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Degger [83]
2 years ago
8

8. Question You should never drive: When weather conditions are not ideal. When road conditions are not ideal. Without having ma

de the necessary
All of the above.​
Engineering
1 answer:
LiRa [457]2 years ago
6 0
The answer is All of the above
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How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0
3 years ago
Consider a pan of water being heated (a) by placing it on an electric range and (b) by placing a heating element in the water. W
Brrunno [24]

Answer:

Method B is the more efficient way of heating the water.

Explanation:

Method B is more efficient because by placing a heating element in the water as in described in method B, the heat that is lost to the surroundings is minimized which implies that more heat is supplied directly to the water. Therefore, more heating is achieved with a lesser amount of electrical energy input. Whereas placing the pan on a range means more heat losses to the surrounding and as such it will take a longer time for the water to heat up and also take more electrical energy.

7 0
3 years ago
A particle moves along a circular path of radius 300 mm. If its angular velocity is θ = (2t) rad/s, where t is in seconds, deter
uysha [10]

Answer:

4.83m/s^{2}

Explanation:

For a particle moving in a circular path the resultant  acceleration at any point is the vector sum of radial and the tangential acceleration

Radial acceleration is given by a_{radial}=w^{2}r

Applying values we get  a_{radial}=(2t)^{2}X0.3m

Thus a_{radial}=1.2t^{2}

At time = 2seconds a_{radial}= 4.8m/s^{2}

The tangential acceleration is given by a_{tangential} =\frac{dV}{dt}=\frac{d(wr)}{dt}

a_{tangential}=\frac{d(2tr)}{dt}

a_{tangential}= 2r

a_{tangential}=0.6m/s^{2}

Thus the resultant acceleration is given by

a_{res} =\sqrt{a_{rad}^{2}+a_{tangential}^{2}}

a_{res} =\sqrt{4.8^{2}+0.6^{2}  } =4.83m/s^{2}

8 0
3 years ago
Riverdale is the best show ever yes or no I will give you all my points if you answer yes ???
Verizon [17]

only the first season is good it went downhill ever since

6 0
2 years ago
Read 2 more answers
A hypothetical metal alloy has a grain diameter of 1.7 ´ 10-2 mm. After a heat treatment at 450°C for 250 min the grain diameter
Yuliya22 [10]

Answer:1103 minutes/66180 seconds

Explanation:

FIRST STEP: is to FIND the value of K using the formula below;

K= d^n - d^n(o)/t .....................(1).

Parameters given from the question d^n = 4.5×10^-2 mm, d^n(o)= 1.7 × 10^-2 mm, t= 250 minutes(min) and n= 2.1.

Slotting in the parameters into the equation (1) above,then;

(4.5×10^-2)^2.1 - (1.7×10^-2)^2.1/ 250

= 5.2 × 10^-6 mm^2.1/min.

SECOND STEP: with the value of K from the second step, we can use it to calculate the required time based on the diameter. Therefore, equation (1) becomes;

t= d^2.1 - d^2.1(o)/ K

(8.7×10^-2)^2.1 - (1.7×10^-2)^2.1/ 5.2 × 10^-6

= 1,103 minutes.

5 0
3 years ago
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