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max2010maxim [7]
3 years ago
8

5. The disks, or pads, that exist between bones as gliding joints are made of which substance?

Physics
1 answer:
DerKrebs [107]3 years ago
5 0

Answer:

cartilage

Explanation:

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A net force of 50 newtons is applied to a 20 kilogram cart that is already moving at 1 m/s the final speed of the cart was 3 m/s
valina [46]

Answer:

0.8 seconds

Explanation:

F=ma

Let x be the seconds the force is applied.

m = 20kg

F = 50 Newtons (kg*m/sec^2)

acceleration, a, is provided for x seconds to increase the speed from 1 m/s to  3 m/s, an increase of 2m/s

Let's calculate the acceleration of the cart:

F=ma

(50 kg*m/s^2) = (20kg)*a

a = 2.5 m/s^2

---

The acceleration is 2.5 m/s^2.  The cart increases speed by 2.5 m/s every second.  

We want the number of seconds it takes to add 2.0 m/sec to the speed:

(2.5 m/s^2)*x = 2.0 m/s

x = (2.0/2.5) sec

x = 0.8 seconds

7 0
3 years ago
Lactic acid is produced during low-moderate exercise <br> True <br> False
Marizza181 [45]
The answer to your question is true.
3 0
4 years ago
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A LETTER FROM THE LORAX
KengaRu [80]

Answer:

Explanation:

You could try to say how helpful they are what they are and what they do

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3 years ago
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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
4 years ago
A cellist tunes the C string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.5
Dahasolnce [82]

Answer:

a

\lambda  = 1.18 \  m

b

v  =  77.172 \  m/s

c

T  = 151.41 \  N

Explanation:

From the question we are told that

   The frequency is  f =  65.4 \  Hz

   The  length of the vibrating string is  L  =  0.590 \  m

   The  mass is  m  =  15.0 \ g  =  0.015 \  kg

Generally the wavelength is mathematically represented as

           \lambda =  2 *  L

=>        \lambda  =  2 *   0.590

=>         \lambda  = 1.18 \  m

Generally the wave speed is  

          v  =  \lambda  *  f

=>       v  =  1.18 * 65.4

=>       v  =  77.172 \  m/s

Generally the tension on the wire is mathematically represented as

        T  =  v^2  *  \frac{ m }{L }

=>      T  =  77.172 ^2  *  \frac{  0.015  }{0.590}

=>      T  = 151.41 \  N

7 0
3 years ago
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