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3 years ago

Confirm if this is correct or not. If it isn't correct, please correct it.

Physics

1 answer:

kow [346]3 years ago

3 0

Answer:

d = 421.83 m

Explanation:

It is given that,

Height, h = 396.9 m

Horizontal speed, v = 46.87 m/s

We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,

$396.9=0\times t+\dfrac{1}{2}\times 9.8 t^2\\\\t=9\ s$

Now d is the distance covered by the cannonball. So,

$d=vt\\\\d=46.87\times 9\\\\d=421.83\ m$

Hence, this is the required solution.

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If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?

Paraphin [41]

Answer:

Option C. 70 Ω

Explanation:

Data obtained from the question include:

Resistor (R) = 20 Ω

From diagram given ABOVE, we observed the following

1. R and R are in parallel connections.

2. 2R and 2R are in parallel connections.

3. 4R and 4R are in parallel connections.

Next, we shall determine the equivalent resistance in each case.

This is illustrated below:

1. Determination of the equivalent resistance for R and R parallel connections.

R = 20 Ω

Equivalent R = (R×R) /(R+R)

Equivalent R = (20 × 20) /(20 + 20)

Equivalent R = 400/40

Equivalent R = 10 Ω

2. Determination of the equivalent resistance for 2R and 2R parallel connections.

R = 20 Ω

2R = 2 × 20 = 40 Ω

Equivalent 2R = (2R×2R) /(2R+2R)

Equivalent 2R = (40 × 40) /(40 + 40)

Equivalent 2R = 1600/80

Equivalent 2R = 20 Ω

3. Determination of the equivalent resistance for 4R and 4R parallel connections.

R = 20 Ω

4R = 4 × 20 = 80 Ω

Equivalent 4R = (4R×4R) /(4R+4R)

Equivalent 4R = (80 × 80) /(80 + 80)

Equivalent 4R = 6400/160

Equivalent 4R = 40 Ω

Thus, the equivalence of R, 2R and 4R are now in series connections. We can obtain the equivalent resistance in the circuit as follow:

Equivalent of R = 10 Ω

Equivalent of 2R = 20 Ω

Equivalent of 4R = 40 Ω

Equivalent =?

Equivalent = Equivalent of (R + 2R + 4R)

Equivalent = 10 + 20 + 40

Equivalent = 70 Ω

Therefore, the equivalent resistance between point A and B is 70 Ω.

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A hollow cast-iron cylinder 4m long, 300mm outer diameter, and thickness of metal 50mm is subjected to a central load on the top

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Here, the calculated Magnitude of the load P is 2945.2 kN, the Longitudinal strain produced is 0.0005 and the decrease in length is 2 mm.

Given,

Length, L = 4 m

Outer diameter, D = 300mm, D= 0.3 m

Thickness, t = 50 mm, t = 0.05 m

Stress produced, σ = 75000 kN/m²

Young's modulus for cast iron, E = 1.5 x 10⁸ kN/m²

Calculating the diameter of the cylinder,

Diameter of cylinder, d = (D) – (2t) = 0.3 –( 2 × 0.05)

d= 0.2 m

(i) Magnitude of the load P:

Using the relation, σ =P/A

P = σ × A = 75000 × π /4 (D² – d² )

P= 75000 × π/4 (0.3² – 0.2²)

P= 75000 × π/4 (0.09 – 0.04)

P = 2945.2 kN

Hence, Magnitude of the load P is 2945.2 kN.

(ii) Longitudinal strain produced, e :

Using the relation, Strain, (e) = stress/E

e= 75000/(1.5 x 10⁸)= 0.0005

Hence, the Longitudinal strain produced is 0.0005.

(iii)Total decrease in length, dL:

The total decrease in length can be calculated using the strain as the ratio of change in length to the original length is known as Strain.

Strain = change in length/original length

e= dL/L

0.0005 = dL/4

dL = 0.0005 × 4m = 0.002m=2mm

Hence,the decrease in length is 2 mm.

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