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olya-2409
6 days ago
6

At one instant of time a rocket is traveling in outer space at 2500m/s and is exhausting fuel at a rate of 100 kg/s. If the spee

d of the fuel as it leaves the rocket is 1500 m/s, relative to the rocket, the thrust is:
Physics
1 answer:
Keith_Richards [1K]6 days ago
7 0

Thrust is quantified as a reaction force, in accordance with Newton's third law. When a system accelerates or expels mass in one direction, this resulting mass generates a force of equal strength but in the opposite direction on that system. This relationship can be expressed mathematically as:

T = v\frac{dm}{dt}

Where:

v = velocity of the exhaust gases as perceived from the rocket.

\frac{dm}{dt}= Change in mass over time

The provided data is as follows:

v = 1500m/s

\frac{dm}{dt} = 100kg/s

After substitution, we obtain:

T = 1500*100

\therefore T = 1.5*10^5N

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Consider the uniform electric field \vec{E} =(4000~\hat{j}+3000~\hat{k})~\text{N/C} ​E ​⃗ ​​ =(4000 ​j ​^ ​​ +3000 ​k ​^ ​​ ) N/
kicyunya [1011]

Answer:

Electric flux is calculated as \phi=31562.63\ Nm^2/C

Explanation:

We start with the given parameters:

The electric field impacting the circular surface is E=(4000j+3000k)\ N/C

Our objective is to ascertain the electric flux passing through a circular region with a radius of 1.83 m situated in the xy-plane. The area vector is oriented in the z direction. The formula for electric flux is expressed as:

\phi=E{\cdot}A

\phi=(4000j+3000k){\cdot}Ak

Applying properties of the dot product, we calculate the electric flux as:

\phi=3000\times Ak

\phi=3000\times \pi (1.83)^2

\phi=31562.63\ Nm^2/C

Consequently, the electric flux for the circular area is \phi=31562.63\ Nm^2/C. Thus, this represents the required answer.

4 0
1 day ago
A small block of mass 200 g starts at rest at A, slides to B where its speed is vB=8.0m/s,vB=8.0m/s, then slides along the horiz
serg [1189]

Answer

Data provided:

mass of the block = 200 g = 0.2 Kg

Velocity at A = 0 m/s

Velocity at B = 8 m/s

distance of slide = 10 m

height of the block = 4 m

calculation for the block's potential energy

    P = m g h

    P = 0.2 x 9.8 x 4

    P = 7.84 J

kinetic energy calculated as

    KE = \dfrac{1}{2}mv^2

    KE = \dfrac{1}{2}\times 0.2 \times 7.84^2

    KE =6.14 J

Work done = P - KE

work = 7.84 - 6.14

work = 1.7 J

b) using the formula v² = u² + 2 a s

   0 = 8² - 2 x a x 10

   a = 3.2 m/s²

ma - μ mg = 0

 \mu = \dfrac{a}{g}

 \mu = \dfrac{3.2}{9.8}

 \mu = 0.327

7 0
8 days ago
The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2
serg [1189]

Respuesta:

Explicación:

Al analizar esta pregunta, considera el movimiento circular. Primero, determina la máxima fuerza que puede aplicarse al hilo. F = mg, entonces F = (10)(10) = 100 N. Luego, calcula la aceleración centrípeta de la masa de 0.5 kg, a = F/m, así que a = 100/.5 = 200 m/s². En la hoja de ecuaciones, usa la fórmula a (aceleración centrípeta) = v²/r, por lo que 200 = v²/2; por consiguiente, v = 20 m/s. ¡Espero que esto sea útil!

0 0
12 days ago
A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. 1.84 x 103 J
Yuliya22 [1153]
<span>The work done corresponds to the potential energy that the person acquires while ascending the stairs.
work = potential energy acquired = mgh
W = 75kg * 9.8m/s² * 2.50m = 1837.5 J</span>
4 0
6 days ago
A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards
Ostrovityanka [942]

Answer:

b. The loop's current consistently flows in a counterclockwise direction.

Explanation:

As a magnet descends through a wire loop, it generates an induced current within that loop. This induced current arises due to the magnet's movement, leading to a variation in magnetic flux. Lenz's law states that the induced current will act to counteract the change that produces it. In this scenario, the only feasible resistance to the magnet’s fall is through inducing a similar pole on the loop to counteract its downward motion. An induced current that circulates counterclockwise in the wire loop mimics the polarity of a northern pole, thereby repelling the magnet's descent. Furthermore, as the magnet passes the wire loop, this induced north pole will seek to attract the magnet's south end in an effort to halt its downward progression.

3 0
2 days ago
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