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VashaNatasha [74]
3 years ago
8

Help ;-;

Physics
1 answer:
nasty-shy [4]3 years ago
3 0

Answer:

all qn 1,2,3 have same answer ,. Yes,. hope it helps

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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
What is the magnitude of the vector described below
Romashka [77]

Answer:

D. Miles per hour

Explanation:

Have a Good Day

5 0
2 years ago
What has to increase in order for an object to accelerate?
KengaRu [80]

Answer:

Answer: B. Explanation: For an object to accelerate the force on it must be increased. According to Newton's second law of motion.

Explanation:

I do Accelerate to good luck

7 0
2 years ago
Read 2 more answers
A superball and a clay ball are dropped from a height of 10cm above a tabletop. They have the same mass 0.05kg and the same size
Olenka [21]

Answer:4.08 N

Explanation:

Given data

superball dropped from a height of 10  cm

Mass of ball\left ( m\right )=0.05kg

time of contact\left ( t\right )=34.3\times 10^{-3} s

Now we know impulse =Force\times time\ of\ contact=Change in momentum

F_{average}\times t=m\left ( v-(-v)\right )

and velocity at the bottom is given by

v=\sqrt{2gh}

F_{average}\times 34.3\times 10^{-3}=0.05\left ( 1.4-(-1.4)\right )

F_{average}=4.081N

5 0
3 years ago
A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the tables edge in the
GREYUIT [131]

A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the edge of the table in the horizontal direction,

  • t= 0.45 seconds.
  • V=2.22m/s
  • VT=4.95 m/s

This is further explained below.

<h3>What is its speed when it hits the floor...?</h3>

Generally, the equation for motion is mathematically given as

S= ut + 0.5at²

Therefore

y = Voy t + 0.5gt^2

1 = 0.5x 98 x 6²

1=4.9t^2

t=\sqrt{0.2041 }

t= 0.45 seconds.

b) Horizontal motions are uniform.

V=Horizontal displacement/time

V=1/0.45

V=2.22m/s

C)

Vx: 2.22 m/s At bottom,

Vy² = Voy² + 2as

Vy² = 2x95x1

Vy² = 19.6

Total velocity

VT=\sqrt{( 2.22 m/)^2+19.6}

VT=4.95 m/s

Read more about  Arithmetic

brainly.com/question/22568180

#SPJ1

4 0
2 years ago
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