The human body station 2

Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 13.9 revolutions/s. Find the linear speed v

Lostsunrise [7]

Answer:

the linear speed of the car is 28.83 m/s

Explanation:

Given;

radius of the car, r = 0.33 m

angular speed of each tire, ω = 13.9 rev/s = 13.9 x 2π = 87.35 rad/s

The linear speed of the car is calculated as;

V = ωr

V = 87.35 rad/s x 0.33 m

V = 28.83 m/s

Therefore, the linear speed of the car is 28.83 m/s

3 0

3 years ago

A 0.31 kg cart on a horizontal frictionless track is attached to a string. The string passes over a disk-shaped pulley of mass 0

Stella [2.4K]

To solve this problem it is necessary to apply the concepts related to Newton's second law and its derived expressions for angular and linear movements.

Our values are given by,

$M_{cart} = 0.31kg\\m_{pulley} = 0.08kg\\r_{pulley} = 0.012m\\F = 1.1N\\$

If we carry out summation of Torques on the pulley we will have to,

$F_2*d-F_1*d = I \alpha$

Where,

I = Inertia moment

$\alpha =$ Angular acceleration, which is equal in linear terms to a/r (acceleration and radius)

The moment of inertia for this object is given as

$I = \frac{1}{2} mr^2$

Replacing this equations we have know that

$(F_2 - F_1)d = (\frac{1}{2}(m_{pulley})r^2) (\frac{a}{r})$

$F_2 - F_1 = \frac{1}{2}m_{pulley} \frac{F_1}{M_{cart}}$

$F_2 = (1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))F_1$

Or

$F_1 = \frac{F_2}{(1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))}$

Replacing our values we have that

$F_1 = \frac{1.1}{(1 + (0.5)(\frac{0.08}{0.31}))}$

$F_1 = 0.974 N$

Therefore the tension in the string between the pulley and the cart is 0.974 N

6 0

4 years ago

Which equations represent the relationship between wavelength and frequency for a sounds wave

ioda

Waves can be described using a number of different characteristics of a wave. Wavelength and frequency are two such characteristics. The relationship between wavelength and frequency is that the frequency of a wave multiplied by its wavelength gives the speed of the wave

6 0

3 years ago

A car traveling at 60 mph has how much more energy than a car going at 20 mph? How many times does the kinetic energy of a car i

irinina [24]

K.E. increases by 9 times

Explanation:

The kinetic energy of a car is given by:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the car

v is its speed

From this definition, we see that the kinetic energy depends on the square of the velocity. Assuming that both cars have same mass, m, the kinetic energy of the first car is:

$K_1 = \frac{1}{2}m(60)^2$

while the kinetic energy of the second car is

$K_2 = \frac{1}{2}m(20)^2$

if we calculate the ratio, we get

$\frac{K_1}{K_2}=\frac{(60)^2}{(20)^2}=3^2 =9$

5 0

3 years ago

Read 2 more answers

A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius

alisha [4.7K]

Answer:

I2 = 3.076 A

Explanation:

In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:

$B=\frac{\mu_oI}{2R}$ (1)

I: current in the wire

R: radius of the wire

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:

$B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}$ (2)

I1: current of the first ring = 8A

R1: radius of the first ring = 0.078m

I2: current of the second ring = ?

R2: radius of the first second = 0.03m

To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:

$\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A$

The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.

5 0

3 years ago