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2 years ago

How to find yield strength of a load vs deflection?

Physics

1 answer:

liraira [26]2 years ago

6 0

Σ/ε
σ = F/A
ε = ΔL/L
F = force
A = area
L = lenght
ΔL = |old lenght - new lenght|

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The gravitational acceleration is 9.81 m/s2 here on Earth at sea level. What is the gravitational acceleration at a height of 35

azamat

To solve this problem it is necessary to apply the definition of severity of Newtonian laws in which it is specified that gravity is defined by

$g= \frac{GM}{R^2}$

Where

G= Gravitational Constant

M = Mass of Earth

R= Radius from center of the planet

According to the information we need to find the gravity 350km more than the radius of Earth, then

$g_{ss} = \frac{GM}{R+h^2}$

$g_{ss} = \frac{6.67*10^{-11}*5.972*10^{24}}{(6371*10^3+350*10^3)^2}$

$g_{ss} = 8.82m/s^2$

Therefore the gravitational acceleration at 350km is $8.82m/s^2$

5 0

3 years ago

El Niño occurs every 2 to 12 years. true false

Novosadov [1.4K]

True
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2 years ago

Read 2 more answers

For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number repr

leonid [27]

<u>Answer:</u> The weight of the object is 29.4 N

<u>Explanation:</u>

To calculate the weight of the object, we use the equation:

$W=m\times g$

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = $9.8m/s^2$

Putting values in above equation, we get:

$W=3kg\times 9.8m/s^2\\\\W=29.4N$

Hence, the weight of the object is 29.4 N

6 0

3 years ago

Ask Questions Action and reaction force pairs are all around you, but they aren't always obvious. Write down

NARA [144]

Answer:

Writing with a pencil. The pencil pushes on the paper. The paper pushes on the pencil.

Explanation:

Newton's third law.

5 0

2 years ago

A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled

hjlf

Answer:

1.84 m

Explanation:

For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.

So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,

a = g + rω²

= 9.8 m/s² + 1.10 m × (18.85 rad/s)²

= 9.8 m/s² + 390.85 m/s²

= 400.65 m/s²

Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.

v² = u² + 2a(h₂ - h₁)

So, v² - u² = 2a(h₂ - h₁)

h₂ - h₁ = (v² - u²)/2a

h₂ = h₁ + (v² - u²)/2a

substituting the values of the variables into the equation, we have

h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)

h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²

h₂ = 1.3 m + 0.54 m

h₂ = 1.84 m

7 0

2 years ago