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Jobisdone [24]
3 years ago
8

A motor car is travelling at 144km/h in a 90km/h speed zone .The driver suddenly sees speed camera 80m ahead before the camera c

an read the cars speed .
Calculate the deceleration required to comply with the speed limit before being caught by the camera
Engineering
1 answer:
Kruka [31]3 years ago
4 0

The deceleration required to comply with the speed limit before being caught by the camera is 6.094 meters per square second.

Let assume that the motor car <em>decelerates</em> at <em>constant</em> rate. Given the <em>initial</em> and <em>final</em> speeds (v_{o}, v), in meters per second, and <em>travelled</em> distance (s), in meters, the deceleration (a), in meters per square second, is determined by this formula:

a = \frac{v^{2}-v_{o}^{2}}{2\cdot s}

If we know that v_{o} = 25\,\frac{m}{s}, v = 40\,\frac{m}{s} and s = 80\,m, then the deceleration required to comply with the speed limit is:

a = \frac{\left(25\,\frac{m}{s} \right)^{2}-\left(40\,\frac{m}{s} \right)^{2}}{2\cdot (80\,m)}

a = -6.094\,\frac{m}{s^{2}}

The deceleration required to comply with the speed limit before being caught by the camera is 6.094 meters per square second.

We kindly invite to check this question on uniform accelerated motion: brainly.com/question/12920060

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Answer: Option (b)

<u>Explanation:</u>

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This act provides some basic requirements for EPA Regulation such as follows;

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6 0
3 years ago
Question 1 A design team completes their high-fidelity prototype of a responsive website. Before they hand off designs to the en
MissTica

A question the design team should answer before handing off the designs is: are the designs a true representation of the intended end user experience?

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In conclusion, the design team should answer whether or not the designs are a true representation of the intended end user experience before handing off the designs.

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5 0
2 years ago
What is the difference between a refrigeration cycle and a heat pump cycle?
sukhopar [10]

Answer:

In refrigeration cycle heat transfer from inside refrigeration

In heat pump cycle heat transfer from environment

Explanation:

heat cycle is mechanical process use for cool the temperature but

In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator

5 0
3 years ago
What are the de Broglie frequencies and wavelengths of (a) an electron accelerated to 50 eV (b) a proton accelerated to 100 eV
DaniilM [7]

Answer:

(a) De-Brogie wavelength is 0.173 nm and frequency is 2.42 x 10^16 Hz

(b) De-Brogie wavelength is 2.875 pm and frequency is 4.8 x 10^16 Hz

Explanation:

(a)

First, we need to find velocity of electron. Since, it is accelerated by electric potential. Therefore,

K.E of electron = (1/2)mv² = (50 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 8 x 10^(-18) J

Mass of electron = m = 9.1 x 10^(-31) kg

Therefore,

v² = [8 x 10^(-18) J](2)/(9.1 x 10^(-31) kg)

v = √1.75 x 10^13

v = 4.2 x 10^6 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(9.1 x 10^(-31) kg)(4.2 x 10^6 m/s)

<u>λ = 0.173 x 10^(-9) m = 0.173 nm</u>

The frequency is given as:

Frequency = f = v/λ

f = (4.2 x 10^6 m/s)/(0.173 x 10^(-9) m)

<u>f = 2.42 x 10^16 Hz</u>

(b)

First, we need to find velocity of proton. Since, it is accelerated by electric potential. Therefore,

K.E of proton = (1/2)mv² = (100 eV)(1.6 x 10^-19 J/1 eV)

(1/2)mv² = 1.6 x 10^(-17) J

Mass of proton = m = 1.67 x 10^(-27) kg

Therefore,

v² = [1.6 x 10^(-17) J](2)/(1.67 x 10^(-27) kg)

v = √1.916 x 10^10

v = 1.38 x 10^5 m/s

Now, the de Broglie's wavelength is given as:

λ = h/mv

where,

h = Plank's Constant = 6.626 x 10^(-34) kg.m²/s

Therefore,

λ = (6.626 x 10^(-34) kg.m²/s)/(1.67 x 10^(-27) kg)(1.38 x 10^5 m/s)

<u>λ = 2.875 x 10^(-12) m = 2.875 pm</u>

The frequency is given as:

Frequency = f = v/λ

f = (1.38 x 10^5 m/s)/(2.875 x 10^(-12) m)

<u>f = 4.8 x 10^16 Hz</u>

6 0
3 years ago
. Two rods, with masses MA and MB having a coefficient of restitution, e, move
GarryVolchara [31]

Answer:

a) V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}

V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}

b) U_A = 3.66 m/s

V_B = 4.32 m/s

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let U_A be the initial velocity of rod A

Let U_B be the initial velocity of rod B

Let V_A be the final velocity of rod A

Let V_B be the final velocity of rod B

Using the principle of conservation of momentum:

M_AU_A + M_BU_B = M_AV_A + M_BV_B............(1)

Coefficient of restitution, e = \frac{V_B - V_A}{U_A - U_B}

V_A = V_B - e(U_A - U_B)........................(2)

Substitute equation (2) into equation (1)

M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B..............(3)

Solving for V_B in equation (3) above:

V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}....................(4)

From equation (2):

V_B = V_A + e(U_A -U_B)......(5)

Substitute equation (5) into (1)

M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))..........(6)

Solving for V_A in equation (6) above:

V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}.........(7)

b)

M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?

Rod A is said to be at rest after the impact, V_A = 0 m/s

Substitute these parameters into equation (7)

0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s

To calculate the final velocity, V_B, substitute the given parameters into (4):

V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s

c) Impulse, I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)

I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))

I = 0 kg m/s^2

d) %\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%

%\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%

% \triangle KE = -47.85 \%

7 0
3 years ago
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