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

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

Read 2 more answers

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

5 0

3 years ago

Read 2 more answers

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)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

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

⇒A₀ $e^{\frac{-bt}{2m}}$ =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