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3 years ago

I only need help with the two bottom questions! Please help though!!!

Physics

1 answer:

kompoz [17]3 years ago

3 0

If the rabbits died out, the other organisms would be likely to die. (If that were their main source of food.)

producer- something that produces food for other organisms (and is consumed). ex.grass

primary consumers- eat the producers and like first. ex. rabbit
secondary consumer- eats the primary consumers

Scavenger - is basically an organism that scavenges for food ex. vulture.

good luck!

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Can black wholes be an exit to the universe,why or why not? (just curious).

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A rock is launched at angle theta=53.2∘ above the horizontal from an altitude of ℎ=182 km with an initial speed ????0=1.61 km/s.

Mariulka [41]

Answer:

The rock's final speed at the required altitude will be 42.24 m/s.

Explanation:

Let's start by finding the initial vertical speed.

Vertical Speed = 1.61 * Sin (53.2°)

Vertical Speed = 0.8 m/s

We want to know the speed of the rock when it is at an altitude of 91 km.

The total displacement of the rock from its starting position will thus be equal to -91 km

We can use this in the following equation:

$s=u*t+\frac{1}{2} (a*t^2)$

$-91=0.8*t+\frac{1}{2} (-9.8*t^2)$

t = 4.3918 seconds

Thus it takes 4.3918 seconds to reach the required altitude. We can now find the speed as follows:

$V=U+at$

$V=0.8+(-9.8)*(4.3918)$

$V = -42.24$

Thus the rock's final speed at the required altitude will be 42.24 m/s.

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3 years ago

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LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

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