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3 years ago

Convert 68 centimeters into feet.

Physics

2 answers:

FromTheMoon [43]3 years ago

6 0

Answer:

1 foot

Explanation:

your answer is 1.6

andreev551 [17]3 years ago

3 0

Answer:

Explanation:

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A shell is fired from the ground with an initial speed of 1.74 ✕ 103 m/s at an initial angle of 58° to the horizontal.

ValentinkaMS [17]

Answer:

a) horizontal range s=277671.77 m

b) time the shell is in motion 301.143 s

Explanation:

Is a parabolic movement so the velocity have two components:

$v = 1.74 x 10^{3} ( \frac{m}{s} )$

$\alpha = 58$ °

$v_{y} =v*Sen (\alpha )\\v_{x} =v*Cos (\alpha )$

$v_{y} =1740*Sen (58 )\\v_{x} =1740*Cos (58)$

$v_{y} = 1475.603 \frac{m}{s}\\v_{x} = 922.059 \frac{m}{s}\\$

$t= \frac{2*v_{y} }{g}$

$t= \frac{2*1475.603 }{9.8}$

$t= 301.143 s$

$v= \frac{s}{t} \\s= v_{x}*t$

$s= 922.059 \frac{m}{s} * 301.143 s$

$s = 277671.77 m$

4 0

4 years ago

Plates move apart at ____ boundaries.

Keith_Richards [23]

<span>The correct answer is C. divergent. Divergent boundaries are something like lines found between two tectonic plates. These constructive boundaries, or extensional boundaries, are formed and exist between two tectonic plates that move away from each other. Convergent would thus be where they meet. This is all according to the tectonic theory.</span>

5 0

3 years ago

Read 2 more answers

A digital stop-clock measures time in minutes and seconds.

ad-work [718]

Answer:

01:20 (1 minute 20 seconds)

Explanation:

subtract the starting time from the stopping time to get the time used: 02:10 - 00:50 = 01:20

4 0

3 years ago

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 2.33 m away from a waterfall 0.488 m in heigh

Ulleksa [173]

Answer:

5.68 m/s

Explanation:

The motion of the salmon is the same as a projectile: it is launched with an initial speed $u$ at an angle of $\theta=38^{\circ}$ above the horizontal.

The motion of the salmon consists of two indipendent motion:

- Along the horizontal direction, it is a uniform motion with constant velocity

$v_x = u cos \theta$

So that the distance travelled is

$d=v_x t = u cos \theta t$ (1)

- Along the vertical direction, it is a uniformly accelerated motion with constant acceleration downward, so the vertical displacement is

$y = u sin \theta t - \frac{1}{2}gt^2$ (2)

where g is the acceleration of gravity.

We know the following:

- The horizontal distance travelled by the salmon to reach the waterfall is

d = 2.33 m

- The vertical distance travelled is the height of the waterfall,

y = 0.488 m

From (1) we get:

$t=\frac{d}{u cos \theta}$

And substituting into (2), we can solve the equation to find t, the time at which the salmon reaches the waterfall:

$y = u sin \theta \frac{d}{u cos \theta} - \frac{1}{2}gt^2 = d tan \theta - \frac{1}{2}gt^2\\t = \sqrt{\frac{2(d tan \theta - y)}{g}}=\sqrt{\frac{2(2.33 tan 38^{\circ}-0.488)}{9.81}}=0.521 s$

And then, we can use eq.(1) again to find the initial speed, u:

$u=\frac{d}{cos \theta t}=\frac{2.33}{cos(38^{\circ})(0.521)}=5.68 m/s$

5 0

3 years ago

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$