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Alenkinab [10]
3 years ago
9

Two rods are made from materials that have different Young's moduli. The rods are constructed to have the same length and same c

ross-sectional area. The same force is hung from the end of each rod. The rod with the smaller Young's modulus will stretch more than the rod with the larger Young's modulus. Is this true or false?
Physics
1 answer:
ANTONII [103]3 years ago
5 0

Answer:

This is true,the rod with smaller elastic modulus will stretch more than larger elastic modulus.

Explanation:

σ=E*ε

ε=δ/L

σ=E*δ/L

δ=(σ*L)/E

σ=F/A

δ=(F*L)/(A*E)

As Force,Area and Length is same

δ∞1/E

From the expression as E increase δ will be small,so there will be more stretch for smaller elastic modulus.

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Describe the composition of the nucleus of an atom.
gavmur [86]

Answer:

Explanation:the atom consists of a tiny nucleus at its center which is surrounded by a moving electrons. The nucleus contains a positively charged proton equal in size with the negatively charged electrons . The nucleus also may contain neutrons which have the same mass with the protons but no charge is neutral.

4 0
2 years ago
Read 2 more answers
What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t
Gekata [30.6K]

Start by facing East. Your first displacement is the vector

<em>d</em>₁ = (225 m) <em>i</em>

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

<em>d</em>₂ = (350 m) <em>j</em>

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )

<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>

The net displacement is

<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃

<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>

and its magnitude is

|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

7 0
3 years ago
A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
sp2606 [1]

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

          x = v₀² / 2a

let's calculate

          x = 2²/(2 0.8)

         x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

7 0
2 years ago
Fa car's power output is increased, its efficiency:
Inessa [10]
It’s solved by using a pretty standard formula for efficiency.

4 0
3 years ago
A speedboat moves on a lake with initial velocity vector v1,x = 8.57 m/s and v1,y = -2.61 m/s, then accelerates for 6.67 seconds
KengaRu [80]
First, you find the velocity at each component. The general equation is:

a = (v2 - v1)/t

a,x = (v2,x - v1,x)/t
-0.105 = (v2,x - 8.57)/6.67
v2,x = 7.87 m/s

a,y = (v2,y - v1,y)/t
0.101 = (v2,y - -2.61)/6.67
v2,y = -1.94 m/s

To find the final speed, find the resultant velocity by taking the hypotenuse.

v^2 = (v2,x)^2 + (v2,y)^2
v^2 = (7.87)^2 + (-1.94)^2
v = 8.1 m/s
3 0
3 years ago
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