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Allushta [10]
1 year ago
10

a rocket with a mass of 4kg accelerates up from the ground at a rate of 20ft m/s^2, what is the drag force acting on the rocket

if the thrust of the rockets is 150n
Physics
1 answer:
Maru [420]1 year ago
8 0

The drag force acting on the rocket is 80N.

<h3>Give an explanation of drag force?</h3>

The divergence in velocity between the fluid and the item, also known as drag, exerts a force on it. Between the liquid and the solid object, there should be motion. Drag is absent in the absence of motion.

The air molecules are more compressed (pushed together) on the surfaces that are facing the front while being more dispersed (spread out) on the surfaces facing the back. Turbulent flow, which occurs when air layers split from the surface and start to swirl, is what causes this.

The drag force acting on the rocket F = ma

Given,

m = 4kg, a = 20ftm/s²

Substituting m and a values in the above formula,

The drag force acting on the rocket F = 4×20

The drag force acting on the rocket F = 80N.

To know more about drag force visit:

brainly.com/question/15144984

#SPJ4

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If a bicycle travels 1200m, and it takes 300seconds, what is its speed in m/s.
ch4aika [34]
Answer: 4

Explanation: speed= distance/time
7 0
3 years ago
You are moving a wagon with a friend's help you push on the left side of the wagon with 25 of force while your friend pulls from
Pavel [41]

Answer:

10N to the left side towards you

Explanation:

The net force is the resultant force that acts on a body.

Force is a push or pull on a body.    

 Push to left side  = 25N

 Pull to the right  = 15N

Net force  = Push to left side   -  Pull to the right  = 25N  - 15N

 Net force  = 10N to the left side towards you

The net force is therefore 10N to the left side towards you

5 0
3 years ago
A spherical balloon is made from a material whose mass is 3.30 kg. The thickness of the material is negligible compared to the 1
stellarik [79]

Answer:

563712.04903 Pa

Explanation:

m = Mass of material = 3.3 kg

r = Radius of sphere = 1.25 m

v = Volume of balloon = \frac{4}{3}\pi r^3

M = Molar mass of helium = 4.0026\times 10^{-3}\ kg/mol

\rho = Density of surrounding air = 1.19\ kg/m^3

R = Gas constant = 8.314 J/mol K

T = Temperature = 345 K

Weight of balloon + Weight of helium = Weight of air displaced

mg+m_{He}g=\rho vg\\\Rightarrow m_{He}=\rho vg-m\\\Rightarrow m_{He}=1.19\times \frac{4}{3}\pi 1.25^3-3.3\\\Rightarrow m_{He}=6.4356\ kg

Mass of helium is 6.4356 kg

Moles of helium

n=\frac{m}{M}\\\Rightarrow n=\frac{6.4356}{4.0026\times 10^{-3}}\\\Rightarrow n=1607.85489

Ideal gas law

P=\frac{nRT}{v}\\\Rightarrow P=\frac{1607.85489\times 8.314\times 345}{\frac{4}{3}\pi 1.25^3}\\\Rightarrow P=563712.04903\ Pa

The absolute pressure of the Helium gas is 563712.04903 Pa

3 0
3 years ago
Suppose the velocity of an object moving along a line is positive. are position, displacement, and distance traveled equal? expl
Pavlova-9 [17]

No, the object's displacement and distance travelled will be equal, but since the initial position is unknown, the object's position might not match up with its displacement and distance travelled.

We cannot assert that the displacement or distance equals the position because the initial position is not provided. We could reach a different conclusion if the starting position had been zero because the distance from zero is equal to the position.

Find more on velocity at : brainly.com/question/11347225

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4 0
2 years ago
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t
Bond [772]

Answer:

\frac{D_{jet}}{D_{prop}}=2.865

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

D=\frac{1}{2}CpAv^2\\

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\   \frac{D_{jet}}{D_{prop}}=2.865

4 0
3 years ago
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