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Gelneren [198K]

2 years ago

8

A particle is located on the x axis at x = 2.0 m from the origin. A force of 25 N, directed 30° above the x axis in the x-y plan

e, acts on the particle. What is the torque about the origin on the particle?

1 answer:

nata0808 [166]2 years ago

7 0

Answer:

25 Nm

Explanation:

Parameters given

Distance of the particle from the origin, r = 2.0 m

Force acting on particle, F = 25 N

Angle of force, θ = 30°

The torque acting on a particle is given as:

τ = r * F * sinθ

Where r = radius of axis of rotation. This is the same as the position of the particle on the x axis.

Therefore, the torque will be:

τ = 2 * 25 * sin30

τ = 25 Nm

The torque acting on the particle is 25 Nm.

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After an arrow is shot, is the force unbalanced or balanced? BRAINLY.

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Answer:

The force is unbalanced

Explanation:

After an arrow is shot, the force acting on the arrow is unbalanced. The resulting net force gives the arrow an initial acceleration which wanes with time and the body is brought to rest.

The net force acting on an arrow is not zero and this indicates that the forces acting on the arrow is unbalanced.

If the force is balanced, the arrow is expect to continue in uniform motion but that is not the case as air resistance has massive impact on this body.

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2 years ago

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

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Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

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2 years ago

A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp

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Answer:

$W_2=\sqrt{\frac{3}{5} }W_1$

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

$I_1 =\frac{2}{5}MR^2$

$K_1 = \frac{1}{2}IW_1^2$

So, replacing, we get that:

$K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

At the same way, the moment of inertia and kinetic energy for second ball is:

$I_2 =\frac{2}{3}MR^2$

$K_2 = \frac{1}{2}IW_2^2$

So:

$K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2$

Then, $K_2$ is equal to $K_1$ , so:

$K_2 = K_1$

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$\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2$

Finally, solving for $W_2$ , we get:

$W_2=\sqrt{\frac{3}{5} }W_1$

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Answer:

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The maximum value of Sin2θ is 1.

It means 2θ = 90

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Thus, the range is maximum when the angle of projection is 45 degree.

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2 years ago