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iogann1982 [59]
3 years ago
9

Three mountain climbers set out to climb a mountain from the same altitude and all arrive at the same location at the top. Mount

ain climber A took a long gradual slope to the top, B went a steeper but shorter path, and C tackled the sheer straight side to the top. Assume all three climbers have the same mass. Which climber gained the greatest potential energy?A. AB. BC. CD. all did the same amount of work
Physics
1 answer:
Vikki [24]3 years ago
8 0

Answer:All did same amount of work

Explanation:

All three mountain climbers took different routes to reach the peak. There is a constant gravity force acting on the object i.e. gravity force. Gravity force is conservative i.e. work done by a conservative force is independent of the path followed.

Therefore work done by all the mountain climbers is the same i.e. all did the same amount of work.

                               

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Identify the conditions for an elastic collision in a closed system. Check all that apply.
s344n2d4d5 [400]

Answer:

In an elastic collision:

  • There is no external net force acting. Thus, Momentum before and after collision is equal. Momentum remains conserved.
  • Total energy always remains conserved as energy cannot be created nor destroyed. It can change from one form to another.
  • There is no lost due to friction in elastic collision. So the kinetic energy is also conserved.
  • Velocities may change after collision. If the masses are equal, the velocities interchange.

When one object is stationary:

Final velocity of object 1:

v₁ = (m₁ - m₂)u₁/(m₁ +m₂)

Final velocity of object 2:

v₂ = (2 m₁ u₁)/(m₁+m₂) =

  • Objects do not stick together in elastic collision. They stick together in inelastic collision.
  • One object may be stationary before the elastic collision.

Thus, conditions for an elastic collision:

  • Energy is conserved.
  • Velocities may change.
  • Momentum is conserved.
  • Kinetic energy is conserved.
  • One object may be stationary before the elastic collision.
7 0
3 years ago
Read 2 more answers
How does the value of the electrostatic force vary with the distance between them?
MariettaO [177]
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
5 0
2 years ago
Which system of measurement has been standardized worldwide for scientific uses?
belka [17]
<h2>Answer: The Systeme international (International System of Units)</h2>

The International System of Units (SI) is used in almost every country in the world (<em>except Burma, Liberia and the United States</em>).

This system was created in 1960 by the 11th General Conference of Weights and Measures in France and is made up of seven basic units:  

-Ampere (electric current)

-Kelvin (temperature)

-Second (time)

-Meter (length)

-Kilogram (mass)

-Candela (luminous intensity)

-Mol (amount of substance) *added to the system in 1971

Plus an unlimited number of derived units from the main ones.

7 0
3 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
3 years ago
A COPPER WIRE OF LENGTH 4M AND AREA OF CROSS-SECTION 1.2CM^2 IS STRECHED WITH A FORCE 4.8×10^3 N . IF YOUNGS MODULAS FOR COPPER
VARVARA [1.3K]

Stress = force / area

= 4.8 x 10³ / 1.2 x 10^-4

= 4 x 10⁷ N /m²

YOUNGS MODULAS = stress / strain

= 4 x 10⁷ / 1.2 x 10^11

= 3.3 x 10^-4

INCREMENT OF LENGTH = longitudinal length x intitial length

= ( 3.3 x 10^-4 ) x 4

= 13.2 x 10^-4 m

= 13.2 mm

mark ‼️ it brainliest if it helps you ❤️

6 0
3 years ago
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