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Lynna
14 days ago
8

For years, space travel was believed to be impossible because there was nothing there Rockets could push off in space in order t

o provide the propulsion necessary to accelerate this inability to provide propulsion is because..
a) a space is a void of air so the rockets have nothing to push off.
b) gravity is absent in space.

c)a space is a void of wit so there is no air resistance in space.

d) nonsense! rockets do accelerate in space and have been able to do so for a long time.
​
Physics
1 answer:
Maru [2.3K]14 days ago
7 0

Rockets do indeed accelerate in the vacuum of space, something that has been possible for an extended period.

Response: Option D

Clarification:

A rocket's design aims to reach its destination by consuming fuel onboard. Hence, external forces are not necessary for motion in space. According to Newton's third law, action and reaction forces propel rockets free from Earth's gravity.

The remaining thrust needed for the rocket's journey in space comes from the combustion of fuel in its chambers. The fuel's combustion in designated compartments propels the rockets onward. Therefore, the belief that space exploration was impossible previously due to lack of propulsion mechanisms is a complete misconception.

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The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes
Sav [2230]

mass₃<mass₁=mass₅<mass₂=mass₄

Explanation:

Data points:-

1. mass:  m      speed: v

2. mass: 4 m   speed: v

3. mass: 2 m   speed: ¼ v

4. mass: 4 m   speed: v

5. mass: 4 m   speed: ½ v

We know that the formula for Kinetic energy (KE) is ½ mv²

Where m represents the mass of the object

           v represents the object's velocity

<psubstituting the="" given="" values="" for="" mass="" and="" speed="" from="" previous="" data:="">

The KE of Body 1(mass₁) = ½*m*v²             = mv²/2

KE of Body 2(mass₂) = ½*4m*v²         = 2mv²

KE of Body 3(mass₃) = ½*2m*(1/4v)²  =  mv²/16

KE of Body 4(mass₄) = ½*4m*v ²        =  2mv ²

KE of Body 5(mass₅) = ½*4m*(1/2v)²  =   mv²/2

</psubstituting>
6 0
17 days ago
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
Softa [2035]

Answer:

57.94°

Explanation:

We understand that the formula for flux is

\Phi =E\times S\times COS\Theta

where Ф represents flux

           E indicates electric field

           S denotes surface area

        θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

                          S=\pi \times 0.057^{2}

                         COS\Theta =\frac{\Phi }{S\times E}

 =   \frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}

 =0.5306

 θ=57.94°

4 0
19 days ago
(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i
kicyunya [2264]

Response:

a) P = 370.993\,kPa, b) F = 25.948\,kN

Clarification:

a) The absolute pressure at a depth of 27.5 meters is:

P = P_{atm} + P_{man}

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 370.993\,kPa

b) The force applied by the water is:

F = (P - P_{atm})\cdot A

F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}

F = 25.948\,kN

5 0
8 days ago
Read 2 more answers
A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4
Yuliya22 [2438]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

R=\sqrt{P^2+Q^2+2PQCos\theta}

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

\theta= 90^o

R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70

(Cos90^o=0),(sin 90^o=1)

\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0
1 month ago
In space, astronauts don’t have gravity to keep them in place. That makes doing even simple tasks difficult. Gene Cernan was the
Keith_Richards [2263]

Newton's First Law: A body remains in its current state of motion or at rest unless a force acts upon it.

Newton's Second Law: Motion changes are proportional to the applied force and oriented in the same direction.

Newton's Third Law: Every action has a corresponding and opposite reaction.

Tasks that would be challenging to perform in orbit include:

-operating a valve

-navigating on foot

-attempting to take a shower

-remaining still


4 0
7 days ago
Read 2 more answers
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