11. A collection of a small group of factors that are related is known as a(n)

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago

if you're shopping for a rack switch, what component on the switch tells you it can be mounted to a rack?

Tatiana [17]

if you're shopping for a rack switch, the component on the switch that tells you it can be mounted to a rack is the:

Rack ears

Rack ears are L-shaped objects that can be used to hold a rack switch firmly to the support walls. Rack ears are often found at the front panel of the rack which is to be mounted.

When a person purchases a rack switch and finds the rack ears there, it is a signal that the item can be secured firmly to rails. Sometimes, the rack ears also appear as extensions.

Learn more about rack ears here:

brainly.com/question/13318148

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6 0

2 years ago

Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re

Amiraneli [1.4K]

After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

Learn more about Momentum here:

brainly.com/question/1042017

#SPJ4

4 0

2 years ago

Which of the following occurs with both a cold front and a mountain breeze?

BigorU [14]

Answer:

b warm air rises

Explanation:

5 0

3 years ago

For an atom to be neutral the amount of ? and? must be equal

maks197457 [2]

Answer: Well first of all, for an Atom to be neutral, it shall have an equal number of the protons and the neutrons. So given that, the protons are able to be positively charged and since the electrons are negatively charged (Protons and Electrons are completely different from each other) then that means the Algebric are aggregated of charges because it equal to zero. So overall, the 'neutral atom' has to be neutral, and how? Protons must equal the number of electrons or else it wouldn't be considered neutral.

Hopefully this helps you very much!! :-) Have a great daayy!

6 0

3 years ago

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