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3241004551 [841]
3 years ago
9

How does a pie chart help form conclusion

Physics
1 answer:
daser333 [38]3 years ago
3 0

Its easy to create a visual and see major differences between data. If something takes up a large portion of the pie, its easy for you to tell just by looking at it
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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}.

6 0
3 years ago
When a positively charged body touches a neutral body, the neutral body will
PtichkaEL [24]
Become more positive instead of negative
7 0
3 years ago
Calculate the speed of an 8.0 × 104 kg airliner with a kinetic energy of 1.1 × 109 J.
Mazyrski [523]
KE = ½mv² 
<span>1.1 x 10^9 = ½ x 8 x 10^4 v² 
v² = 2.75 x 10^4 
v ≈ 165.8 m/s</span>
5 0
3 years ago
On a windless day, two identical balls P and Q are dropped from the top and the middle of a tower at exactly the same time as sh
Kruka [31]

Answer: time is the same

Explanation: the distance(H) is the same in each case .

we drop the balls , no drag force using basic kimnematics

y =gt*t/2  , yo=0 , vo=0 , y=H  , so : t= sqrt(2H/g)

comment: if distance H starts to grow....we could begin to note a difference  because of gravity g is smaller as we go up

3 0
3 years ago
A 2 kilogram mass is lifted 4 meters above the ground what is the change it gravitational energy
lianna [129]

Answer:

78.4 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its location in the gravitational field.

The change in gravitational potential energy of an object is given by:

\Delta U=mg\Delta h

where:

m is the mass of the object

g=9.8 m/s^2 is the acceleration due to gravity

\Delta h is the change in height of the object

For the mass in  this problem, we have:

m = 2 kg is the mass

\Delta h = 4 m is the change in height

So, its change in gravitational energy is:

\Delta U=(2)(9.8)(4)=78.4 J

6 0
3 years ago
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