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oee [108]
3 years ago
14

A 1100 kg automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a

constant acceleration of 5.0 m/s2. At the same instant a 2000 kg truck, traveling at a constant speed of 7.0 m/s, overtakes and passes the automobile. A)How far is the com of the automobile-truck system from the traffic light at t=3.0 s? B)What is the speed of the com then?
Physics
1 answer:
Viktor [21]3 years ago
6 0

Answer:

a ) 21 m b) 7 m/s

Explanation:

For this case, we can consider for our reference system, that point zero (this is, at the rest of the traffic signal) is when X₀ = 0, and direction of movement is from left to right (positive sign)

For the car:  

a= acceleration is constant = 5 m/s² m1 , mass = 1100 Kg , V = not constant

This is an uniformly accelerated rectilinear movement, and applicable formulas are:

V = V₀ + at , X = X₀ + V₀t + 1/2at²

For the truck:

V = speed is constant = 7 m/s , mass = 2000 Kg, a = 0

This is an uniform rectilinear movement, and the applicable formula is:

X = X₀ + Vt

a) At t = 3.0 sec, we can use the formula for the track, considering that X₀ = 0 (when it pass the rest of the traffic signal)

X = 0 + (7 m/s)x(3 s) = 21 m  

So, at 3 sec, truck will be at 21 m from the rest of traffic light and, as truck has a constant speed; at that exact second; truck and car will be together as a com, so both will be 21 m away from point zero

b) At the exact time (3 sec, or in distance words, 21 m) car and truck will be together as a com, so they will both have the exact speed, hence, speed of car at that point will be 7 m/s

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Design a modified priority encoder that receives an 8-bit input, A7:0and produces a 3-bit output, Y2:0. Y indicates the most sig
cupoosta [38]

Answer:

Explanation:

The first image that I attached to this solution is the diagram of a truth table for an 8 to 3 bit encoder.

The second image  gives a sketch of the schematic.

The Boolean expression for the priority encoder including its zero inputs is defined in the third image attached.

Below is a snippet of the code for an 8 to 3 bit Priority encoder:

library IEEE;

use IEEE.STD_LOGIC_1164.all;

entity encoder8_3 is

   port(

       din : in STD_LOGIC_VECTOR(7 downto 0);

       dout : out STD_LOGIC_VECTOR(2 downto 0)

        );

end encoder8_3;

architecture encoder8_3_arc of encoder8_3 is

begin

   dout <= "000" when (din="10000000") else

           "001" when (din="01000000") else

           "010" when (din="00100000") else

           "011" when (din="00010000") else

           "100" when (din="00001000") else

           "101" when (din="00000100") else

           "110" when (din="00000010") else

           "111";

end encoder8_3_arc;

3 0
3 years ago
A furniture shop refinishes chairs. Employees use one of two methods to refinish each chair. Method I takes 0.5 hours and the ma
satela [25.4K]

Answer:

method I = 88 chairs

method II = 62 chairs

Explanation:

This problem can be modeled by a system of two linear equations.

Define x as the number of chairs refinished by method I and y by method II

The sum of hours spent on both methods should equal 199 and the sum of total material cost should equal $1226, therefore:

0.5x + 2.5y = 199

9x+7y = 1226

Multiplying the first equation by -18 and adding it to the second equation we can solve for the value of y:

9x+7y +(-9x - 45)= 1226+(-3582)\\y=\frac{2356}{38} = 62\\

We can now apply the value of y found to the first equation and solve for x:

0.5x+2.5*62 = 199\\x=\frac{199 - (2.5*62)}{0.5} = 88

Therefore, they should refinish 88 chairs with method I and 62 chairs with method II

4 0
3 years ago
Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with
IgorLugansk [536]

(a) 2.5\cdot 10^{-6}eV

The energy of a photon is given by:

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

For the microwave photon,

\lambda=50.00 cm = 0.50 m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV

(b) 2.5 eV

For the energy of the photon, we can use the same formula:

E=\frac{hc}{\lambda}

For the visible light photon,

\lambda=500 nm = 5 \cdot 10^{-7}m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV

(c) 2500 eV

For the energy of the photon, we can use the same formula:

E=\frac{hc}{\lambda}

For the x-ray photon,

\lambda=0.5 nm = 5 \cdot 10^{-10}m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV

6 0
3 years ago
The product of the frequency and the wavelength of a wave equals the
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Answer:

b. speed of a wave

Explanation:

The speed of a wave is defined as the product between the wave's frequency and the wave's wavelength:

v=\lambda f

where

\lambda is the wavelength of the wave

f is the frequency

Therefore, we see that this matches the definition listed in choice B:

b. speed of the wave.

The other options are:

a. number of waves passing a point in a second.  --> frequency

c. distance between wave crests.  --> wavelength

d. time for one full wave to pass. --> period

5 0
3 years ago
A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T
Nady [450]
The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse
6 0
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