m.
snow_tiger [21]
The formula for accelerational displacement is at^2/2, so we know that 3.9t^2/2 = 200, or 3.9t^2 = 400. t = \sqrt{400/3.9} \approx 10.12739367, at = v, so 3.9 * 10.12739367 \approx 39.5  m/s
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3 years ago
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An AWG No.16 line cord with type HPN insulation has an amp airy of
Lostsunrise [7]

12 amp is your answer

5 0
2 years ago
A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad
Korolek [52]

Answer:

9.21954 m/s

54 m/s²

Angle is zero

Explanation:

r = Radius of arm = 1.5 m

\omega = Angular velocity = 6 rad/s

The horizontal component of speed is given by

v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s

The vertical component of speed is given by

v_v=2\ m/s

The resultant of the two components will give us the velocity of hammer with respect to the ground

v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s

The velocity of hammer relative to the ground is 9.21954 m/s

Acceleration in the vertical component is zero

Net acceleration is given by

a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2

Net acceleration is 54 m/s²

As the acceleration is towards the center the angle is zero.

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2 years ago
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