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devlian [24]
3 years ago
7

A moderate force will break an egg. However, an egg droppedon the road usually breaks while one dropped on the grass usuallydoes

nt break. THis is because the egg dropped on thegrass............???
A. has a greater change in momentum
B. has a lesser change in momentum
C. the time interval for stoppin is greater
D. the time interval for stoppin is less
Physics
1 answer:
Fed [463]3 years ago
6 0

Answer:

C. the time interval for stopping is greater.

Explanation:

As the egg falls onto the grass, it takes a a greater amount of time for it to stop, and thus the force that is being applied to it is in increments; there is never enough force applied on the egg for it to break. That's why the egg doesn't break when it lands on the grass.

In contrast, when the egg is dropped on the road, <em>all of the force that is being applied by the road on the egg is in the tiny interval when the egg touches the road</em>, That force is large enough to break the egg because it is being applied in a tiny amount of time. That's why the egg dropped on the road breaks.

<em>So here is the rule of thumb: if you don't want to break your things but still want to drop them, drop them such that it takes  some for them to stop—because force will applied to them gradually. </em>

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Answer:

More work done with less power

The increase in gravitational energy is the same as the height which is a function of gravitational energy is the same in both cases

Explanation:

Climbing the mountain in zigzag pattern is easier because

1. The time it takes to climb increases so that the required power or rate of doing work decreases

2. Climbing in zigzag pattern affords the use of leverages by the sides

3. Similar mechanical power gain and efficiency from using a drive screw instead of a nail to fasten items together can be achieved

The increase in gravitational energy is the same gravitational energy ~ mass × gravity ×height

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What does the model below represent?
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Answer:

B. a chemical reaction

Explanation:

2CO + O2 → 2C02

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Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc
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Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

E =  \frac{V}{L}

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

J = \frac{E}{p}

Substitute (V/L)  for E in the above equation of current density.

J = \frac{V}{pL} ------(1)

Substitute iR for V in equation (1)

J = \frac{iR}{pL} ------(2)

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

p = \frac{RA}{L}

A = \frac{pL}{R}

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A  and λ is use for (m/L) in the eqn above

\lambda = d\frac{p}{\frac{R}{L} } ------(3)

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³  for d and 1.69 × 10⁸ Ω .m

\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1

c) Using the equation (2) current density for aluminum cable is,

J = \frac{iR}{pL}

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

\lambda = d\frac{p}{\frac{R}{L} }

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

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Answer:

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Explanation:

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I = ∫r² dm

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thus

I = ∫r²dm = ∫r²2ρπh r dr =2ρπh ∫r³ dr = 2ρπh (R⁴/4- 0⁴/4)= ρπhR⁴ /2= mR²/2

replacing values

I = mR²/2= 0.017 kg * (0.06 m)²/2 = 3.06 *10⁻⁵ kg*m²

from Newton's second law applied to rotational motion

τ= Iα , where τ=net torque and α= angular acceleration

since the angular velocity ω is related with the angular acceleration through

ω= ωo + α*t → α =(ω-ωo)/t =  (21 rad/s-0)/0.8 s = 26.25 rad/s²

therefore

τ= Iα= 3.06 *10⁻⁵ kg*m²*26.25 rad/s² = 8.03* 10⁻⁴ N*m

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