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ser-zykov [4K]
3 years ago
9

Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.

Physics
1 answer:
ozzi3 years ago
3 0

Answer:

The thrust is 6\times 10^6\ N

Explanation:

Given that,

Mass of gas, m=1.5\times 10^3\ kg

The rate at which the gas is expelling, \dfrac{dv}{dt}=4\times 10^{3}\ m/s

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

F=\dfrac{p}{t}

Also, p = mv

F=\dfrac{mv}{t}

So,

F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N

So, the thrust is 6\times 10^6\ N

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The summer camps had a field trip from the campus to Fragrance Hill. They traveled at an average speed of 65 km/h in the first 2
ludmilkaskok [199]

Answer:

Explanation:

They traveled this distance in 2 parts, essentially. Part 1 had an average speed for a certain number of hours, part 2 had an average speed for a certain number of hours, and those 2 parts taken together took them a distance of 364 km. In equation form, that looks like this:

km/hr part 1 + km/hr part 2 = 364 km

Now we need to find each part on the left side of that equation. Part 1 first:

We traveled 65 km/hr for 2 hours, so that took us

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We traveled 78 km/hr the second part for x hours, so that took us

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130 km + 78x km = 364 km and subtracting 130 km from both sides:

78x km = 234 km and dividing by 78 km:

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7 0
3 years ago
This is an example of which scientific practice?
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3 years ago
A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro
Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia =  1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0
3 years ago
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