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stiks02 [169]
2 years ago
10

What simple machine makes up most of the joints in your body

Physics
1 answer:
Vladimir [108]2 years ago
3 0
Levers
Most machines in your body are levers that consist of bones and muscles. Tendons and muscles pull on bones, making them act as levers. Joint, near where tendon is attached to bone acts as fulcrum. Muscles produce input force.
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A solid cylinder of mass 3.0 kg and radius 0.2 m starts from rest at the top of a ramp, inclined 15°, and rolls to the bottom wi
Lera25 [3.4K]

Answer:

v = 4.4271 m/s

Explanation:

Given

m = 3 Kg

R = 0.2 m

∅ = 15°

h = 1.5 m

g = 9.8 m/s²

v = ?

Ignoring frictional losses, at the bottom of the plane

Total kinetic energy is  =  Potential Energy at the top of plane

Using Law of conservation of energy we have

U = Kt + Kr

m*g*h = 0.5*m*v² + 0.5*I*ω²

knowing that

Icylinder = 0.5*m*R²

ω = v/R

we have

m*g*h = 0.5*m*v² + 0.5*(0.5*mR²)*(v/R)² = 0.75*m*v²

⇒  v = √(g*h/0.75) = √(9.8 m/s²*1.5 m/0.75)

⇒  v = 4.4271 m/s

3 0
3 years ago
Determine the stopping location of the prize wheel. At this moment it is centered on the number 11. It is spinning at a rate of
Goshia [24]

Answer:

The wheel spins for 11.43s. The number cannot be determined.

Explanation:

We know that initial velocity is 124.4 rpm = 13.027137522 rad/s, acceleration is -1.14 rad/s^2, and final velocity is assumed to be 0 rad/s. We are asked to find t and displacement. We use the equation \omega=\omega_0+\alpha t where \omega=0 rad/s, \omega_0=13.027 rad/s, \alpha=-1.14rad/s^2, and t=?.

Rearrange the equation to obtain \frac{\omega-\omega_0}{\alpha}=t. Plug in the numbers and solve to obtain t=11.43s.

The number of the wheel cannot be determined as we do not know the placement of numbers on the wheel.

5 0
3 years ago
What is the potential energy for and object with mass of 15 Kg and 20 m above the ground?
vaieri [72.5K]

Answer:

3000J

Explanation:

The gravitational energy for any object is given by E=mgh

where m is mass, g is gravitational field strength anf h is height above ground

5 0
3 years ago
What would happen to the apparent change in mass if the direction of the current is reversed?
pantera1 [17]
When you reverse the direction of the current, the current loop generated by the magentic field is revered.
8 0
3 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
2 years ago
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