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Mashcka [7]
3 years ago
14

A rocket weighs 9800N (opposing force) what is it mass? What netforce moves the rocket? What applied force gives it a vertical a

cceleration of 4m/sec
Physics
1 answer:
Slav-nsk [51]3 years ago
6 0

For the first part of this question, consider that "weight" can be described as mass x acceleration of gravity. Weight is expressed in Newtons. To solve for mass in this case, simply divide 9800N by 9.8m/s^2 (Earth's gravitational acceleration). This will give you a mass of 1000 kg. This mass is moved due to the net force supplied by the normal force from the rocket "pushing" off of Earth.

For the second part, we will use the equation F = ma, which is Newton's second law. For this, we know the m, or mass, is 1000 kg. Also, we know the a, or acceleration, will be 4 m/s^2. To solve for force, we will multiply both of these values. This gives a force of 4000 N. I hope this clears things up!

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a. Define Valency. b. Write the symbol and valency of the following : (i) Oxygen (ii) potassium 7. a. Magnesium burns in Oxygen
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6 0
3 years ago
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A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo
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Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2

E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

E = 9 J

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3 0
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The current through a 10 ohm resistor connected to a 120 volt power supply is
Leno4ka [110]

Answer:I=12 A

Explanation:

Given

Resistance R=10 \Omega

Voltage V=120 V

According to ohm's law current through a conductor is directly proportional to the voltage applied.

V\propto I

V=IR

where V=Voltage

I=Current

R=resistance

I=\frac{V}{R}

I=\frac{120}{10}

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