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jek_recluse
3 days ago
15

Exactly one pound of bread dough is placed in a baking tin. The dough is cooked in an oven at 350°F, releasing a wonderful aroma

of freshly baked bread during the cooking process. Is the mass of the baked loaf greater than, less than, or the same as the one pound of original dough?
Physics
1 answer:
Softa [2.6K]3 days ago
7 0
The mass of the baked loaf will be lower than that of the dough.
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To avoid an accident, a driver steps on the brakes to stop a 1000-kg car traveling at 65km/h. if the braking distance is 35 m, h
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Initially, we need to calculate the acceleration required for the car to halt from its initial speed based on the distance traveled. This can be done using the formula,
                                             2ad = Vf² - Vi²
where a represents acceleration, d is distance, and Vf and Vi are the final and initial speeds respectively. Plugging in the known quantities,
           2(a)(35 m) = (0 m/s)² - ((65 km/h) x (1000 m/ 1 km) x (1 hr / 3600 s))²
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1 month ago
An arthroscope can be inserted into a knee joint through a very tiny incision to help a doctor examine the knee. A light is tran
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Example Response: The technology referenced is fiber optics. Their small size and flexibility allow doctors to visualize areas that would otherwise require surgical intervention to examine.

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17 days ago
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
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Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ \frac{du}{dy}

therefore

= µ \frac{v}{h}............1

substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

Area = \frac{\pi }{4} d^{2}

area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

3 0
1 month ago
A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha
serg [3228]

Answer:

a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}

Explanation:

Data provided

initial velocity v₀=20 cm/s at time t=3s

final velocity vf=0 at time t=8 s

Required

Average Acceleration for the interval from 3s to 8s

Solution

Acceleration can be defined as the first derivative of velocity concerning time

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Answer:

The potential energy tied to the charge diminishes.

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Explanation:

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