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baherus [9]
2 years ago
15

Object A has a mass of 5 kg and a velocity of 6 m/s to the east while Object 1 point B has a mass of 12 kg and velocity 0.6 m/s

also to the east. What is the momentum of the system? (Let east be positive)
help plss I got family
Physics
1 answer:
Olin [163]2 years ago
8 0

Answer:

Momentum of system = 37.2 Kgm/s.

Explanation:

<u>Given the following data;</u>

  • Mass A = 5 kg
  • Velocity A = 6 m/s
  • Mass B = 12 kg
  • Velocity B = 0.6 m/s

To find the momentum of the system;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

<u>For object A;</u>

Momentum A = 5 * 6

Momentum A = 30 Kgm/s

<u>For object B;</u>

Momentum B = 12 * 0.6

Momentum B = 7.2 Kgm/s

Next, we would determine the momentum of this system using the formula;

Momentum of system = Momentum A + Momentum B

Substituting the values into the formula, we have;

Momentum of system = 30 + 7.2

<em>Momentum of system = 37.2 Kgm/s.</em>

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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
A lamp draws a current of 0.50 A when it is connected to a 120 V source? What is the resistance of the lamp?
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Given,
Current (I) = 0.50A
Voltage (V) = 120 volts
Resistance (R) =?
We know that:-
Voltage (V) = Current (I) x Resistance (R)
→Resistance (R) = Voltage (V) / Current (I)
= 120/0.50
= 24Ω
∴ Resistance (R) = 24Ω
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<h3>What is a momentum?</h3>
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To know more about momentum, refer:

brainly.com/question/1042017

#SPJ4

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