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ch4aika [34]
3 years ago
5

Rocket engineers use newton's third law during launch. identify the action force.

Physics
2 answers:
AfilCa [17]3 years ago
7 0

Answer:

Option (A)

Explanation:

Newton's third law states that, for every action force there is an equal and opposite reaction force.

Action and reaction forces acts on two different bodies.

In case of rocket propulsion, the action force is due to the exhaust of gases pushing downward on the earth and due to the reaction force the rocket moves upwards in the space.

rosijanka [135]3 years ago
6 0
The answer is A. Newton's third law of motion states that for every action, there is an equal and opposite reaction. A rocket exerts a large force on the gas that is in the rocket chamber (action). The gas thus exerts a large reaction force forward on the rocket (reaction). The large reaction force is called thrust.
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A ball dropped from a bridge takes three seconds to reach the water below how far is the bridge above the water?
tatuchka [14]

<u>Given that:</u>

Ball dropped from a bridge at the rate of 3 seconds

Determine the height of fall (S) = ?

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                          a= g =9.81 m/s  (since free fall)

                            S = 0+ 1/2 × 9.81 × 3²

                          <em> S = 44.145 m</em>

<em>44.145 m far is the bridge from water</em>

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3 years ago
When and where was the worst hurricane in the united states
Andre45 [30]
It was in Texas on September 8, 1900.
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3 years ago
One of the largest planes ever to fly, and the largest to fly frequently, is the Ukrainian-built Antonov An-124 Ruslan. The grav
Lerok [7]

Answer:

m = 236212 [kg]

Explanation:

The potential energy can be determined by means of the product of mass by gravity by height. In this way, we have the following equation.

P=m*g*h\\

where:

P = potential energy = 3360000000 [J]

m = mass [kg]

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Now, we can clear the mass from the equation above:

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4 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

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We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation           -<span>v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
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