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vfiekz [6]
2 years ago
14

Explain how each of the following factors affects resistance through a wire:

Physics
1 answer:
Nostrana [21]2 years ago
5 0

Answer:

C. length

Explanation:

i hope it helps

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Bagaimana cara untuk melatih kemampuan melempar tangkap yang baik dalam bola basket​
anyanavicka [17]

Answer:

pergi ke pertandingan sepak bola dan amati

7 0
2 years ago
The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper
-BARSIC- [3]

To solve this problem we will apply the normal distribution, with which we will obtain the probability that the given event will occur. Concepts such as the mean and standard deviation will be present throughout the solution of the problem. Increasing or decreasing the average would change the location or center point of the curve. The change in the standard deviation would lead to the change in the dispersion of the data. As the standard deviation increases, the curve would become flatter.

Let X be the output voltage of power supply

X∼N (5,0.02^2)

A

The lower and upper specifications for voltage are 4.95 V and 5.05 V, respectively

P(4.95

P(4.95

P(4.95

P(4.95

P(4.95

Hence probability that a power supply selected at random will conform to the specifications on voltage is 0.9876

8 0
3 years ago
Question 7 (1 point)
Temka [501]

Answer:

Forms over water, warm humid air mass, it's a polar air mass

Explanation: I think that's right sorry if it's not..

GL! :)

4 0
2 years ago
Select the correct answer.
skelet666 [1.2K]

Answer:

B is the best answer for the question

6 0
3 years ago
Read 2 more answers
A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet
Gnom [1K]

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

6 0
3 years ago
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