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2 years ago

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A cliff diver on an alien planet dives off of a 32 meter tall cliff and lands in a sea of hydrochloric acid 1.20 seconds later.

Assuming that the extra-terrestrial’s initial speed was zero, what is the free fall acceleration of this strange world?

1 answer:

Cloud [144]2 years ago

3 0

Answer:

44.4m/s^2

Explanation:

Use the formula...S = ut + 1/2at^2

where...S = 32m...u = 0m/s....t = 1.20s

32 = (0)(1.20) + 0.5(1.20^2)a

;Acceleration of free fall = 44.4m/s^2

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In 1923, the United States Army (there was no U.S. Air Force at that time) set a record for in-flight refueling of airplanes. Us

dybincka [34]

Answer:

1.95m/s

Explanation:

Please view the attached file for the detailed solution.

The following were the conversion factors used in order to express all quatities in SI units:

$1 gallon=0.00378541m^3\\1 inch=0.0254m\\1 minute=60s$

6 0

2 years ago

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac

GREYUIT [131]

Answer:

The flux through the surface of the cube is $2.314\ Nm^{2}/C$

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = $8.0\times 10^{- 2}\ m$

Volume Charge density, $\rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}$

Now,

To calculate the electric flux:

$\phi = \frac{q}{\epsilon_{o}}$ (1)

where

$\phi$ = electric flux

$\epsilon_{o} = 8.85\times 10^{- 12}\ F/m$ = permittivity of free space

Volume Charge density for the given case is given by the formula:

$\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}$ (2)

Volume of cube, $V = a^{3}$

Thus

$V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}$

Thus from eqn (2), the total charge is given by:

$q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}$

$q = 2.048\times 10^{-11}\ F = 20.48\ pF$

Now, substitute the value of 'q' in eqn (1):

$\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C$

5 0

2 years ago

An elementary particle of mass m completely absorbs a photon, after which its mass is 1.01m. (a) what was the energy of the inco

sdas [7]

A.) We use the famous equation proposed by Albert Einstein written below:

E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
c is the speed of light equal to 3×10⁸ m/s

Substituting the value:

E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules

b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.

8 0

2 years ago

A measuring cylinder contains 60cm3 of oil at 0 celcius. When a piece of ice was roped into the cylinder it sank completely in o

mariarad [96]

Answer:

$S_i=\frac{9}{10} =0.9$

Explanation:

Given:

volume of oil in the cylinder, $V_o=60\ cm^2$

volume of the oil level when the ice is immersed, $V=90\ cm^3$

the volume level of oil when the ice melted, $V'=87\ cm^3$

<u>Now, therefore the volume of ice:</u>

$V_i=V-V_o$

$V_i=90-60$

$V_i=30\ cm^3$

<u>Now the volume of water:</u>

$V_w=V'-V_o$

$V_w=87-60$

$V_w=27\ cm^3$

As we know that the relative density is the ratio of density of the substance to the density of water.

<u>So, the relative density of ice:</u>

$S_i=\frac{\rho_i}{\rho_w}$ .....................(1)

as we know that density is given as:

$\rm \rho=\frac{mass}{volume}$

now eq. (1)

$S_i=\frac{m}{V_{i}}\div \frac{m}{V_w}$

where, m = mass of the water or the ice which remains constant in any phase

$S_i=\frac{V_w}{V_i}$

$S_i=\frac{27}{30}$

$S_i=\frac{9}{10} =0.9$

7 0

2 years ago

A steel cable 1.25 in. in diameter and 50 ft long is to lift a 20-ton load without permanently deforming. What is the length of

Over [174]

Answer:

50.0543248872 ft

Explanation:

F = Load = 20 ton = $20\times 2000\ lb$

d = Diameter = 1.25 in

$L_1$ = Initial length = 50 ft

$L_2$ = Final length

A = Area = $\dfrac{\pi}{4}d^2$

Y = Young's modulus = $30\times 10^6\ psi$

Young's modulus is given by

$Y=\dfrac{FL}{A\Delta L}\\\Rightarrow Y=\dfrac{FL_1}{\dfrac{\pi}{4}d^2(L_2-L_1)}\\\Rightarrow L_2=\dfrac{4FL_1}{Y\pi d^2}+L_1\\\Rightarrow L_2=\dfrac{4\times 40000\times 50}{30\times 10^6\times \pi\times 1.25^2}+50\\\Rightarrow L_2=50.0543248872\ ft$

The length during the lift is 50.0543248872 ft

6 0

2 years ago