Answer:
E=0.036 V/m
Explanation:
Given that
Resistivity ,ρ=2.44 x 10⁻⁸ ohms.m
d= 0.9 mm
L= 14 cm
I = 940 m A = 0.94 A
We know that electric field E
E= V/L
V= I R
R=ρL/A
So we can say that
E= ρI/A
Now by putting the values

E=0.036 V/m
Answer:
Vx = 35.31 [km/h]
Vy = 18.77 [km/h]
Explanation:
In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.
![v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h]](https://tex.z-dn.net/?f=v_%7Bx%7D%20%3D%2040%2Acos%2828%29%5C%5CV_%7Bx%7D%20%3D%2035.31%20%5Bkm%2Fh%5D)
In order to find the vertical component, we must use the sine function of the angle.
![V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h]](https://tex.z-dn.net/?f=V_%7By%7D%3D40%2Asin%2828%29%5C%5CV_%7By%7D%20%3D%2018.77%20%5Bkm%2Fh%5D)
Heat required to melt 0.05 kg of aluminum is 28.7 kJ.
<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>
The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.
The formula to be used is given below:
- Heat required = mass * heat capacity * temperature change
Assuming the aluminum sheet was at room temperature initially.;
Room temperature = 25 °C
Melting point of aluminum = 660.3 °C
Temperature difference = (660.3 - 25) = 635.3 903
Heat capacity of aluminum = 903 J/kg/903
Heat required = 0.05 * 903 * 635.3
Heat required = 28.7 kJ
In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.
Learn more about heat capacity at: brainly.com/question/21406849
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Answer:
Acceleration, 
Explanation:
It is given that,
Initial velocity of the car, u = 10 m/s (in right)
Final velocity of the car, v = -5 m/s (in left)
Time taken, t = 10 s
Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :



So, the acceleration of the car is
. Hence, this is the required solution.
Answer:
68 db
Explanation:
Since now instead of one two dogs are barking simultaneously close to each other, therefore we take n =2.
Ignoring interference effects, the barking of two dogs result in a higher level of intensity which is given by,
β(db)=10×㏒(2)
=3 db
So, a reasonable estimate for the raised Intensity Level is: 65db+3db = 68db