2 ways in which a weather satellite can orbit the earth

Imagine two fixed charges on the x axis. Charge one is +q and is located to the left of charge two which is equal to -4q. Where

givi [52]

Answer: B)To the left of the charges.

Explanation: between the charges the electric field will not cancel but will be added since electric field lines from both charges point in the same direction. To the right of the charge the -4q will take over as it’s strength overcomes the strength of the +q charge. At this point the magnitude of +q will never reach a magnitude strong enough to cancel the -4q. To the left, it is further away from -4q and is closer to +q and electric field lines point in different direction

7 0

3 years ago

Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation d

Schach [20]

Answer:

Work out = 28.27 kJ/kg

Explanation:

For R-134a, from the saturated tables at 800 kPa, we get

$h_{fg}$ = 171.82 kJ/kg

Therefore, at saturation pressure 140 kPa, saturation temperature is

$T_{L}$ = -18.77°C = 254.23 K

At saturation pressure 800 kPa, the saturation temperature is

$T_{H}$ = 31.31°C = 304.31 K

Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.

Thus, $q_{reject}$ = $h_{fg}$ = 171.82 kJ/kg

We know COP of heat pump

COP = $\frac{T_{H}}{T_{H}-T_{L}}$

= $\frac{304.31}{304.31-254.23}$

= 6.076

Therefore, Work out put, W = $\frac{q_{reject}}{COP}$

= 171.82 / 6.076

= 28.27 kJ/kg

8 0

3 years ago

A baby born more than 3 weeks before the due date is considered

natulia [17]

Fetal because you are still in the womb and not fully developed

6 0

2 years ago

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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

In 2-3 complete sentences, explain why the needle on a compass always points in the direction of magnetic north.

Alex17521 [72]

The needle on a compass always points in the direction of magnetic north because of the magnetic poles of earth. the compass is essentially a magnet itself, so the southern pole of the compass is attracted to the northern pole of earth.

8 0

3 years ago

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