Who designs each page or screen with which a user interacts?

A driver traveling in her 16-foot SUV at the speed limit of 30 mph was arrested for running a red light at 15th and Main, an int

Lynna [10]

Answer:

(a) Yes

(b) 102.8 ft

Explanation:

(a)First let convert mile per hour to feet per second

30 mph = 30 * 5280 / 3600 = 44 ft/s

The time it takes for this driver to decelerate comfortably to 0 speed is

t = v / a = 44 / 10 = 4.4 (s)

given that it also takes 1.5 seconds for the driver reaction, the total time she would need is 5.9 seconds. Therefore, if the yellow light was on for 4 seconds, that's not enough time and the dilemma zone would exist.

(b) At this rate the distance covered by the driver is

$s = v_0t + \frac{at^2}{2}$

$s =44*1.5 + 44(4.4) - \frac{10*4.4^2}{2} = 162.8 (ft)$

Since the intersection is only 60 feet wide, the dilemma zone must be

162.8 - 60 = 102.8 ft

4 0

3 years ago

On a ship the price gift is 24 euros .What is the difference in the price on a day when the exchange rate is £1=2378

bija089 [108]

Answer:

The answer is " $\$29.7072$ ".

Explanation:

Please find the complete question in the attachment file.

As the original prices are 24 euros, $\$ 1.2378$ must be multiplied by 24.

$\to 1.2378 \times 24 = \$29.7072$

Therefore the new price value will be $\$ 29.7072$

4 0

3 years ago

Determine the slopes and deflections at points B and C for the beam shown below by the moment-area method. E=constant=70Gpa I=50

inn [45]

Answer:

hello your question is incomplete attached below is the complete question

answer :

Slopes : B = 180 mm , C = 373 mm

Deflection: B = 0.0514 rad , C = 0.077 rad

Explanation:

Given data :

I = 500(10^6) mm^4

E = 70 GPa

The M / EI diagram is attached below

Deflection angle at B

∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI

= 1800 / ( 500 * 70 ) = 0.0514 rad

slope at B

ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI

= 6300 / ( 500 * 70 ) = 0.18 m = 180 mm

Deflection angle at C

∅C = ∅CA = [ 1800 + 300*3 ] / EI

= 2700 / ( 500 * 70 )

= 2700 / 35000 = 0.077 rad

Slope at C

ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]

= 13050 / 35000 = 373 mm

3 0

3 years ago

What are two factors that determine the thermal energy of a substance

Sav [38]

Mass and chemical composition

6 0

3 years ago

1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are rem

BARSIC [14]

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁ = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

$\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

$\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}$

$P_2= \dfrac{1060}{5^{1.33}} = 124.65 \ kPa$

$\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

$\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2} \right$

$5^{0.33} = \left \dfrac{T_1}{T_2} \right$

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3× $c_v$ ×(T₂ - T₁)

$c_v$ = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86 kJ/(kg·K)

We use cv given by (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3× $c_v$ ×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = $U_g = h_g -pV_g$

$h_g$ = 2777.12 kJ/kg

$V_g$ = 0.194349 m³/kg

p = 1000 kPa

$U_{g1}$ = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, $U_{g1}$ + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

3 0

3 years ago

Who designs each page or screen with which a user interacts?​

Who designs each page or screen with which a user interacts?