Why do stars appear to move across the sky?

Digiron [165]

Answer:

i) acceleration from B to D is 0, because the velocity is constant (stays the same)

ii) whatever units of distathat might be, we can calculate the number:

for 4 time-steps (2 to 6) the velocity is 6 per time step, that makes 24 distance units in these 4 time steps. it's the same the area underneath the graph.

there is also the vertical line from 0 to 2. we can calculate that distance like the area of a triangle with 2*6 / 2 = 6

the total distance from 0 to D is therefore 30

4 0

3 years ago

El tubo de entrada que suministra presión de aire para operar un gato hidráulico tiene 2 cm de diámetro. El pistón de salida es

Bess [88]

Answer:

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

Explanation:

Asumiendo que la presión ( $P$ ), medida en pascales, tiene una distribución uniforme sobre la superficie del pistón, se calcula a partir de la siguiente expresion:

$P = \frac{F}{A}$

Donde:

$F$ - Fuerza motriz, medida en newtons.

$A$ - Área del pistón, medida en metros cuadrados.

La fuerza motriz es equivalente al peso del automóvil. El área del pistón ( $A$ ), medido en metros cuadrados, es determinado por:

$A=\frac{\pi}{4}\cdot D^{2}$

Donde $D$ es el diámetro del pistón, medido en metros.

Si $D = 0.32\,m$ y $F =17,640\,N$ , entonces la presión neumática es:

$A = \frac{\pi}{4}\cdot (0.32\,m)^{2}$

$A \approx 0.080\,m^{2}$

$P = \frac{17,640\,N}{0.080\,m^{2}}$

$P = 220,500\,Pa$

La presión neumática para levantar un automóvil de 17,640 newtons es 220,500 pascales.

8 0

3 years ago

A 115-turn circular coil of radius 2.71 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coi

tester [92]

Answer:

80.6 mV

Explanation:

Parameters given:

Number of turns, N = 115

Radius of coil, r = 2.71 cm = 0.0271m

Time taken, t = 0.133s

Initial magnetic field, Bin = 50.1 mT = 0.0501 T

Final magnetic field, Bfin = 90.5 mT = 0.0905 T

Induces EMF is given as:

EMF = [(Bfin - Bin) * N * A] / t

EMF = [(0.0905 - 0.0501) * 115 * pi * 0.0271²] / 0.133

EMF = (0.0404 * 115 * 3.142 * 0.0007344) / 0.133

EMF = 0.0806 V = 80.6 mV

3 0

3 years ago

A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei

Alenkinab [10]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

$F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\$

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

$F = (150\ N)Sin\ 20^o$

8 0

3 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

4 years ago