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3 years ago

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Write Newton's 3rd law. How would this law relate to a rocket ship taking off from the earth? Would this law affect the rocket s

hip much once it is in space (be sure to explain your answer)?

1 answer:

irina1246 [14]3 years ago

6 0

Answer:

Part A

Newton's 3rd law states that action and reaction are equal and opposite, mathematically, we have;

$F_A$ = - $F_B$

Where;

$F_A$ = The action force

$F_B$ = The reaction force

Part B

The law indicates that the force with which a rocket ship uses in taking off from the Earth, $F_A$ is equal in magnitude, and opposite in direction to the reaction force of the Earth to the motion of the rocket, (-) $F_B$

Part C

The law is a universal law, and it will also affect the rocket ship in space, as the force of the jet from the exhaust is directed towards Earth while in space, the rocket is propelled deeper into space

Explanation:

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Answer:

a. applied science

Explanation:

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3 years ago

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The cable holding a 2125 kg elevator has a maximum strength of 21,750 N. What is the maximum upward acceleration the cable can g

vivado [14]

Answer:

10.23m/s^2

Explanation:

GIven data

mass of elevator = 2125 kg

Force= 21,750 N

Required

The maximum acceleration upward

F= ma

a= F/m

a=21,750/2125

a= 10.23m/s^2

Hence the acceleration is 10.23m/s^2

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3 years ago

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3 years ago

An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be

Inessa05 [86]

Explanation:

The given data is as follows.

Inner wheel Radius = 20 m,

Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

Angular velocity of automobile = w = $\frac{V}{R}$

where, V = linear velocity of automobile m/min,

R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

u = linear velocity of inner wheel

and, u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

w =

= $\frac{u}{(R - d)}$

u = $V \times \frac{(R - d)}{R}$

= $V \times (1 - \frac{d}{R})$

If radius of wheel is r it will cover distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus, u = $2\pi rn$ = $V \times (1 - \frac{d}{R})$

Now, rotation per minute of inner wheel is calculated as follows.

n = $\frac{V}{2 \pi r \times (1 - \frac{d}{R})}$

= $\frac{V}{2 \pi r \times (1 - \frac{0.75}{20})}$ (since 2d = 1.5m given, d = 0.75m),

= $\frac{V}{r} \times 0.1532$

So, rotation per minute of outer wheel; n' =

= $\frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}$

= $\frac{V}{r} \times 0.1651$

5 0

4 years ago

An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (

Crazy boy [7]

Answer:

Acceleration, $a=9.36\times 10^{18}\ m/s^2$

Explanation:

It is given that,

Speed of electron, $v=8\times 10^6\ m/s$

Charge on an electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

Magnetic field, $B=5.5i-3.7j$

Magnitude, $|B|=\sqrt{5.5^2+(-3.77)^2}=6.66\ T$

Magnetic force is given by :

$F=qvB$

Also, F = ma

$a=\dfrac{qvB}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 8\times 10^6\times 6.66}{9.1\times 10^{-31}}$

$a=9.36\times 10^{18}\ m/s^2$

So, the acceleration of the electron is $9.36\times 10^{18}\ m/s^2$ . Hence, this is the required solution.

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3 years ago