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sweet-ann [11.9K]
3 years ago
9

Having a stronger negative or positive change on one side is called what ?

Physics
1 answer:
aivan3 [116]3 years ago
5 0
The answer is action potential. This usually happens when one side of the ion has too much negative or positive charge and by having this, if one of them is stronger in the other side, that charge will likely move on the other side in order for the ion to be balanced and to prevent the other side to be more stronger.
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List Five examples from daily life in which you see periodic motion caused by a pendulum
Basile [38]

Answer:

by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.

Explanation:

5 0
3 years ago
If two balls collide with each other, they will move apart at the same speed if
denis-greek [22]
A.
if you have seen a newton's cradle this will make sense.

in order for both of them to travel at the same speed, the balls need to have the same mass and the speed to begin with tocontinue to travel at the same speed because mass can affect the impact of the force on the balls by each other, causing each ball to have different speeds.
7 0
4 years ago
Read 2 more answers
.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum accelerat
AlexFokin [52]

Answer:

5.51 m/s^2

Explanation:

Initial scale reading = 50 kg  

assume the greatest scale reading = 78.09 kg

<u>Determine the maximum acceleration for these elevators</u>

At rest the weight is = 50 kg

Weight ( F ) = mg = 50 * 9.81 = 490.5 N<u> </u>

<u> </u>At the 10th floor weight = 78.09 kg

Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N

F = change in weight

Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)

50 * a = 275.61

Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2

3 0
3 years ago
One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel
Gekata [30.6K]

Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

v_{He-4} = -1.57 \cdot 10^{7} m/s  

b) E_{He-4} = 8.23 \cdot 10^{-13} J

E_{U-235} = 1.41 \cdot 10^{-14} J

 

Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg  

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg            

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}    (1)

Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}  

Solving the above equation for v_{U-235} we have:

v_{U-235} = 2.68 \cdot 10^{5} m/s

And by entering that value into equation (1):

v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s                        

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

3 0
3 years ago
4 applications of elasticity​
svet-max [94.6K]

Answer:

elasticity

1.price elasticity of demand

2.income elasticity of demand

3.cross elasticity of demand

4.elasticity of supply

Explanation:

1. price elasticity of demand is the degree to which the effective desire for something changes as its price changes. In general, people desire things less as those things become more expensive.

2. income elasticity of demand is the responsiveness of the quantity demanded for a good to a change in consumer income. It is measured as the ratio of the percentage change in quantity demanded to the percentage change in income.

3. cross elasticity of demand or cross-price elasticity of demand measures the responsiveness of the quantity demanded for a good to a change in the price of another good, ceteris paribus.

4.price elasticity of supply is a measure used in economics to show the responsiveness, or elasticity, of the quantity supplied of a good or service to a change in its price.

8 0
3 years ago
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