Please help on this one someone?

What is the main function of the spongy bone'

storchak [24]

Oooooo there's a spongy bone? that's cool! Lol okay okay, I will research it and help you out.

Here's what I found:

Cancellous bone, also known as spongy or trabecular bone, is one of the two types of bone tissue found in the human body. ... It is very porous and contains red bonemarrow, where blood cells are made.

7 0

3 years ago

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Is the muzzle velocity a scalar or vector quantity?

Vlad [161]

It had better be a vector, otherwise there's be no excuse for calling it a "velocity". It would just be the muzzle speed.

8 0

3 years ago

According to the scientific observation that energy is conserved in a "closed system,” no machine can do work on its surrounding

tensa zangetsu [6.8K]

I believe it is theory

7 0

4 years ago

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A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

4 0

3 years ago

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It takes a minimum distance of 41.14 m to stop a car moving at 11.0 m/s by applying the brakes (without locking the wheels). Ass

padilas [110]

Answer:

Find the time it took for the car to stop at 11.0m/s

V = deltax/t

t = 41.14/11.0 = 3.74s

Now find at what rate it was decelerating, so find the acceleration during that interval of time.

vf = vi + at

-11.0m/s = a3.74s

a = -2.94m/s^2

The acceleration is negative because is pulling the car towards its opposite direction to make it stop.

Now find how much time it would take for the car to stop at 28.0m/s but with the same acceleration, the car is the same so its acceleration to stop the car will remain the same.

vf = vi + at

0 = 28.0 - 2.94t

t = 9.52

Once the time is obtained, you can find the final position, xf, by plugging the time acceleration and velocity values.

xf = 0 + (28m/s)(9.52s) + 1/2(-2.94)(9.52s)^2

xf = 266.6m - 133.23m = 133m

8 0

4 years ago