What will happen if you rub two different materials together?

Without the wheels, a bicycle frame has a mass of 8.29 kg. Each of the wheels can be roughly modeled as a uniform solid disk wit

trapecia [35]

Answer:

69.66 Joule

Explanation:

mass of bicycle frame, mf = 8.29 kg

mass of wheel, mw = 0.820 kg

radius, r = 0.343 m

velocity, v = 3.6 m/s

There are two wheels in the bicycle.

There are two types of kinetic energy of the system one is kinetic energy of rotation and another is rotational kinetic energy.

$K = \frac{1}{2}m_{f}v^{2}+ 2\times \frac{1}{2}m_{w}v^{2}+ 2\times \frac{1}{2}I_{w}\omega^{2}$

$K = \frac{1}{2}m_{f}v^{2}+m_{w}v^{2}+ \frac{1}{2}\times m_{w}v^{2}$

$K = \frac{1}{2}m_{f}v^{2}+ \frac{3}{2}\times m_{w}v^{2}$

$K = \frac{1}{2}\times 8.29\times 3.6^{2}+ \frac{3}{2}\times 0.820\times 3.6^{2}$

K = 69.66 J

3 0

4 years ago

Help please!!! Physics circular motion

svp [43]

Answer:

2. 3.1415 m/s

3. 0.63m

4. 0.006 m/s^2

Explanation:

2. v=(2*pi*r))/T, put in values and solve.

3. Circumference=2*pi*radius, radius is 0.1m, plug in and solve.

4. a=(v^2)/r, (don't worry about converting to meters, since units are the same in kilometers, they will cancel out) plug in values and solve.

That's it!

6 0

3 years ago

Two circular loops of wire, each containing a single turn, have the same radius of 5.10 cm and a common center. The planes of th

Fofino [41]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

A distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 n

maksim [4K]

Answer:

A) receding from the earth

B) $3.078x10^6m/s$

Explanation:

A) receding from the earth

The wavelength went from 434.1nm to 438.6nm, there was an increase in wavelength (also knowecn as redshift due to the doppler efft), this increase is due to the fact that the source that emits the radiation (the distant galaxy) is moving away and therefore the light waves it emits are "stretched", causing us to see a wavelength greater than the original.

B) $3.078x10^6m/s$

to calculate the relative speed we use the following formula:

$v_{rel}=c(1-\frac{\lambda_{1}}{\lambda_{2}} )$

where $c$ is the speed of light: $c=3x10^8m/s$

$\lambda_{1}$ is the wavelength emited by the source, and

$\lambda_{2}$ is the wavelength measured on earth.

we substitute all the values and do the calculations:

$v_{rel}=(3x10^8m/s)(1-\frac{434.1nm}{438.6nm} )\\\\v_{rel}=(3x10^8m/s)(1-0.98974)\\\\v_{rel}=(3x10^8m/s)(0.01026)\\\\v_{rel}=3.078x10^6m/s$

the relative speed is: $3.078x10^6m/s$

5 0

3 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago