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Musya8 [376]
3 years ago
12

A mountain climber ascends a mountain to its peak. The peak is 12,470 ft above sea level. The climber then descends 80 ft to mee

t a fellow climber. Find the climber's elevation above sea level after meeting the other climber. A. -12,390 ft. B. 12,550 ft. C. 11,670 ft. D. 12,930 ft
Physics
1 answer:
o-na [289]3 years ago
8 0
The answer is 12,390 ft.

At first, a climber is at 12,470 <span>ft above sea level. But then, he goes down 80 ft to meet a fellow climber. So, this simply needs to be distracted:
12,470 ft - 80 ft = 12,390 ft
This is the elevation </span>above sea level at which he meet the other climber.
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Which three statements describe electromagnetic waves?
o-na [289]

Answer:

It is most likely option A B and C

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3 years ago
Read 2 more answers
A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th
ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = 9.8\ m/s^2

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

                Kinetic energy K.E = \frac{1}{2}mv^2.

From above we know that -

   Kinetic energy at the bottom of the cliff = potential energy at a height h

                 \frac{1}{2}mv^2=\ mgh

                ⇒ v^2=\ 2gh

                ⇒ h=\ \frac{v^2}{2g}

                ⇒ h=\ \frac{(24.2)^2}{2\times 9.8}

                ⇒ h=\ 29.88\ m

Hence, the height of the cliff is 29.88 m

             


5 0
3 years ago
A potentialiterence of 12 volts is induced across a straight wire 0.20 meters long as it is moved at a constant speed of 3.0 met
shtirl [24]

Strength of the magnetic field: 20 T

Explanation:

For a conductive wire moving perpendicular to a magnetic field, the electromotive force (voltage) induced in the wire due to electromagnetic induction is given by

\epsilon=BvL

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For the wire in this problem, we have:

\epsilon=12 V (induced emf)

L = 0.20 m (length of the wire)

v = 3.0 m/s (speed)

Solving  for B, we find the strength of the magnetic field:

B=\frac{\epsilon}{vL}=\frac{12}{(0.20)(3.0)}=20 T

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

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