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2 years ago

A 55 kg roller skater is at rest on a flat skating rink, a 198 N horizontal force is needed to set the skater in motion.

Physics

1 answer:

Rufina [12.5K]2 years ago

8 0

Answer:

Explanation:

To get the person Moving you have to overcome the static (means not moving) friction coefficient. U(static)

To get the person going at the same speed you have to overcome the kinetic friction coefficient. U(Kinetic)

Force to get him moving is 198 N. Force = ma = U(static)Mg

combining the 2 equations you get 198N = U(static)* 55kg *9.8m/s^2 Solve for U(static)

Same equation to keep him moving except with the dynamic force and the dynamic U

175N= U(kinetic)*55kg*9.8m/s^2 Solve (U dynamic)

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Calculate their densties in Si unit.<br>200mg,0.0004m

Kryger [21]

Question: calculate their densties in Si unit.

200mg,0.0004m³

Answer:

0.5 kg/m³

Explanation:

Applying,

D = m/V........................ Equation 1

Where D = density, m = mass, V = volume.

From the question,

Given: m = 200 mg = (200/1000000) kg = 2.0×10⁻⁴ kg, V = 0.0004 m³ = 4.0×10⁻⁴ m³

Substitute these values into equation 1

D = (2.0×10⁻⁴ kg)/(4.0×10⁻⁴)

D = 2/4

D = 0.5 kg/m³

Hence the density in S.I unit is 0.5 kg/m³

3 0

2 years ago

Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each ar

liberstina [14]

Answer:

∑ τ =0, L₀ = $L_{f}$

Explanation:

In a circular turning movement, when the arms are extended and then contracted in two possibilities:

- They are lowered the force of gravity is what pulls them, the tension of the muscle becomes zero to allow this movement.

In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.

- Another possibility is that the arms have stuck to the body, in this case the person's muscles perform the force, this force is horizontal and the displacement is the horizontal of the center of mass of the arms from the extended position to the contracted

In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant

∑ τ =0

L₀ = $L_{f}$

I₀ w₀ = I w

4 0

3 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

tia_tia [17]

Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

$a_{c}=\omega ^{2}r$

where:

$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

$\omega=72.72x10^{-6} rad/s^{2}$

So the centripetal acceleration will be:

$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$