The diameter of a lead pipe was measured to be 2.40 centimeters. its diameter in units of meters is _______

Why does moon appear to have alot of phases

Bingel [31]

It depends on the angle of the earth, and our point of view. A full moon would occur if we were to be right in front of it.

4 0

3 years ago

What is the weight of a 20kg stone?

worty [1.4K]

Answer: w = 200N

Explanation:

w = mg

m = 20kg

g = 10m/s^ (approximately)

w = 20kg . 10m/s^

w = 200N

5 0

3 years ago

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In 2017, the company SpaceX became the first private company to send supplies to the International Space Station with a reusable

pav-90 [236]

Answer:

Approximately $3.98\; \rm m \cdot s^{-2}$ .

Assumption: air resistance on the rocket is negligible. Take $g = \rm 9.81\; m \cdot s^{-2}$ .

Explanation:

By Newton's Second Law of Motion, the acceleration of the rocket is proportional to the net force on it.

$\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ .

Note that in this case, the uppercase letter $\rm M$ in the units stands for "mega-", which is the same as $10^6$ times the unit that follows. For example, $\rm 1\; Mg = 10^6\; g$ , while $\rm 1\; MN = 10^6\; N$ .

Convert the mass of the rocket and the thrust of its engines to SI standard units:

The standard unit for mass is kilograms: $\displaystyle m = \rm 552\; Mg = 552 \times 10^6\; g \times \frac{1\; \rm kg}{10^3\; g} = 552 \times 10^3 \; kg$ .
The standard for forces (including thrust) is Newtons: $\text{Thrust} = \rm 7.61 \; MN = 7.61 \times 10^6\; N$ .

At launch, the velocity of the rocket would be pretty low. Hence, compared to thrust and weight, the air resistance on the rocket would be pretty negligible. The two main forces that contribute to the net force of the rocket would be:

Thrust (which is supposed to go upwards), and
Weight (downwards due to gravity.)

The thrust on the rocket is already known to be $\rm 7.61 \times 10^6\; N$ . Since the rocket is quite close to the ground, the gravitational acceleration on it should be approximately $9.81\; \rm m \cdot s^{-2} = 9.81 \; N \cdot kg^{-1}$ . Hence, the weight on the rocket would be approximately $9.81\; \rm N \cdot kg^{-1} \times 552 \times 10^3\; kg = 5.41412\times 10^6\; N$ .

The magnitude of the net force on the rocket would be

$\begin{aligned}&\text{Thrust} - \text{Weight} \\ &= 7.61 \times 10^6\; \rm N - 5.41412\times 10^6\; N \\ &\approx 2.19 \times 10^6\; \rm N\end{aligned}$ .

Apply the formula $\displaystyle \text{Acceleration} = \frac{\text{Net Force}}{\text{Mass}}$ to find the net force on the rocket. To make sure that the output (acceleration) is in SI units (meters-per-second,) make sure that the inputs (net force and mass) are also in SI units (Newtons for net force and kilograms for mass.)

$\begin{aligned}\displaystyle &\text{Acceleration} \\ &= \frac{\text{Net Force}}{\text{Mass}} \\ &= \frac{2.19 \times 10^6\; \rm N}{552 \times 10^3\; \rm kg} \\ &\approx \rm 3.98\; \rm m \cdot s^{-2}\end{aligned}$ .