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2 years ago

If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?

1 answer:

6 0

The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
$g=9.81 m/s^2$

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
$y(t) = \frac{1}{2}gt^2$
And if we substitute t=2.7 s, we find the distance covered:
$y(t)= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m$

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You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg i

Kisachek [45]

Answer:

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

Initial work done by the spring, where c is the compression = 0.28m:

$W_s = \frac{1}{2}kc^2$

Work lost to friction:

$W_f =\mu mg(c-x)$

Energy:

$E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2$

(a) Solve for v:

$v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}$

(b) Solve $\frac{dv}{dx}=0$ for x:

$\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}$

$\frac{dv}{dx}=0$ if:

$\mu g-\frac{k}{m}x = 0$

$x_{max} = \frac{\mu gm}{k}$

$v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}$

6 0

3 years ago

A hot-air balloon plus cargo has a mass of 308 kg and a volume of 2910 m3 on a day when the outside air density is 1.22 kg/m3. T

lesya [120]

9514 1404 393

Answer:

1.114 kg/m³

Explanation:

The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...

2910d +308 = 2910×1.22

Solving for d, we find ...

2910d = 2919(1.22) -308

d = 1.22 -308/2910

d ≈ 1.114 . . . kg/m³

The density of the hot air is about 1.114 kg/m³.

6 0

3 years ago

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent

tensa zangetsu [6.8K]

Answer:

(a): The car's relative position to the base of the cliff is x= 32.52m.

(b): The lenght of the car in the ir is tfall= 1.78 sec.

Explanation:

Vo= 0

V= ?

d= 50m

h= 30m

a= 4 m/s²

t= √(2*d/a)

t= 5 sec

V= a*t

V= 20 m/s

Vx= V * cos(24º)

Vx= 18.27 m/s

Vy= V* sin(24º)

Vy= 8.13 m/s

h= Vy*t + g*t²/2

clearing t:

tfall= 1.78 sec (b)

x= Vx * tfall

x= 32.52 m (a)

4 0

3 years ago

Calculate the wavelength of a photon having 3.26 x 10^-19 joules of energy

bekas [8.4K]

The energy of a photon is given by:
$E=hf$
where h is the Planck constant and f is the photon frequency.
We know the energy of the photon, $E=3.26 \cdot 10^{-19} J$ , so we can rearrange the equation to calculate the frequency of the photon:
$f= \frac{E}{h}= \frac{3.26 \cdot 10^{-19}J}{6.6 \cdot 10^{-34}Js}=4.94 \cdot 10^{14}Hz$

And now we can use the following relationship between frequency f, wavelength $\lambda$ and speed of light c to find the wavelength of the photon:
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{4.94 \cdot 10^{14} Hz}=6.07\cdot 10^{-7} m=607 nm$

8 0

2 years ago

The space shuttle fleet was designed with two booster stages. If the second stage provides a thrust of 73 kilo-newtons and the

Tresset [83]

Answer:

m = 81281.5 pounds.

Explanation:

Given that,

Force, F = 73 kN

Acceleration of the space shuttle, a = 16000 mi/h²

1 miles/h² = 0.0001241 m/s2

16000 mi/h² = 1.98 m/s²

We need to find the mass of the spacecraft.

According to Newton's second law,

F = ma

m is mass of the spacecraft

$m=\dfrac{F}{a}\\\\m=\dfrac{73\times 10^3\ N}{1.98\ m/s^2}\\\\m=36868.68\ kg$

Since, 1 kg = 2.20462 pounds

m = 81281.5 pounds

Hence, the mass of the spacecraft is 81281.5 pounds.

8 0

2 years ago