The force on a rocket moving with a veloctiy 300 m/s is 210N. The rate of consumption of fuel of rocket is

Options

A. 0.7 kg/s
B. 1.4 kg/s
C. 0.07 kg/s
D. 10.7 kg/s

Answer

To find the rate of consumption of fuel of the rocket, we can use Newton's second law of motion and the principle of conservation of momentum. The force acting on the rocket can be expressed in terms of the rate of change of momentum.

### Step-by-Step Solution:

1. **Understand the relationship between force, mass flow rate, and velocity:**
The force \( F \) acting on the rocket can be expressed as:
\[ F = \frac{dm}{dt} \cdot v \]
where:
- \( F \) is the force acting on the rocket (in Newtons),
- \( \frac{dm}{dt} \) is the rate of change of mass (mass flow rate of fuel, in kg/s),
- \( v \) is the velocity of the rocket (in m/s).

2. **Rearrange the equation to find the mass flow rate:**
We can rearrange the equation to solve for \( \frac{dm}{dt} \):
\[ \frac{dm}{dt} = \frac{F}{v} \]

3. **Substitute the known values:**
Given:
- \( F = 210 \, \text{N} \)
- \( v = 300 \, \text{m/s} \)

Substitute these values into the equation:
\[ \frac{dm}{dt} = \frac{210 \, \text{N}}{300 \, \text{m/s}} \]

4. **Calculate the rate of consumption of fuel:**
\[ \frac{dm}{dt} = \frac{210}{300} = 0.7 \, \text{kg/s} \]

### Final Answer:
The rate of consumption of fuel of the rocket is \( 0.7 \, \text{kg/s} \).

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