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Lostsunrise [7]
3 years ago
13

A ball is launched horizontally from the top of a cliff with an initial velocity of 20 m/s. If the ball strikes the ground after

five seconds, then (a) how high was the cliff and (b) how far from the base of the cliff will the ball land?
Physics
1 answer:
stich3 [128]3 years ago
6 0

Answer: you are the imposter you killed red then vented in admin

Explanation:

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aleksley [76]
Electric pumps are not useful they clog
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Pls help . Which process produces the energy that is used in photosynthesis?
levacccp [35]

Answer: Nuclear fusion

Explanation:

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3 years ago
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If a range hood removes contaminants at a flow rate of F liters of air per​ second, then the percent P of contaminants that are
podryga [215]

Explanation:

The given value of P is as follows.

           P = 1.06F + 22.18,           10 \leq F \leq 70

or,       P' = 1.06

As p' is defined and non-zero. Hence, only critical points are boundary points.

For F = 10, the value of P will be calculated as follows.

         P = 1.06 \times 10 + 22.18

            = 32.78

For F = 70, the value of P will be calculated as follows.

        P = 1.06 \times 70 + 22.18

           = 96.38

Therefore, the minimum value of P is 32.78 and maximum value of P is 96.38.

3 0
2 years ago
How do I solve the following problem?
lukranit [14]
Use the impulse-momentum theorem.

Ft = mv

Substitute your known values:

(120kg)(20m/s) = F(1.5)

F= 1600N

Hope this helps!
7 0
3 years ago
A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30 Pa moves through air which has a temperature
Fofino [41]

Answer:

Density change, Δρ = 2.4 × 10⁻⁴ kg/m³

Temperature Change, ΔT = 0.0258 K

Velocity Change, Δc = 0.0148 m/s

Explanation:

For sound waves moving through the air,

Pressure and Temperature varies thus

(P₀/P) = (T₀/T)^(k/(k-1))

Where P₀ = initial pressure of air = 101KPa = 101000 Pa

P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa

T₀ = initial temperature of air = 30°C = 303.15 K

T = final temperature of air = ?

k = ratio of specific heats = Cp/Cv = 1.4

(101000/101030) = (303.15/T)^(1.4/(1.4-1))

0.9990703 =(303.15/T)^(3.5)

Solving This,

T = 303.1758 K

ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K

Density can be calculate in two ways,

First method

Δρ = ρ - ρ₀

P₀ = ρ₀RT₀

ρ₀ = P₀/RT₀

R = gas constant for air = 287 J/kg.k

where all of these are values for air before the wave propagates

P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K

ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³

ρ = P/RT

P = 101030 Pa, T = 303.1758K

ρ = 101030/(287×303.1758) = 1.1611115 kg/m³

Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³

Second method

(ρ₀/ρ) = (T₀/T)^(1/(k-1))

Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.

Velocity Change

c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s

c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s

Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s

QED!

5 0
3 years ago
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