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3 years ago

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During heavy rain, a section of a mountainside measuring 2.3 km horizontally (perpendicular to the slope), 0.75 km up along the

slope, and 1.4 m deep slips into a valley in a mud slide. Assume that the mud ends up uniformly distributed over a surface area of the valley measuring 0.52 km x 0.52 km and that the mass of a cubic meter of mud is 1900 kg. What is the mass of the mud sitting above a 5.5 m2 area of the valley floor?

1 answer:

Jobisdone [24]3 years ago

6 0

Answer:

93328 kg

Explanation:

$\rho$ = Density of mud = 1900 kg/m³

Volume of the cuboid is given by

$V=2300\times 750\times 1.4\\\Rightarrow V=2415000\ m^3$

Volume is also given by

$V=Al\\\Rightarrow l=\dfrac{V}{A}\\\Rightarrow l=\dfrac{2415000}{520\times 520}\\\Rightarrow l=8.9312\ m$

The length of the cuboid is 0.89312 m

For the small volume

$V_a=al\\\Rightarrow V_a=5.5\times 8.9312\\\Rightarrow V_a=49.12\ m^3$

Mass is given by

$m=\rho V_a\\\Rightarrow m=1900\times 49.12\\\Rightarrow m=93328\ kg$

The mass of the mud is 93328 kg

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A car is going south on I-69 at 33 m/s (74 mph). The car has good brakes so its maximum braking acceleration is – 8.5 m/s^2 . Tr

dmitriy555 [2]

Answer:

1.69515 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-33^2}{2\times -8.5}\\\Rightarrow s=64.06\ m$

The distance between the traffic and the car after braking is 120-64.06 = 55.94 m

Time = Distance / Speed

$\text{Time}=\frac{55.94}{33}\\\Rightarrow \text{Time}=1.69515\ seconds$

The reaction time cannot be more than 1.69515 seconds

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3 years ago

Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra

Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

$\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h$

with this time, we know can now calculate the distance at which B is:

$\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi$

and applying Pythagoras:

$\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}$

Now substituting all the values in (3) and solving for $\displaystyle{\frac{ds}{dt} }$ we get:

$\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}$

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3 years ago

Estimate frequency of vibration of your arm. Let the length of the arm be 0.57 m. Consider the arm as a simple pendulum and assu

skad [1K]

Answer:

0.80865 Hz

1.23662 seconds

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

l = Length of arm = 0.57 m

Length of simple pendulum is given by

$L=\dfrac{2}{3}l\\\Rightarrow L=\dfrac{2}{3}\times 0.57\\\Rightarrow L=0.38\ m$

The frequency is given by

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.81}{0.38}}\\\Rightarrow f=0.80865\ Hz$

The frequency is 0.80865 Hz

The time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{0.80865}\\\Rightarrow T=1.23662\ s$

The time period is 1.23662 seconds

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Zigmanuir [339]

Answer:

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Explanation:

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