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4 years ago

15

Is molten lava heterogeneous

1 answer:

zhenek [66]3 years ago

4 0

Yes, considering lava accommodates rocks that have collapsed of cave walls it is heterogeneous.

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Two cars, one of mass 1100 kg, and the second of mass 2500 kg, are moving at right angles to each other when they collide and st

Pie

Answer:

$v_{f}=17.47 m/s$

Explanation:

Let's use the conservation of momentum to solve it.

$p_{initial}= p_{final}$ (1)

The total initial momentum will be: $m_{1}v_{1i}+m_{2}v_{2i}$
The total final momentum will be: $m_{1}v_{1f}+m_{2}v_{2f}$ , but as they stick together after the collision, v1f = v2f = vf.

So we can rewrite (1), using the above information:

$m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{f}+m_{2}v_{f}$

$m_{1}v_{1i}+m_{2}v_{2i}=v_{f}(m_{1}+m_{2})$

$v_{f}=\frac{m_{1}v_{1i}+m_{2}v_{2i}}{m_{1}+m_{2}}$

$v_{f}=\frac{1100\cdot 14+2500\cdot 19}{1100+2500}$

Finally, the magnitude of the velocity of the wreckage of the two cars immediately after the collision is:

$v_{f}=17.47 m/s$

I hope it helps you!

5 0

3 years ago

What term is used for a weather condition characterized by lack of precipitation over a long period of time?

Marrrta [24]

Answer:

Explanation:

<h2><u><em>Drought</em></u></h2>

4 0

3 years ago

Read 2 more answers

Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at

dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

The amplitude of the lateral force is $F = 25 \ N$

The frequency is $f = 1 \ Hz$

The mass of the bridge per unit length is $\mu = 2000 \ kg /m$

The length of the central span is $d = 144 m$

The oscillation amplitude of the section considered at the time considered is $A = 75 \ mm = 0.075 \ m$

The time taken for the undriven oscillation to decay to $\frac{1}{e}$ of its original value is t = 6T

Generally the mass of the section considered is mathematically represented as

$m = \mu * d$

=> $m = 2000 * 144$

=> $m = 288000 \ kg$

Generally the oscillation amplitude of the section after a time period t is mathematically represented as

$A(t) = A_o e^{-\frac{bt}{2m} }$

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question

$\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=> $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

=> $e^{-1} =e^{-\frac{b6T}{2m} }$

=> $-\frac{3T b}{m} = -1$

=> $b = \frac{m}{3T}$

Generally the amplitude of the section considered is mathematically represented as

$A = \frac{n * F }{ b * 2 \pi }$

=> $A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }$

=> $n = A * \frac{m}{3} * \frac{2\pi}{25}$

=> $n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}$

=> $n = 1810 \ people$

3 0

3 years ago

Calculate the Reynolds number for an oil gusher that shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter

Alenkasestr [34]

Answer:

$Re=1992.24$

Explanation:

Given:

vertical height of oil coming out of pipe, $h=25\ m$

diameter of pipe, $d=0.1\ m$

length of pipe, $l=50\ m$

density of oil, $\rho = 900\ kg.m^{-3}$

viscosity of oil, $\mu=1\ Pa.s$

Now, since the oil is being shot verically upwards it will have some initial velocity and will have zero final velocity at the top.

<u>Using the equation of motion:</u>

$v^2=u^2-2gh$

where:

v = final velocity

u = initial velocity

Putting the respective values:

$0^2=u^2-2\times 9.8\times 25$

$u=22.136\ m.s^{-1}$

<u>For Reynold's no. we have the relation as:</u>

$Re=\frac{\rho.u.d}{\mu}$

$Re=\frac{900\times 22.136\times 0.1}{1}$

$Re=1992.24$

6 0

3 years ago

PLEASE HELP !!! ASAP ! Jill and Scott both road their bikes for 30 minutes. Jill traveled 5 kilometers and Scott traveled 10. Ca

Arturiano [62]

Ok first of all we have to convert in km/hr=Speed

So 30min are 0.5hours

Then we use S= distance/time = 5km/0.5 hrs = 10km/hr —> Jill speed

Then S= d/t = 10km/0.5hrs = 20km/hr —> Scott speed

So Scott rode faster

3 0

3 years ago

Read 2 more answers