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nexus9112 [7]
3 years ago
15

What voltage is needed to push 3 amps of current through a 1500-watt hair dryer ??

Physics
1 answer:
RideAnS [48]3 years ago
8 0
V = P/I .............
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What is 2,000m to km​
lesantik [10]

Answer: 2km

Explanation:

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S Four point charges each having charge Q are located at the corners of a square having sides of length a. Find expressions for(
Ket [755]

The total electric potential at the center of the square due to the four charges is V  = √2Q/πÈa.

<h3>What do you mean by electric potential? </h3>

The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.       

V = kq/r

Where V represents electric potential, K is coulomb constant, q  is Charge and r is distance between any  two around charge to the point charge.

Electric potential at O due to four charges is given by,

V = 4KQ/ r

where, r = √2a/2 = a/√2

V = 4k × Q√2/a

V  = √2Q/πÈa

The total electric potential at the center of the square due to the four charges is V  = √2Q/πÈa.

To learn more about electric potential refer to:

brainly.com/question/12645463

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3 0
2 years ago
Explain in details the concept of economic geography​
vova2212 [387]

Answer:

In the words of Hartshorn and Alexander: “Economic Geography is the study of the spatial variation on the earth’s surface of activities related to producing, exchanging and consuming goods and services. Whenever possible the goal is to develop generalizations and theories to account for these spatial variations.”

Explanation:

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3 years ago
How fast should a rocket ship move for its mass to be dilated to 167% of its rest<br>mass?​
andrey2020 [161]

Answer:

The two forces acting on rockets at the moment of launch are the thrust upwards and the weight downwards. Weight is the force due to gravity and is calculated (at the Earth’s surface) by multiplying the mass (kilograms) by 9.8.The resultant force on each rocket is calculated using the equation resultant force = thrust – weight.

Hopefully, this answer helps you! :)

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3 years ago
A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the
andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

F = mg

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

\sum \tau = 0

Regarding the forces we have,

3Mg-1mg=0

Re-arrange to find M,

M = \frac{m}{3}

M = \frac{50}{3}

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Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

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