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3 years ago

11

A ball is dropped from a 150-m tall building. Neglecting air resistance, what will the speed of the ball be when it reaches the

ground 5.5 seconds later?

1 answer:

Lerok [7]3 years ago

7 0

54 m/s is the answer

You might be interested in

A rock is dropped off a cliff and falls the first half of the distance to the ground in 2.0 seconds. how long will it take to fa

son4ous [18]

Answer:

The answer is 0.83 seconds.

Explanation:

The formula of free fall is following:

$h=1/2*g*t^2$

Where g=9.8 m/s^2 and t=2 seconds, the rock takes:

$h=1/2*9.8*2^2=19.6$

19.6 meters. This is the half distance of the cliff. The whole distance is 39.2 meters. So it takes:

$39.2=1/2*9.8*t^2\\t^2=8\\t=2.83$

2.83 second to fall down completely. The rock takes the second half of the cliff in 0.83 seconds

8 0

3 years ago

A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca

igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

$f=mg\ sin\theta$

$f=1100\ kg\times 9.8\ m/s^2\ sin(4)$

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0

3 years ago

PLS HELP ME!!! Explain frame of reference. How is it different when you are riding inside a car or standing by the highway?

RSB [31]

Answer:

So when you are riding in a car everything you see is from the cars frame. Where if you are standing by the highway it is from your point of view.

Explanation:

So lets say someone sees a apple core outside the bus it looks like the apple is traveling the opposite way of the bus but to the man standing by the highway would say it is stationary.

4 0

3 years ago

What is a process that returns to its beginning and repeat it self in the same sequence

jasenka [17]

Answer:

a sequence

Explanation: was that an option?, or was it not a multiple choice question ?

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