PHYSICS HIGH SCHOOL

A ball is thrown into the air with an initial upward velocity of 60 ft/s. Its height (h) in feet after t seconds is given by the function h = –16t² + 60t + 6. What will the height be at t = 3 seconds?A.) 35 feet
B.) 40 feet
C.) 42 feet
D.) 45 feet

Answers

Answer 1
Answer: To answer the question, directly substitute the value given for t to the given function of height.
                     h = -16t² + 60t + 6
                     h = (-16)(3 s)² + (60)(3) + 6 = 42
Therefore, the height of the ball after 3 seconds is 42 ft. 
Answer 2
Answer:

Answer:

The height of the ball when it hits t = 3 seconds will be 42 feet.

Explanation:

The user above me provides a very helpful explanation, so I am only writing this in his/her/their honor for his/her/their work :D

h = -16t² + 60t + 6

h = (-16)(3 s)² + (60)(3) + 6 = 42

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Tamika inflates and ties off a balloon in her classroom and then takes it outside on a cold, winter day. Immediately, she notices that the balloon has gotten a little bit smaller. Why does this occur?A) because there are fewer molecules in the balloon outside
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Answer: Option (C) is the correct answer.

Explanation:

When the balloon was placed inside Tamika's classroom then the temperature was high as compared to the temperature outside the classroom.

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Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.200W/(m⋅K) ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1.60m in diameter having a layer of fat 3.90cm thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at 2.80∘C during hibernation, while the inner surface of the fat layer is at 30.9∘C a) What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 51.4W ? b) How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 51.4W ?

Answers

A) The temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 51.4W is; T_i = 38.52°C

B) The thickness of the layer contained within the fur so that the bear loses heat at a rate of 51.4 W is; t = 13.41 cm

We are given;

Diameter of sphere; d = 1.6 m

Radius of sphere; r = d/2

r = 1.6/2

r = 0.8 m

Thickness of bear; t = 3.9 cm cm = 0.039 m

Outer surface Temperature of fur; T_h = 2.8 ∘C

Inner surface Temperature of fat;T_f = 30.9 ∘C

Thermal conductivity of fat; K_f = 0.2 W/m⋅k

Thermal conductivity of air; K_a = 0.024 W/m⋅k

A) To find the temperature at the fat-inner fur boundary when heat loss is 51.4 W, we will use the heat current formula;

H = K_f•A(T_f - T_i)/t

Where;

A is area = 4πr²

A = 4π × 0.8²

A = 8.04 m²

T_i is the temperature we are looking for

H is heat loss = 51.4

t is thickness

Making T_i the subject gives;

T_i = (T_f × H × t)/(K_f × A)

T_i = (30.9 × 51.4 × 0.039)/(0.2 × 8.04)

T_i = 38.52°C

B) We want to find the thickness of the layer contained within the fur. Thus, we will use K_a instead of K_f. Let us make t the subject in the heat current formula to get;

t = (K_a•A(T_i - T_h)/H

t = (0.024 × 8.04 × (38.52 - 2.8))/51.4

t = 0.1341 m

t = 13.41 cm

Read more at; brainly.com/question/14548124

Answer:

Explanation:

Using the equation

H = Q/t = k A ( T hot - T cold) / L

where H is the rate of heat loss = 51.4 W, T cold be temperature of the outer surface, A is the surface area of the fat layer which is a model of sphere ( surface area of a sphere ) = 4πr² where diameter = 1.60 m

radius = 1.60 m / 2 = 0.80 m

A = 4 × 3.142 × ( 0.8²) = 8.04352 m²

making T cold subject of the formula

T cold =  T hot -   (HL)/(KA)  = 30.9° C - ( 51.4 W × 3.9 × 10⁻² m) / ( 0.2 W/mK × 8.04352 m² ) =  30.9° C - 1.25 ° C = 29.65° C

b) The thickness of air layer for the bear to lose heat t a rate of 51.4 W

thermal conductivity of air is 0.024 W/mK and rearranging the earlier formula

L = \frac{kA( T HOT - T COLD )}H} = (0.024 W/ m K × 8.04352 m²) ( 29.65° C - 2.8°C) / 51.4 W = 0.101 m = 10.1 m

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