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3 years ago

How does a bicycles kinetic energy change when it slows down?

Physics

1 answer:

-Dominant- [34]3 years ago

7 0

The kinetic energy decreases

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The chart below lists the densities of various gemstones. If a gemstone has a mass of 6.24 g and a volume of 1.98 cm cubed, whic

xxMikexx [17]

Answer:

C. Garnet

Explanation:

density = mass/volume

density = 6.24g/1.98cm^3

density = 3.15 g/cm^3 (Garnet)

Hope this helps!

5 0

2 years ago

A spring with spring constant 13.1 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

Alenkinab [10]

Answer:

Explanation:

spring constant, K = 13.1 N/m

22 oscillations in 20 seconds

time taken to complete one oscillation is called time period.

T = 20 / 22 second = 0.909 seconds

(a) let m be the mass.

The formula for the time period is

$T = 2\pi \sqrt{\frac{m}{K}}$

$m = \frac{T^{2}K}{4\pi ^{2}}$

$m = \frac{0.909^{2}\times 13.1}{4\pi ^{2}}$

m = 0.275 kg

(b) maximum speed, v = ω A = 2π A / T

v = ( 2 x 31.4 x 0.1) / 0.909

v = 0.691 m/s

6 0

3 years ago

The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 s, what are the (a) di

nlexa [21]

Answer and Explanation:

Let:

$x(t)=Acos(\omega t+ \phi)$

The equation representing a simple harmonic motion, where:

$x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time$

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

$v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)$

Also, you may know these fundamental formulas:

$f=\frac{\omega}{2 \pi} \\\\T=\frac{2 \pi}{\omega}$

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

$x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m$

(b)

$v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s$

(c)

$a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2$

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

$\phi = \frac{\pi}{5}$

(e)

$f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz$

(f)

$T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s$

8 0

2 years ago

Read 2 more answers

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$