All categories

3 years ago

What is the correct definition of a fracture?

Physics

1 answer:

MA_775_DIABLO [31]3 years ago

4 0

The correct definition of a fracture is break in the bone

Explanation:

When nay injury results in the breaking or causing any cracking in the bones of any parts then this will lead to fracture. When the injury caused is near the ligament or tissue in which the bone is connected or attached then it will lead to an avulsion fractures. Thus this will lead to the pulling of bone form the original position thereby leading more pain in the spot associated with the fracture.

Sports people are the victims of this type of fracture. Fracture may occur anywhere mostly legs,hands,ankle,hip and elbow. sometimes it may be in finger, shoulder,knee,etc. The main symptoms that are associated with fracture includes, selling, inability in moving the fractured part or pain associated when trying to move that part, Loss of the affected part's function,etc.

You might be interested in

Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.

Dennis_Churaev [7]

Answer:

$\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}$

Explanation:

The rotational kinetic energy when the cylinder is with the rope is:

$E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2$

where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:

$E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})}$ (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

$I_c=\frac{1}{2}MR^2\\\\I_r=mR^2$

Finally, by replacing in (1):

$\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}$

hope this helps!!

7 0

3 years ago

A skateboarder traveling with an initial velocity 9.0 meters per second,

meriva

Answer:

25m/s

Steps:

First, The equation v= u + a * t shows us what we need to find, (the finale velocity).

Second, we substitute the values given:

v= 9m/s + 4m/s2 * 4s

Last, We calculate the values:

Multiply 4m/s2 * 4s = 16m/s

Add 9m/s + 16m/s

Answer: 25m/s

Hope this helps :)

4 0

3 years ago

A proton and an electron are moving in the +x direction in a magnetic field in the +z

Cloud [144]

Answer:

Explanation:

A proton and electron are moving in the positive x direction, this shows that their velocity will be in the positive x direction

V = v•i

Magnetic field Is the positive z direction

B = B•k

A. For proton.

Proton has a positive charge of q

Direction of force on proton

Force is given as

F = q(v×B)

F = q( v•i × B•k)

F = qvB (i×k)

From vectors i×k = -j

F = -qvB •j

Then, for the positive charge, the force will act in the negative direction of the y-axis

B. For electron

Electron has a negative of -q

Direction of force on proton

Force is given as

F = q(v×B)

F = -q( v•i × B•k)

F = -qvB (i×k)

From vectors i×k = -j

F = --qvB •j

F = qvB •j

Then, for the negative charge, the force will act in the positive direction of the y-axis

5 0

3 years ago

A sled is moving at a constant speed down a surface inclined at 45 degrees with the horizontal and travels 30meter in 4 seconds.

Vika [28.1K]

Constant speed along the inclined surface = 30 m / 4 s = 7.5 m/s

Vertical speed = inclined speed * sin(45) = 7.5 *√2 / 2 = 5.3 m/s

Answer: 5.3 m/s

8 0

3 years ago

PLEASE HELP

hichkok12 [17]

The correct answer is 195.6 N

Explanation:

Different from the mass (total of matter) the weight is affected by gravity. Due to this, the weight changes according to the location of a body in the universe as gravity is not the same in all planets or celestial bodies. Moreover, this factor is measured in Newtons and it can be calculated using this simple formula W (Weight) = m (mass) x g (force of gravity). Now, leps calculate the weigh of someone whose mass is 120 kg and it is located on the moon:

F = 120 kg x 1.63 m/s2

F= 195.6 N

7 0

3 years ago