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liberstina [14]
3 years ago
10

Calculating Welght What is the weight of a 45 kg box? N

Physics
1 answer:
Julli [10]3 years ago
7 0

Answer:

Im not sure if this is right but here is the answer i think it is....So multiply 45 by 2.205 to get 99.225 pounds. Alternately, to convert from kg to Newtons, use the fact that 1 kg is 9.8 Newtons

Explanation:

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The specific heat of aluminium is 897J/kg k . This values is almost 2.3 times of specific heat of copper .

Aluminium has higher thermal capacity than water

Explanation:

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A pneumatic system consists of several identical 1.5 inch diameter cylinders to lift 450-pound pallets in a warehouse. the gauge
DerKrebs [107]
The applicable equation:

P = F/A

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F = Force or weight
A = surface area

Pressure on each cylinder = (W/n)/A
Where n = number of cylinders. Additionally, pressure in the reservoir is equivalent to the pressure in each cylinder.
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Therefore,
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3 0
3 years ago
Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E mid
algol13

Answer:

The intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

|E_{1}|=k\frac{q_{1}}{d^{2}}

Where:

  • k is Coulomb's constant
  • q(1) is 20.00 μC or 20*10⁻⁶ C
  • d is the distance from q1 to the midpoint (d=10.0 cm)

So, we will have:

|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}

|E_{1}|=1.8*10^{7}\: N/C

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

|E_{2}|=k\frac{q_{2}}{d^{2}}

|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}

|E_{2}|=7.2*10^{6}\: N/C

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C

I hope it helps you!

5 0
3 years ago
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