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nata0808 [166]
2 years ago
14

Give three factors that affected the kinetic energy of the person as she reached the bottom

Physics
1 answer:
soldier1979 [14.2K]2 years ago
4 0

Answer:

As the person moves down the zip wire, her increase in kinetic energy is less than her decrease in gravitational potential energy.

Explanation:

Work is done against the air resistance, causing thermal energy to transfer to the surroundings

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How does the position of an object relate to the energy stored in an object?
Kamila [148]

Answer:

Potential Energy

Explanation:

Potential energy is the energy stored in an object due to it's position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height.

4 0
2 years ago
An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for
AURORKA [14]

Answer:

<em>2 m/s</em>

<em></em>

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

E = Blv

where E is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

l is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

v is the velocity of the fluid through the field = ?

B is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x v

2.88 x 10^-3 = 1.44 x 10^-3 x v

v = 2.88/1.44 = <em>2 m/s</em>

8 0
3 years ago
Two point charges each carrying a charge of + 4.5 E - 6 C are located 4.5 meters away from each other. How strong is the electro
BARSIC [14]

Answer:

0.009 N, repulsive

Explanation:

The electrostatic force between two electric charges is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, we have

q_1 =q_2 = +4.5\cdot 10^{-6}C are the two charges

r = 4.5 m is their separation

Substituting into the equation, we find

F=(9\cdot 10^9 Nm^2 C^{-2})\frac{(+4.5\cdot 10^{-6} C)(4.5\cdot 10^{-6} C)}{(4.5 m)^2}=0.009 N

Moreover, the force is repulsive. In fact, the following rules apply:

- When two charges have same sign, they repel each other

- When two charges have opposite signs, they attract each other

7 0
3 years ago
Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the bea
notka56 [123]

Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, t=0.50\ s

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, v=0\ m/s

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, a=g=-9.8\ m/s^2

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

v=u+at

Where, 'u' is the initial velocity.

Plug in the given values and solve for 'u'. This gives,

0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s

Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s

3 0
3 years ago
Read 2 more answers
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
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