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diamong [38]
3 years ago
14

Which is heavier , 1kg of steel or i kg of cork​

Physics
2 answers:
goblinko [34]3 years ago
6 0

Answer:

Same weight so neither is heavier

LenaWriter [7]3 years ago
5 0

Since their masses are equal, their weights are equal whenever they're both on the same planet.

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A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of th
Ber [7]

Answer:

0.546 ohm / μm

Explanation:

Given that :

N = 1.015 * 10^17

Electron mobility, u = 3900

Hole mobility, h = 1900

Ng = 4.42 x10^22

q = 1.6*10^-19

Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

= 0.01578880889 ohm /cm

Resistivity of germanium :

R = 1 / 2q * sqrt(Ng) * sqrt(u*h)

R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)

R = 1 /0.0001831

R = 5461.4964 ohm /cm

5461.4964 / 10000

0.546 ohm / μm

7 0
3 years ago
HELP
blagie [28]

Answer:

A

Explanation:

hope its right

3 0
3 years ago
What will a spring scale read for the weight of a 57.0-kg woman in an elevator that moves upward with constant speed of 5.0 m/s
Tanzania [10]
The spring scale will read 559 Newton's or 125.7 pounds.
4 0
3 years ago
A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav
uysha [10]
In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.

7 0
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
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