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WARRIOR [948]
3 years ago
6

3. A car has a mass of 2.50 x 10^3 kg. If the force acting on the car is 7.65 x 10^3 N to the

Physics
1 answer:
const2013 [10]3 years ago
6 0

Answer:

3.06m/s² to the east

Explanation:

Given parameters:

Mass of car = 2.5 x 10³kg

Force acting on the car  = 7.65 x 10³N

Unknown:

Acceleration of the car  = ?

Solution:

From Newton's second law of motion:

      Force  = mass x acceleration

   Acceleration  = \frac{Force }{mass}   = \frac{7.65 x 10^{3} }{2.5 x 10^{3} }    = 3.06m/s² to the east

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Car A hits car B (initially at rest and of equal mass) frombehind while going 35 m/s. Immediately after the collision, car Bmove
kolezko [41]

Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2

kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2

Since the masses are equals:

m=ma=mb

For the known velocities, the kinetics energies result:

ki=0.5*mVa_{i}^2

ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m

kf=0.5*mbVb_{f}^2

kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m

The lost energy in the collision is the difference between the initial and final kinectic energies:

kl=ki-kf

kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m

Finally the relation between the lost and the initial kinetic energy:

kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m

kl/ki = 24/49=0.49

7 0
3 years ago
What is the distance from one crest to the next crest?
ella [17]

<span>The distance between wave crests is called wavelength. It is a characteristic shared by waves of all kinds, including ocean waves and sound waves. Wavelength is measured from the highest point, or summit, of one wave's crest to the summit of the next wave's <span>crest</span></span>

<span><span>hope this helps</span></span>

3 0
3 years ago
Which is a likely use for a base? a. as a vitamin in your food b. etching metals for printing c. making foods taste sour d. maki
natta225 [31]
I think the correct answer would be making soaps and detergents. It is the manufacture of soaps and detergents that makes use of a base. It is included in the process of such product. Hope this answers the question. Have a nice day.
5 0
3 years ago
A polarized light that has an intensity I0 = 60.0 W/m² is incident on three polarizing disks whose planes are parallel and cente
nikitadnepr [17]

Answer:

The transmitted intensity through all polarizers is I_3 =41.31 W/m^2

Explanation:

 According to Malu's law the intensity of a polarized light having an initial intensity I_0 is mathematically represented as

               I = I_0cos^2 \theta

Now  considering the polarizer(The polarizing disk) the equation above becomes

          I = I_0 (cos^2 \theta)^n

Where n is the number of polarizers

       Substituting  60.0W/m^2 for the initial intensity 3 for the n and 20° for the angle of rotation

           I_3 = 60 (cos^220)^3

               =41.31 W/m^2

             

     

                         

6 0
3 years ago
Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(
bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude  = 2.20 mm

b) frequency =

     f = \dfrac{\omega}{2\pi}

     f = \dfrac{743}{2\pi}

          f = 118.25 Hz

c) wavelength

        k= \dfrac{2\pi}{\lambda}

        \lambda= \dfrac{2\pi}{k}

        \lambda= \dfrac{2\pi}{7.02}

        \lambda= 0.895\ m

d) speed

         v = \dfrac{\omega}{k}

         v = \dfrac{743}{7.02}

                v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

  T = \dfrac{v^2\ m}{L}

  T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}

      T = 27.87 N

g) Power transmitted by the wave

  P = \dfrac{1}{2}m\ v \omega^2\ A^2

  P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2

      P = 0.438 W

5 0
3 years ago
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