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Vikentia [17]
3 years ago
13

What happens to the Connecticut Energy of a snowball as it rolls across the lawn and gains mass?

Physics
1 answer:
statuscvo [17]3 years ago
6 0
I think it will go down like decrease minus or whatever you call it
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Why do slippery fluids such as oil reduce sliding friction
Gwar [14]
   fluid friction<span> occurs when an object moves through a liquid or gas. the force needed to overcome </span>fluid friction <span>is usually less then that needed to overcome </span>sliding friction<span>. the </span>fluid<span> keeps the surface from making direct contact and thus </span>reducing friction<span>.</span>
6 0
3 years ago
. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

8 0
3 years ago
Analyze how buffers allow you to eat acidic and basic foods without changing your blood pH.
blondinia [14]
Buffers neutralize the acid and the bases
7 0
3 years ago
47. A car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. What
Otrada [13]

Answer: 114 km/h

Explanation:

The formula for determining average speed is expressed as

Average speed = total distance/total time

The car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. It means that the total distance that the car travels is

85 + 200 = 285 km

The total time spent by the car is

0.5 + 2 = 2.5 hours

Therefore,

Average speed = 285/2.5 = 114 km/h

3 0
3 years ago
Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y
WITCHER [35]

Question:

A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm  from the center of the cavity.  

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 3.65\times 10^5N/C

Explanation:

A point charge ,q = -2.14\times 10^{-6} C is located in the center of a spherical cavity of radius , r =6.55\times 10^{-2}  m inside an insulating spherical charged solid.  

The charge density in the solid , d = 7.35 \times 10^{-4}C/m^3.

Distance from the center of the cavity,R =9.5\times 10^{-2 }m

Volume of shell of charge= V  =(\frac{4\pi}{3})[ R^3 - r^3 ]

Charge on the shell ,Q = V \times d'

Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d

Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}

Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}

Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}

Q =1.7745 \times 10^{-6 }C

Electric field at 9.5\times 10^{-2}m due to shellE1  = \frac{k Q}{R^2}

E1 =  \frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}

E1 =1.769\times 10^6 N/C

Electric field at  9.5\times 10^{-2} due to 'q' at center E2 = \frac{kq}{R^2}

E2 =\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}

E2 =2.134\times 10^6 N/C

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

=[  2.134  - 1.769 ]\times 10^6

= 3.65\times 10^5 N/C

8 0
3 years ago
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