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Natalija [7]
3 years ago
10

Which of the following is true about a rigid body in dynamic equilibrium? The body can have translational motion, but it cannot

have rotational motion. The body can have translational motion and rotational motion, as long as its translational and angular accelerations are equal to zero. The body cannot have translational motion, but it can have rotational motion. The body cannot have translational or rotational motion of any kind.
Physics
1 answer:
mr_godi [17]3 years ago
5 0

Answer:

The correct answer is "The rigid body can have rotational and transnational motion, as long as it's transnational and angular accelerations are equal to zero."

Explanation:

A rigid body by definition does not deform when forces act on it. In case of static equilibrium a rigid body cannot have any sort of motion while in case of dynamic equilibrium it can move but with constant velocities only thus having no acceleration weather transnational or angular.

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If you added enough ropes and pulleys to lift any size mass would it be possible to never have to apply a force to move your mas
tatiyna

Answer:

we have to apply force to move something things wants to save their position and if we want to change it

we have too apply force

7 0
3 years ago
Read 2 more answers
The ratio of the lift force L to the drag force D for the simple airfoil is L/D = 10. If the lift force on a short section of th
Yuri [45]

Answer:

R=64.32\ lb\\\\\theta=84.3\°

Explanation:

Given:

Ratio of lift force to drag force is, \frac{L}{D}=10

Lift force on a short section is, L=64\ lb

Magnitude of resultant, R= ?

The angle of 'R' with the horizontal is, \theta=?

We know that, lift force and drag are at right angles to each other. So, the resultant can be computed using Pythagoras theorem.

For calculating 'R', we first compute drag force 'D'.

As per question:

\frac{L}{D}=10\\\\D=\frac{L}{10}=\frac{64\ lb}{10}=6.4\ lb

Now, the magnitude of resultant 'R' is given as:

R=\sqrt{L^2+D^2}

Plug in the given values and solve for 'R'. This gives,

R=\sqrt{64^2+6.4^2}\\\\R=\sqrt{4096+40.96}\\\\R=\sqrt{4136.96}=64.32\ lb

Therefore, the magnitude of the resultant force 'R' is 64.32 lb.

Now, the angle \theta is given as the arctan of the ratio of the lift and drag force.

Therefore,

\theta=\tan^{-1}(L/D)\\\\\theta=\tan^{-1}(10)\\\\\theta=84.3\°

Therefore, the angle made with the horizontal is 84.3°.

6 0
3 years ago
Why does the air pressure inside the tires of a car increase when the car is driven?
Brrunno [24]
Its because particles of air inside it are moving faster and making more pressure plus there is a pressure from the street.
8 0
3 years ago
A jogger accelerates from rest to 4.86 m/s in 2.43 s. A car accelerates from 20.6 to 32.7 m/s also in 2.43 s. (a) Find the magni
Aleonysh [2.5K]

Explanation:

It is given that,

Initially, the jogger is at rest u₁ = 0

He accelerates from rest to 4.86 m, v₁ = 4.86 m

Time, t₁ = 2.43 s

A car accelerates from u₂ = 20.6 to v₂ = 32.7 m/s in t₂ = 2.43 s

(a) Acceleration of the jogger :

a=\dfrac{v-u}{t}

a=\dfrac{4.86\ m/s-0}{2.43\ s}

a₁ = 2 m/s²

(b) Acceleration of the car,

a=\dfrac{v-u}{t}

a=\dfrac{32.7\ m/s-20.6\ m/s}{2.43\ s}

a₂ = 4.97 m/s²

(c) Distance covered by the car,

d_1=u_1t_1+\dfrac{1}{2}a_1t_1^2

d_1=0+\dfrac{1}{2}\times 2\times (2.43)^2

d₁ = 5.904 m

Distance covered by the jogger,

d_2=u_2t_2+\dfrac{1}{2}a_2t_2^2

d_2=20.6\times 2.43+\dfrac{1}{2}\times 4.97\times (2.43)^2

d₂ = 64.73 m

The car further travel a distance of, d = 64.73 m - 5.904 m = 58.826 m

Hence, this is the required solution.

5 0
2 years ago
When asked to name all the forces on a marker sitting in equilibrium on a desk, a student responds: “Since the marker on the des
guapka [62]

Answer:

The forces acting on the pen which is still on the table can have two forces acting on them.  The forces are gravitational force and the equal and opposite force to the gravitational forces.

The equal and opposite forces that is applied on the pen keeps the pen still on the table.

So, the statement that no force is applied on the pen which is kept still on the table is wrong as two forces are applied on the pen.

As both the forces are equal and opposite so it is cancelled and is still.

6 0
3 years ago
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