Ask question

Ask question

All categories

4 years ago

3

Given two integer variables distance and speed, write an expression that divides distance by speed using floating point arithmet

ic, i.e. a fractional result should be produced.

1 answer:

pychu [463]4 years ago

7 0

Answer:

$\dfrac{Distance}{(double) speed}$ .

Explanation:

For the given two integer variables speed and distance ,the expression that divides distance by speed given as $\dfrac{Distance}{(double) speed}$ .

So the expression is $\dfrac{Distance}{(double) speed}$ .

$\dfrac{Distance}{(double) speed}$ it means that when we double the speed of a vehicle then the braking distance travel will become one forth.

You might be interested in

19 million Joules of chemical energy are supplied to a boiler in a power plant. 5 x 106 Joules of energy are lost through unburn

Fiesta28 [93]

Answer:

67.89%

Explanation:

Energy Efficiency= $\frac {U_{energy}}{T_{energy}}\times 100$

Where $U_{energy}$ is useful energy and $T_{energy}$ is the total energy

The useful energy= (19-5-1.1) million= 12.9 million

The total energy= 19 million

$Efficiency=\frac {12.9}{19}\times 100=67.89473684\approx 67.89$

Therefore, the efficiency of the boiler is 67.89%

8 0

3 years ago

Sea B = 5.00 m a 60.0°. Sea C que tiene la misma magnitud que A y un ángulo de dirección mayor que el de A en 25.0°. Sea A ⦁ B =

uranmaximum [27]

Answer:

$\| \vec A \| = 6.163\,m$

Explanation:

Sean A, B y C vectores coplanares tal que:

$\vec A = (\| \vec A \|\cdot \cos \theta_{A},\| \vec A \|\cdot \sin \theta_{A})$ , $\vec B = (\| \vec B \|\cdot \cos \theta_{B},\| \vec B \|\cdot \sin \theta_{B})$ y $\vec C = (\| \vec C \|\cdot \cos \theta_{C},\| \vec C \|\cdot \sin \theta_{C})$

Donde $\| \vec A \|$ , $\| \vec B \|$ y $\| \vec C \|$ son las normas o magnitudes respectivas de los vectores A, B y C, mientras que $\theta_{A}$ , $\theta_{B}$ y $\theta_{C}$ son las direcciones respectivas de aquellos vectores, medidas en grados sexagesimales.

Por definición de producto escalar, se encuentra que:

$\vec A \,\bullet\, \vec B = \|\vec A \| \| \vec B \| \cos \theta_{B}\cdot \cos \theta_{A} + \|\vec A \| \| \vec B \| \sin \theta_{B}\cdot \sin \theta_{A}$

$\vec B \,\bullet\, \vec C = \|\vec B \| \| \vec C \| \cos \theta_{B}\cdot \cos \theta_{C} + \|\vec B \| \| \vec C \| \sin \theta_{B}\cdot \sin \theta_{C}$

Asimismo, se sabe que $\| \vec B \| = 5\,m$ , $\theta_{B} = 60^{\circ}$ , $\vec A \,\bullet \,\vec B = 30\,m^{2}$ , $\vec B\, \bullet\, \vec C = 35\,m^{2}$ , $\|\vec A \| = \| \vec C \|$ y $\theta_{C} = \theta_{A} + 25^{\circ}$ . Entonces, las ecuaciones quedan simplificadas como siguen:

$30\,m^{2} = 5\|\vec A \| \cdot (\cos 60^{\circ}\cdot \cos \theta_{A} + \sin 60^{\circ}\cdot \sin \theta_{A})$

$35\,m^{2} = 5\|\vec A \| \cdot [\cos 60^{\circ}\cdot \cos (\theta_{A}+25^{\circ}) + \sin 60^{\circ}\cdot \sin (\theta_{A}+25^{\circ})]$

Es decir,

$30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})$

$35\,m^{2} = \| \vec A \| \cdot [2.5\cdot \cos (\theta_{A}+25^{\circ})+4.330\cdot \sin (\theta_{A}+25^{\circ}})]$

Luego, se aplica las siguientes identidades trigonométricas para sumas de ángulos:

$\cos (\theta_{A}+25^{\circ}) = \cos \theta_{A}\cdot \cos 25^{\circ} - \sin \theta_{A}\cdot \sin 25^{\circ}$

$\sin (\theta_{A}+25^{\circ}) = \sin \theta_{A}\cdot \cos 25^{\circ} + \cos \theta_{A} \cdot \sin 25^{\circ}$

Es decir,

$\cos (\theta_{A}+25^{\circ}) = 0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A}$

$\sin (\theta_{A}+25^{\circ}) = 0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A}$

Las nuevas expresiones son las siguientes:

$30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})$

$35\,m^{2} = \| \vec A \| \cdot [2.5\cdot (0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A})+4.330\cdot (0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A})]$

Ahora se simplifican las expresiones, se elimina la norma de $\vec A$ y se desarrolla y simplifica la ecuación resultante:

$30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})$

$35\,m^{2} = \| \vec A \| \cdot (4.097\cdot \cos \theta_{A} +2.865\cdot \sin \theta_{A})$

$\frac{30\,m^{2}}{2.5\cdot \cos \theta_{A}+ 4.330\cdot \sin \theta_{A}} = \frac{35\,m^{2}}{4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}}$

$30\cdot (4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}) = 35\cdot (2.5\cdot \cos \theta_{A}+4.330\cdot \sin \theta_{A})$

$122.91\cdot \cos \theta_{A} + 85.95\cdot \sin \theta_{A} = 87.5\cdot \cos \theta_{A} + 151.55\cdot \sin \theta_{A}$

$35.41\cdot \cos \theta_{A} = 65.6\cdot \sin \theta_{A}$

$\tan \theta_{A} = \frac{35.41}{65.6}$

$\tan \theta_{A} = 0.540$

Ahora se determina el ángulo de $\vec A$ :

$\theta_{A} = \tan^{-1} \left(0.540\right)$

La función tangente es positiva en el primer y tercer cuadrantes y tiene un periodicidad de 180 grados, entonces existen al menos dos soluciones del ángulo citado:

$\theta_{A, 1} \approx 28.369^{\circ}$ y $\theta_{A, 2} \approx 208.369^{\circ}$

Ahora, la magnitud de $\vec A$ es:

$\| \vec A \| = \frac{35\,m^{2}}{4.097\cdot \cos 28.369^{\circ} + 2.865\cdot \sin 28.369^{\circ}}$

$\| \vec A \| = 6.163\,m$

8 0

4 years ago

Technician A says that tire markings now include load ratings on tires. Technician B says that they are only included on truck t

Ludmilka [50]

Technician A is correct hope this helps

5 0

3 years ago

Which of the following are true about algorithms? (You may select more than one)

r-ruslan [8.4K]

Have a orderf. Fr

R
F
F
D
Djeiiekee

7 0

4 years ago

Read 2 more answers

In the truss below, the vertical reaction force at A is

Montano1993 [528]

It’s c trust just trust

7 0

3 years ago