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3 years ago

Which of the following people challenged the geocentric model of the solar system?

1 answer:

6 0

"Copernicus"was the one person among the following choices given in the question that <span>challenged the geocentric model of the solar system. The correct option among all the options that are given in the question is the second option. I hope that this is the answer that has come to your desired help.</span>

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A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di

navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

$X(van)=5.65t+154$

$X(driver)=34.4t+\frac{(-2)t^{2} }{2}$

or by rearanging the drivers equation.

$X(driver)=34.4t+t^{2}$

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

$X(van)=X(driver)$

$5.65t+154=34.4t-t^{2}$

$0=t^{2} -(34.4-5.65)t+154$ $0=t^{2} -28.75t+154$

To solve this equation we use the following formulas

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

Where a=1; b=-28.75; c=154

So we get:

$t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s$ $t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s$

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

$V(driver)=V_{0} +at$ }

$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}$ $V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}$

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).

Best of luck

8 0

3 years ago

What type of machine is wire cutter pliers?

Sonja [21]

It seems that you have missed the necessary options for us to answer this question, but anyway, here is the answer. The type of machine that a wire cutter pliers is classified is a simple machine. When we say simple machine, this is the type of machine that is considered basic wherein you need to apply force for it to function. Hope this helps.

3 0

3 years ago

Read 2 more answers

Q 1: Calculate the pressure.

Oksana_A [137]

Answer:

1: 300pa

2: 10cm2

3: 430pa

4: a: 1.6667pa

b:2.5m2

c:20pa

Explanation:

7 0

3 years ago

A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?

jeka57 [31]

Answer:

<h3>The answer is 45 J</h3>

Explanation:

The work done by an object can be found by using the formula

<h3>workdone = force × distance</h3>

From the question

distance = 3 meters

force = 15 newtons

We have

workdone = 15 × 3

We have the final answer as

Hope this helps you

7 0

3 years ago

The oil level in a tank is 2 m above the ground. The tank cover is air tight and the air pressure above the oil surface is 150 k

Andrew [12]

Answer:

a) 24.692 m/s

b) 19.4 m

Explanation:

To calculate the velocity at the nozzle outflow (V2) we use the Bernoulli equation:

$\frac{P1}{pg}+\frac{V1^2}{2g}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}+Z2$

We know that the velocity above the oil surface (V1) and the pressure at the nozzle outflow (P2) are negligible, the height in the exit is zero (Z2) then:

$\frac{P1}{pg}+Z1=\frac{P2}{pg}+\frac{V2^2}{2g}$

a) The velocity (V2) is:

$\frac{P1}{pg}+Z1=\frac{V2^2}{2g}$

$(\frac{P1}{pg}+Z1)(2g)=V2^2$

$V2=[(\frac{P1}{pg}+Z1)(2g)]^{1/2}$

Substituting the known values we can get the velocity at the out:

Atmospheric pressure= 101000 Pa

Oil density= 0.88x(Water density)=0.88(1000kg/m3)=880kg/m3

$V2=[(\frac{150000Pa+101000 Pa}{(880 kg/m3)(9.81m/s)}+2m)(2(9.81m/s2))]^{1/2}$

$V2=24.692 m/s$

b) To calculate the height we have to apply the Bernoulli equation between the outflow and the maximum height (Z3), so:

$\frac{P2}{pg}+\frac{V2^2}{2g}+Z2=\frac{P3}{pg}+\frac{V3^2}{2g}+Z3$

We know that the velocity above the stream (V3) and the pressure at the nozzle outflow (P2) are negligible, the pressure at the top of the stream (P3) is the atmospheric pressure, then:

$\frac{V2^2}{2g}=\frac{P3}{pg}+Z3$

$Z3=\frac{V2^2}{2g}-\frac{P3}{pg}$

Substituting the known values, the height (Z3) is:

$Z3=\frac{(24.692 m/s)^2}{2(9.81 m/s2)}-\frac{101000 Pa}{(9.81 m/s)(880 kg/m3)}$

Z3=Maximum Height=19.376=19.4 m

3 0

4 years ago