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Reika [66]
3 years ago
6

Which has more thermal energy, a 5-kg bowling all that has been resting on a hot driveway for 4 hours on a 350C day, or the same

bowling ball rolling down a lane in an air-conditioned bowling alley? Explain.
Physics
2 answers:
Leviafan [203]3 years ago
6 0

The bowling ball on the driveway has more thermal energy.  It has been absorbing the radiant energy from the sun during those four hours. It is gaining temperature because of getting heated up, In fact, it will burn your hands when you picked it up.  But the ball in second case it has kinetic energy which is different . At the bowling alley, the ball does gain a slight amount of heat from rolling, but it is not hot enough to hurt your hands.

balandron [24]3 years ago
4 0

Answer:

Case 1

Explanation:

Case 1: A 5-kg bowling ball that has been resting on a hot driveway for 4 hours on a 35°C day,

Case 2: A 5-kg bowling ball rolling down a lane in an air-conditioned bowling alley

The question asks which of the two cases above have a higher thermal energy.

Thermal energy refers to the internal energy of objects due to the kinetic energy of its atoms and/or molecules.

This means that there are greater thermal energy in the atoms or molecules of hot objects than cold objects

Using this illustration, we understand that the bowling ball in case 1 has been absorbing the radiant energy from the sun for four hours and apparently, it is hotter than the ball in case 2;

Conclusively, case 1 has a higher thermal energy than 2.

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what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe
ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

8 0
3 years ago
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Acceleration is defined as the rate of change for which of the following
Minchanka [31]
_Award brainliest if helped!
Velocity


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3 years ago
A 0.250 kg fan cart accelerates at 24 cm/ s^2 for 4.5 seconds. what is the net force (fan thrust minus drag and friction) on the
Norma-Jean [14]

Answer:

0.06 N

1.08 m/s

Explanation:

m = mass of the fan cart = 0.250 kg

a = acceleration of the fan cart = 24 cm/s² = 0.24 m/s²

F = Net force on the cart

Net force on the cart is given as

F = ma

F = (0.250) (0.24)

F = 0.06 N

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart

t = time interval = 4.5 s

Using the equation

v = v₀ + a t

v = 0 + (0.24) (4.5)

v = 1.08 m/s

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3 years ago
What is the name given to the initial 150 counts in 2 minutes?
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3 years ago
Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision
Sidana [21]

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

Learn more about conservation of momentum principle

brainly.com/question/14033058

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