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zepelin [54]
3 years ago
9

Please help! If you held a piece of tissue paper at arms length and asked the heavyweight champion of the world to hit the paper

with 50 pounds of force, what would happen? Explain using your understanding of force.
Physics
1 answer:
FinnZ [79.3K]3 years ago
5 0
Force is equal to mass x acceleration. The heavyweight is packing both greater acceleration and mass than the tissue paper. 
When his fist comes in contact with the paper, the tissue reacts by exerting its own opposite force, providing a tiny bit of resistance until the fist breaks it apart and passes through. 
The fist, although by a a very tiny amount, is decelerated (slowed down) by the paper before it breaks through. 
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babymother [125]

Answer:

C

Explanation:

momentum = mass ×velocity

A. mv = 200

B. mv = 300

C. mv= 400

D. mv= 200

highest is C.

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The ________ of a gas can affect the pressure of a gas.
Diano4ka-milaya [45]

d. none of the above

pressure affects volume not the other way round

5 0
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The centre of mass of a metre rule is at the 50cm mark. state what is meant by Centre of mass​
Tanya [424]

Answer:

Centre of mass of any body is a point where all mass of a body is supposed to be concentrated

it lies in geometrical centre....

3 0
3 years ago
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 50.9 V/m. Calculate the intensity ???? of this w
Sliva [168]

Answer:

3.44 W/m²

1.134 J

Explanation:

E₀ = Intensity of electric field = 50.9 V/m

I = Intensity of electromagnetic wave

Intensity of electromagnetic wave is given as

I = (0.5) ε₀ E₀² c

I = (0.5) (8.85 x 10⁻¹²) (50.9)² (3 x 10⁸)

I = 3.44 W/m²

A = Area = 0.0277 m²

t = time interval = 11.9 s

Amount of energy is given as

U = I A t

U = (3.44) (0.0277) (11.9)

U = 1.134 J

5 0
3 years ago
What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J
Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)

and on the surface the current is

I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A

(b)

For Jb = J0(1 - r/R)

I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )

and on the surface

I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi  \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A

and in a simmilar way for Jb

I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi   J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0
3 years ago
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