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Genrish500 [490]
3 years ago
10

PLZ ANSWER THE QUESTION ​

Physics
1 answer:
adelina 88 [10]3 years ago
7 0

Answer:

a

Explanation:

t

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A rocket at rest with a mass of 942 kg is acted on by an average net force of 6,731 N upwards for 21 s. What is the final veloci
Galina-37 [17]

Answer:

Explanation:

Givens

m = 942

F = 6731

t = 21 seconds

vi = 0

vf = ?

Formula

F = m * (vf -  vi )  / t

Solution

6731 = 942*(vf - 0)/21          Multiply both sides by 21

6731 * 21 = 942*vf

141351 = 942*vf                   Divide by 942

141351/942 = vf

vf = 151 m/s

6 0
2 years ago
What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use
marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force F_{1}=18\ N

Force F_{2}=26\ N

Force F_{3}=14\ N

We need to calculate the torque due to force F₁

Using formula of torque

\tau_{1}=-F_{1}d_{1}

\tau_{1}=-F_{1}\times\dfrac{a}{2}

Put the value into the formula

\tau_{1}=-18\times\dfrac{0.2}{2}

\tau_{1}=-1.8\ N-m

We need to calculate the torque due to force F₂

Using formula of torque

\tau_{2}=F_{2}d_{2}

\tau_{2}=F_{2}\times\dfrac{a}{2}

Put the value into the formula

\tau_{2}=26\times\dfrac{0.2}{2}

\tau_{2}=2.6\ N-m

We need to calculate the torque due to force F₃

Using formula of torque

\tau_{3}=F_{3}d_{3}

\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}

Put the value into the formula

\tau_{3}=0.1(14\sin45+14\cos45)

\tau_{3}=1.92\ N-m

We need to calculate the net torque on the square plate

\tau=\tau_{1}+\tau_{2}+\tau_{3}

\tau=-1.8+2.6+1.92

\tau=2.72\ N-m

Hence, The net torque on the square plate is 2.72 N-m.

3 0
3 years ago
The principle of conservation of momentum is most similar to which of newton's laws of motion?
Effectus [21]
Newton’s Thrid Law, which states that for every reaction there is an opposite reaction.
6 0
3 years ago
What planet has orbital plane different than the rest of planets?
bixtya [17]
The sun has orbited along time so when they ask theses questions I give you the right answer I think lol
4 0
3 years ago
A spherical shell is rolling without slipping at constant speed on a level floor. What percentage of the shell's total kinetic e
IgorC [24]

Answer:

41.667 per cent of the total kinetic energy is translational kinetic energy.

Explanation:

As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:

E = K_{t} + K_{r}

Where:

E - Total energy, measured in joules.

K_{r} - Rotational kinetic energy, measured in joules.

K_{t} - Translational kinetic energy, measured in joules.

The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:

Rotational kinetic energy

K_{r} = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}

Translational kinetic energy

K_{t} = \frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2}

Where:

I_{g} - Moment of inertia of the spherical shell with respect to its center of mass, measured in kg\cdot m^{2}.

\omega - Angular speed of the spherical shell, measured in radians per second.

R - Radius of the spherical shell, measured in meters.

After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:

E = \frac{1}{2}\cdot (I_{g} + m\cdot R^{2})\cdot \omega^{2}

In addition, the moment of inertia of a spherical shell is equal to:

I_{g} = \frac{2}{3}\cdot m\cdot R^{2}

Then, total energy is reduced to this expression:

E = \frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2}

The fraction of the total kinetic energy that is translational in percentage is given by the following expression:

\%K_{t} = \frac{K_{t}}{E}\times 100\,\%

\%K_{t} = \frac{\frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2} }{\frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2} } \times 100\,\%

\%K_{t} = \frac{5}{12}\times 100\,\%

\%K_{t} = 41.667\,\%

41.667 per cent of the total kinetic energy is translational kinetic energy.

7 0
3 years ago
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