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Lina20
1 month ago
7

On a day when the water is flowing relatively gently, water in the Niagara River is moving horizontally at 4.5 m/sm/s before sho

oting over Niagara Falls. After moving over the edge, the water drops 53 mm to the water below.
a. If we ignore air resistance, how much time does it take for the water to go from the top of the falls to the bottom?
b. Express your answer to two significant figures and include the appropriate units.
c. How far does the water move horizontally during this time?
d. Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Maru [3.3K]1 month ago
7 0

Answer:

a.3.29 m/s

b.3.3 m/s

c.14.8 m

d.15 m

Explanation:

We have the following specifics:

Initial Horizontal speed=v_x=4.5 m/s

Vertical component of initial speed=v_y=0

Vertical distance=y=-53 m

a.s=ut+\frac{1}{2}gt^2

Applying the formula and considering g as negative results in g=-9.8m/s^2

-53=0-\frac{1}{2}(9.8)t^2

53=4.9t^2

t^2=\frac{53}{4.9} s

t=\sqrt{\frac{53}{4.9}}=3.29m/s

b.Since the hundredths place exceeds 5, 1 is added to the tenths place, while other digits to the left remain unchanged and the digit to the right is replaced by 0.

t=3.3 m/s

c.Horizontal acceleration=a_x=0

x=v_xt=4.5\times 3.29=14.8 m

d.The digit in the tenths place at 8 surpasses 5, thus adding 1 to the units place, with all other digits on the left side unchanged and the right-side digit altered to 0.

Horizontal distance=15 m

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3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag
serg [3582]

Response:

(a) W=-19.25J

(b) W=-52.8J

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J

This value is negative as it expands against the given pressure.

(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:

W=nRTln(\frac{V_1}{V_2} )

The moles are calculated based on the provided mass of argon:

n=6.56g*\frac{1mol}{39.95g}=0.164mol

Consequently, the work amounts to:

W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J

Best regards.

4 0
18 days ago
Taylor places a nail on a bar magnet. The nail sticks to the magnet when lifted up off the table. She touches a paperclip to the
ValentinkaMS [3465]
Initially, the magnetic domains within the nail were oriented in various directions before coming into contact with the bar magnet. Upon Taylor touching the nail to the bar magnet, the magnetic fields of those domains became aligned, thus transforming the nail into a temporary magnet.
5 0
1 month ago
Read 2 more answers
Which letter correctly identifies the part of the hydrologic cycle that is most directly affected by impervious building materia
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Infiltration

Explanation:

The feature of the hydrologic cycle most impacted by impervious surfaces like concrete and asphalt is infiltration.

  • Infiltration is crucial within the hydrologic cycle.
  • Concrete and similar materials hinder water absorption into the ground.
  • This adversely affects existing groundwater systems.
  • A rise in surface runoff can occur, leading to potential flooding.
  • Infiltration plays a pivotal role in the water cycle.
  • It supplies water to plant roots and replenishes groundwater reserves.
  • Impervious surfaces disrupt this natural process.

learn more:

Biogeochemical cycle

3 0
1 month ago
In the past, salmon would swim more than 1130 km (700 mi) to spawn at the headwaters of the Salmon River in central Idaho. The t
Keith_Richards [3271]

Answer:

E_t_o_t_a_l=7.603MJ

Explanation:

The overall energy expenditure of the salmon, which corresponds to its swimming upstream effort, W, is linked to its specific mechanical power. Mechanical \ power calculated per unit mass can be derived from the following equation:

\frac{p}{m}=\frac{1}{m}|\frac{dW}{dt}|=2W/kg\\\frac{1}{2}|\frac{W}{22\times 24 \times 60\times 60}|=2\\\\W=7.603MJ

As a result, the total energy utilized during the 22-day journey is 7.603 MJ

6 0
1 month ago
slader A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of
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Response: 800N

Clarification:

Provided data:

Ball mass = 0.8kg

Contact duration = 0.05 seconds

Final and initial speed = 25m/s

The average force exerted by the ball on the wall can be calculated using the following relationship:

Force (F) = mass (m) * average acceleration (a)

a= (initial velocity (u) + final velocity (v))/t

m = 0.8kg

u = v = 25m/s

t = contact time of the ball = 0.05s

Thus,

a = (25 + 25) ÷ 0.05 = 1000m/s^2

Hence,

The average force magnitude (F)

F=ma

m = ball mass = 0.8

a = 1000m/s^2

F = 0.8 * 1000

F = 800N

6 0
1 month ago
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