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1 month 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:

1 answer:

Keith_Richards [3.2K]1 month 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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An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

kicyunya [3294]

Answer:

Estimate of the sample's volume: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Mean density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
The volume of the cord is considered negligible.

Explanation:

Overall volume of the sample

The magnitude of the buoyant force equals $\rm 17.50 - 11.20 = 6.30\; N$ .

This also corresponds to the weight (weight, $m \cdot g$ ) of the water displaced by the object. To determine the mass of the displaced water from its weight, apply the formula: divide weight by $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assuming the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To find the volume of the displaced water, use the formula: divide mass by density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assuming the cord's volume is negligible, since the sample is completely submerged in water, its volume should equal the volume of the displaced water.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

Mean Density of the sample

Average density can be calculated by the mass divided by volume.

To compute the mass of the sample from its weight, utilize the formula: divide by $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume from the previous section can be utilized.

Lastly, divide mass by volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

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12 days ago

A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. W

ValentinkaMS [3465]

The force due to electricity on the charge is calculated by multiplying the charge by the intensity of the electric field:
$F=qE$
in our scenario, where $q=3 \mu C= 3 \cdot 10^{-6} C$ and $E=2 kN/C=2000 N/C$ , resulting in the force of
$F=(3 \cdot 10^{-6} C)(2000 N/C)=0.006 N$

Initially, the kinetic energy of the particle is at zero (as it remains stationary), which means its final kinetic energy is equal to the work performed by the electric force over a distance of x=4 m:
$K(4 m)=W=Fd=(0.006 N)(4 m)=0.024 J$

8 0

29 days ago

A motorist enters a freeway at 45 km/h and accelerates uniformly to 99 km/h. From the odometer in the car, the motorist knows th

kicyunya [3294]

a) 19440 km/h²; b) 10 seconds.

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1 month ago

A cliff diver on an alien planet dives off of a 32 meter tall cliff and lands in a sea of hydrochloric acid 1.20 seconds later.

inna [3103]

Answer:

44.4m/s^2

Explanation:

Utilize the equation...S = ut + 1/2at^2

where...S = 32m...u = 0m/s....t = 1.20s

32 = (0)(1.20) + 0.5(1.20^2)a

; The acceleration due to gravity is 44.4m/s^2

3 0

1 month ago

A ball collides elastically with an immovable wall fixed to the earth’s surface. Which statement is false? 1. The ball's speed i

Maru [3345]

Answer:

Statements 4, 6 & 7 are incorrect.

Explanation:

In any elastic collision, the overall momentum vector sum of the system remains zero.

In this scenario, an elastic collision occurs between the ball and a stationary wall. The ball's velocity will consistently revert after the impact, leading to a change in direction of momentum.

The initial momentum of the ball is represented as:

$p=m.v$

where:

m = mass of the ball

v = initial velocity of the body

post-collision for the elastic interaction:

$p=m.(-v)$

Here, the momentum changes solely in direction, thus contradicting statement 7.

During the impact, both the ball and the wall exert forces on each other that are equal and opposite. The wall remains motionless, while the ball is influenced by the wall's reaction force, performing work on it, which contradicts statement 4.

Given that this collision is elastic, the ball's form and dimensions do not alter.

The previous points clearly indicate that not all provided statements hold true, thus violating statement 6.

4 0

25 days ago