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KengaRu [80]
3 years ago
8

A penny is dropped from the Statue of Liberty

Physics
1 answer:
Brrunno [24]3 years ago
5 0

Answer:

a) The velocity is 2.94m/s

b) 0.441

Explanation:

a) Assume gravity is 9.8m/s^2

Use the equation below to solve for the velocity at 0.30 seconds

vf=vi+at,

vf =unknown velocity  vi= initial velocity vi=0m/s  a= 9.8m/s^2 t=0.30seconds

Step 1: Substitute the variables with the knowns

vf=0m/s+(9.8m/s^2)*0.30seconds

Step 2: Solve

vf=2.94m/s

b)

Use the equation below to solve for the displacement at 0.30 seconds

x=vit+\frac{1}{2} at^{2}

Step 1: Substitute the same variables with the knowns

x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}

Note that vi*t=0 as vi=0m/s

Step 2: Solve

x=0.441m

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duplication being slightly more expensive.

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point) A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is
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Answer:

(a) 86.65 J

(b) 149.65 J

Solution:

As per the question:

Diameter of the pool, d = 12 m

⇒ Radius of the pool, r = 6 m

Height of the pool, H = 3 m

Depth of the pool, D = 2.5 m

Density of water, \rho_{w} = 1000\ kg//m^{3}

Acceleration due to gravity, g = 9.8\ m/s^{2}

Now,

(a) Work done in pumping all the water:

Average height of the pool, h = \frac{H + D}{2}

h = \frac{3 + 2.5}{2} = 2.75\ m

Volume of water in the pool, V = \pi r^{2}h = \pi \times 6^{2}\times 2.75 = 311.02\ m^{3}

Mass of water, m_{w} = \frac{\rho_{w}}{V}

m_{w} = \frac{1000}{311.02} = 3.215\ kg

Work done is given by the potential energy of the water as:

W = m_{w}gh = 3.215\times 9.8\times 2.75 = 86.65\ J

(b) Work done to pump all the water through an outlet of 2 m:

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Height, h = 2.75 + 2 = 4.75

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7 0
3 years ago
The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in
maria [59]

Answer: (a). Resistance = 0.4286ohms and Current (I) = 7A

(b). Resistance (R) = 0.027 ohms and Current (I) = 111.1A

(c). Resistance (R) = 0.1071 ohms and Current (I) = 28A

Explanation:

From the question, given that;

ρ = 1.5*10-2ῼ.m

Lo = 7cm = 0.07m

V = 3V

From the formula R = ρL/A, where A is the area of cross section, L is the length of material and ρ is the resistivity.

(A)

L = 4Lo and A = 2Lo*Lo

R = ρL/A

R = ρ4Lo/(2Lo*Lo)

R = 2ρ /Lo = 2*1.5*10-2/0.07

R = 0.4286 ῼ

From this the current becomes;

I = V/R = 3/ 0.4286 = 6.99 = 7A

(B)

L = Lo and A = 4Lo * 2Lo

R = ρL/A

R = ρLo/ (4Lo*2Lo) after eliminating Lo from both sides we get,

R = ρ/8Lo = 1.5*10-2 / 8*0.07

R = 0.027

Current (I) = V/R = 3/0.027 = 111.1A

(C)

L = 2Lo and A = Lo * 4Lo

R = ρL/A

R = ρ2Lo/ (Lo*4Lo) eliminating Lo from both sides we get,

R = ρ/2Lo = 1.5*10-2 / 2*0.07 = 0.1071

The current becomes;

I = V/R = 3/0.1071 = 28A

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