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meriva
3 years ago
6

For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro

m 1828 to1892 to make lead shot for pistols and rifles and molded shot for cannons and other instruments of warfare. Molten lead was dropped from the top of the 82.15 m tall tower into a vat of water. During its free fall, the lead would form a perfectly spherical droplet and solidify. Determine the velocity of the droplet right before it hits the ground.
Physics
1 answer:
Mariulka [41]3 years ago
5 0

Answer:

The velocity of the droplet right before it hits the ground is 40.08 m/s.

Explanation:

To determine the velocity of the droplet right before it hits the ground,

From one of the equations of kinematic for free fall motions,

v = u + gt

Where v is the final velocity

u is the initial velocity

g is acceleration due to gravity (take g = 9.8 m/s²)

and t is time

For the question, v is the velocity of the droplet right before it hits the ground.

u = 0 m/s (Since the molten lead was dropped from rest)

Therefore,

v = gt

First, we will determine the time t

Also, from one of the equations of kinematic for free fall motions,

h = ut + 1/2(gt²)

u = 0 m/s

From the question, the molten lead was dropped from the top of the 82.15 m tall tower, therefore

h = 82.15

Hence,

82.15 = 0×t + 1/2 (9.8 × t²)

82.15 = 1/2 (9.8 × t²)

82.15 = 4.9 t²

t² = 82.15/4.9

∴ t = 4.09 secs

Now, for the velocity v, of the droplet right before it hits the ground,

Recall

v = gt

Then,

v = 9.8 × 4.09

v = 40.08 m/s

Hence, the velocity of the droplet right before it hits the ground is 40.08 m/s.

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natta225 [31]
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5- Transferred, collide
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5 0
3 years ago
Suppose that the average U.S. household uses 15600 kWh (kilowatt‑hours) of energy in a year. If the average rate of energy consu
sweet [91]

Answer:

a) 166.4 s

b) (2.155 × 10⁷) s

Explanation:

15600 KWh for a year,

1 year consists of 365 × 24 hours = 8760 hours.

So, the power consumed in a year for an average household = (Energy/time)

= (15600/8760) = 1.781 KW = 1781 W

a) If the average rate of energy consumed by the house was instead diverted to lift a 1.80 × 10 3 kg car 16.8 m into the air, how long would it take

The power required for this lifting = (mgh/t)

m = 1800 kg

g = 9.8 m/s²

h = 16.8 m

t = ?

P = 1781 W

1781 = (1800×9.8×16.8)/t

t = (296,352/1781)

t = 166.4 s

b) how long would it take to lift a loaded Boeing 747 airplane, with a mass of 4.05 × 10 5 kg , to a cruising altitude of 9.67 km

The power required for this lifting = (mgh/t)

m = 405000 kg

g = 9.8 m/s²

h = 9.67 km = 9670 m

t = ?

P = 1781 W

1781 = (405000×9.8×9670)/t

t = (38,380,230,000/1781)

t = 21,549,820 s = (2.155 × 10⁷) s

Hope this Helps!!!

8 0
3 years ago
What is Average rainfall in the desert?
Vadim26 [7]

Answer:

less than 10 inches, or 25 centimeters, of precipitation a year.

Explanation:

PS. the precipitation recieved in the desert can be in the form of either rain or snow.

7 0
3 years ago
Read 2 more answers
A) Define rest and motion?​
7nadin3 [17]

Answer:

Rest - a body is said to be at rest, if it does not change its position with respect to its surrounding with time. Motion - a body is said to be at motion, if it changes its position with time.

5 0
3 years ago
A point charge Q is located a distance d away from the center of a very long charged wire. The wire has length L >> d and
zzz [600]

Answer:

F = \frac{Qq}{2\pi \epsilon_0 L d}

Explanation:

As we know that if a charge q is distributed uniformly on the line then its linear charge density is given by

\lambda = \frac{q}{L}

now the electric field due to long line charge at a distance d from it is given as

E = \frac{2k\lambda}{d}

E = \frac{q}{2\pi \epsilon_0 d}

now the force on the other charge in this electric field is given as

F = QE

F = \frac{Qq}{2\pi \epsilon_0 L d}

5 0
4 years ago
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