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3 years ago

A lamp draws a current of 20 A when it is connected to a 100 V source? What is the resistance of the lamp?

Physics

1 answer:

S_A_V [24]3 years ago

7 0

Explanation:

R = V / I = 100V / 20A = 5 ohms.

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State three differences between solid friction and viscosity

Mrrafil [7]

Difference between solid friction and viscosity is as follows: Viscosity is proportional to the surface area whereas solid friction is independent of area of solid surfaces in contact. Viscous force on the body depends upon its velocity in the viscous media. But, friction does not depend on the velocity of the body.

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3 years ago

A car has a mass of 1000 km kilometers. The weight of the car is blank newtons use G equals 9.8 in KG for gravity? Help mehhhhh

Schach [20]

The weight of the car in the picture of the computer screen is 9,800 Newton's.

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4 years ago

What is the equation: If F=10 N, a=5 m/s², m=?

Romashka-Z-Leto [24]

Answer:

2 kg

Explanation:

Remember:

F = m * a re-arrange to

F/a = m substitute in the given values

10 / 5 = 2 kg

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2 years ago

half life worksheet answer key 3. What percent of a sample As-81 remains un-decayed after 43.2 seconds

Ivahew [28]

The final mass after decay can be obtained by using under given relation:

half life period of As-81 = 33 seconds

mf = mi x (1/2^n)

= 100 x ( 1/2^(43.2/33))

= 40.4 %

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3 years ago

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago