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Evgesh-ka [11]
4 years ago
15

Determine the amount of work done when a crane lifts a 100-N block form 2m above the ground to 6m above the ground

Physics
1 answer:
Dima020 [189]4 years ago
4 0

Data given:

F=100N

Δx=6m-2m=4m

A=?

Formula needed:

A=F×s

Solution:

A=100N×4m

A=400J

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A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b
musickatia [10]

Answer:

P = 17.28*10⁶ N

Explanation:

Given

L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

A = a*b = (0.54 m)*(0.40 m) = 0.216 m²

then, we apply the formula

ΔL = P*L/(A*E)  ⇒ P = ΔL*A*E/L

⇒  P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒  P = 24624000  N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒  σ = (24624000  N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

So σ > σmax  we use σmax

⇒  P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

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3 years ago
describe the physics modeling process in moving a basketball and say what effects you should overlook​
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Why is the cycle of cellular respiration and photosynthesis important to life for plants
devlian [24]
The cycle of cellular respiration and photosynthesis are important for the life of plants because cellular respiration uses one of the products of photosynthesis (oxygen), and uses it as one of the reactants along with glucose to produce carbon dioxide. These two cycles help maintain a balance for our atmosphere, even though pollution and other factors have disrupted this balance. Hope this helps! :)
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A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106
LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

E_2 = \frac{E_1q_1}{q_2}

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

8 0
3 years ago
Consider two tubes filled with water at the same height, one with fresh water and the other tube with salt water. The pressure i
Olegator [25]

Answer:

B

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3 0
3 years ago
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