Answer: D(t)= 50(4/5)^t
Explanation: If 1/5 of the temperature difference is lost each minute, that means 4/5 of the difference remains each minute. So each minute, the temperature difference is multiplied by a factor of 4/5 (or 0.8).
If we start with the initial temperature difference, 50° Celsius, and keep multiplying by 4/5, this function gives us the temperature difference t minutes after the cake was put in the cooler.
The answer is
Pitch of the buzzer increased (higher tone) as it moves towards the observer
Answer:
temperature and mass
Explanation:
- The higher the temperature of a given quantity of a substance, more is its thermal energy.
- If a substance contains more mass, this also implies that the object has more particles in it . hence, it has high thermal energy.
<em><u>A</u></em><em><u>d</u></em><em><u>d</u></em><em><u>i</u></em><em><u>t</u></em><em><u>i</u></em><em><u>o</u></em><em><u>n</u></em><em><u>a</u></em><em><u>l</u></em><em><u> </u></em><em><u>I</u></em><em><u>n</u></em><em><u>f</u></em><em><u>o</u></em><em><u>r</u></em><em><u>m</u></em><em><u>a</u></em><em><u>t</u></em><em><u>i</u></em><em><u>o</u></em><em><u>n</u></em><em><u> </u></em>:
- Temperature is a measure of the average kinetic energy of the particles of a substance.
- The thermal energy of an object depends on three factors:
- number of molecules in the object
- temperature of the object.
- thermal energy it has.
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
