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soldi70 [24.7K]
4 years ago
13

A 20 N force causes an impulse of 500 N*s. How long was the force applied for?

Physics
1 answer:
yKpoI14uk [10]4 years ago
3 0

Answer:

25sec

Explanation:

Impulse = Force * time (period)

therefore,

500N*s= 20N * x(secs)

divide both sides by 20

results in  500/20=x

x= 25 secs

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What is the force acting on a 10kg object that accelerates from 5 m/s to 20 m/s in 5s?
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Answer:

Option C

Explanation:

v= u + at

20 = 5 + a(5)

15= a(5)

a= 3 m/s²

Force = mass × acceleration

= 10 × 3

= 30 N

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The latent heat of fusion of alcohol is 25 kcal/kg and its melting point is -114 o C. It has a specific heat of 0.60 in its liqu
Elanso [62]

Answer:

Total energy is 170 kJ

Explanation:

Given data:

latent heat of fusion of alcohol is 25 kcal/kg

melting point of alcohol is -114 degree c

specific heat us 0.60 k cal/kg degree c

energy need for 2 kg solid alcohol is

for Melting:

Energy Q is calculated as

Energy, Q = 25 \times 2.0 kg = 50 kJ

Energy required for Heating liquid:

Energy, ΔH = 2.0 kg \times 0.60 \times (100°C) = 120 kJ

Total  energy = (50 kJ + 120 kJ) = 170 kJ

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3 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
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