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AlladinOne [14]
3 years ago
11

Two rowers, who can row at the same speed in still water, head across a river. . . The first rower (Alice) heads straight across

(but of course her boat is pulled downstream by the current). . . The second rower (Bianca) heads slightly upstream, so that with the pull of the current she arrives directly opposite their starting point. . . Who arrives at the other side first?. . A) Alice. B) Bianca. C) It's impossible to tell without more information.
Physics
1 answer:
hjlf3 years ago
4 0
"Alice" is the one among the following choices given in the question that arrives at the other side first. the correct option among all the options that are given in the question is the first option or option "A". the other choices can be easily neglected. i hope that this is the answer that has come to your great help.
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Prof. Kopp is working on an experiment located in an abandoned mine near Duluth, MN. The experiment is located approximately 713
MA_775_DIABLO [31]

Answer:

The average speed of the elevator going down in the abandoned mine is 17.722mph.

Explanation:

If the elevator takes 90 seconds to descend a height of 713m, the average speed of the elevator is:

v_{av}=x_T/t_T=713m/90s=7.922m/s

And if 1m/s is 2.23694mph, the average speed is:

v_{av}=7.922m/s=17.722mph.

8 0
3 years ago
An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over
Tasya [4]

Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

    Em₀ = U₁ + U₂

    Em₀ = m₁ g y₁ + m₂ g y₂

Let's place the reference system at the point where the mass m1 is

     y₁ = 0

    y₂ = y

    Em₀ = m₂ g y = 2 m₁ g y

End point, at height yf = y / 2

    E_{mf} = K₁ + U₁ + K₂ + U₂

    E_{mf} = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g y_{f} + m₂ g y_{f}

Since the masses are joined by a rope, they must have the same speed

     E_{mf} = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g y_{f}

   E_{mf}= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

How energy is conserved

   Em₀ =  E_{mf}

   2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g y_{f}

   2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2

   3/2 v₁² = 2 g y -3/2 g y

   3/2 v₁² = ½ g y

   V₁ = √ (gy / 3)

5 0
2 years ago
What are 7 examples of potential energy
yaroslaw [1]

Answer:

<em>Hewo Otaku Kun Here! (UwU)</em>

Explanation:

1. A rock sitting at the edge of a cliff has potential energy. If the rock falls, the potential energy will be converted to kinetic energy.

2. Tree branches high up in a tree have potential energy because they can fall to the ground.

3. A stick of dynamite has chemical potential energy that would be released when the activation energy from the fuse comes into contact with the chemicals.

4. The food we eat has chemical potential energy because as our body digests it, it provides us with energy for basic metabolism.

5. A stretched spring in a pinball machine has elastic potential energy and can move the steel ball when released.

6. When a crane swings a wrecking ball up to a certain height, it gains more potential energy and has the ability to crash through buildings.

7. A set of double "A" batteries in a remote control car possess chemical potential energy which can supply electricity to run the car.

<em>happy to help!</em>

<em>from: Otaku Kun ^^</em>

8 0
3 years ago
How much heat is needed to raise the temperature of 8g of water by 15oC?
Murrr4er [49]
<h2>Answer: 502.08 J</h2>

Explanation:

The heat (thermal energy) needed in to raise the temperature in a process can be found using the following equation:

Q=m.C.\Delta T   (1)

Where:

Q is the heat

m=8 g is the mass of the element (<u>water</u> in this case)

C is the specific heat capacity of the material. In the case of water is C=4.184\frac{J}{g\°C}

\Delta T=15\°C is the variation in temperature  <u>(which is increased in this case)</u>

Knowing this, let's rewrite (1) with these values:

Q=(8 g)(4.184\frac{J}{g\°C})(15\°C)  (2)

Finally:

Q=502.08 J  

5 0
3 years ago
An object of unknown mass is hung on the end of an unstretched spring and is released from rest. The acceleration of gravity is
lakkis [162]

Answer:

0.33 s

Explanation:

For this case, as the object is hung on the end of an unstretched spring, we can consider this system as a simple pendulum.

For this system, we can determine the period of the motion using the following formula:

T = 2π√(L/g)

Where: T = period (in sec), L = lenght of the spring, g = acceleration of garvity = 9.8 m/s²

By the exact time the object is 2.75 cm before coming to rest, that will be the lenght of the spring we can consider (2.75 cm = 0.0275 m)

Finally:

T = 2π√(0.00275/9.8)

T = 0.33 sec

4 0
3 years ago
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