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3 years ago

High- and low-pressure areas leading to air movement happen when there are differences in:

Physics

2 answers:

Charra [1.4K]3 years ago

7 0

The answer is A. Density

vazorg [7]3 years ago

4 0

I think the correct answer from the choices listed above is option A. High- and low-pressure areas leading to air movement happen when there are differences in density. Higher density means higher pressures and lower density means lower pressure. Hope this answers the question.

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A gasoline engine absorbs 2 500 J of heat and performs 1 000 J of mechanical work in each cycle. The efficiency of the engine is

larisa86 [58]

Answer:

The efficiency of the engine is 40 %

Explanation:

Given;

input energy, = 2500 J

output energy, = 1000 J

The efficiency of the engine is given as;

Efficiency = output energy / input energy

Efficiency = 1000 / 2500

Efficiency = 0.4

Efficiency(%) = 0.4 x 100%

Efficiency = 40 %

Therefore, the efficiency of the engine is 40 %

5 0

3 years ago

Explain retrograde motion and why it confused early astronomers as they viewed the motion of objects in the sky.

Masteriza [31]

Answer:

Explanation:

Retrograde motion in astronomy is, in general, orbital or rotational motion of an object in the direction opposite the rotation of its primary, that is, the central object (right figure). It may also describe other motions such as precession or nutation of an object's rotational axis. Prograde or direct motion is more normal motion in the same direction as the primary rotates. However, "retrograde" and "prograde" can also refer to an object other than the primary if so described. The direction of rotation is determined by an inertial frame of reference, such as distant fixed stars.

3 0

3 years ago

Read 2 more answers

A football player threw a ball upward at an angle or 24 degrees with the horizontal with A velocity of 18 m/s. What are the vert

svet-max [94.6K]

Answer:

$V_y = 16.44\ m/s$

$V_y = 7.32\ m/s$

Explanation:

Given

$\theta = 24$

$Velocity\ (V) = 18m/s$

Required

Determine the vertical and horizontal components

The vertical (Vy) and horizontal (Vx) components is calculated as thus:

$V_x = Vcos\theta$

$V_y = Vsin\theta$

Calculating Vertical Components:

$V_y = Vsin\theta$

$V_y = 18 * sin24$

$V_y = 18 * 0.40673664307$

$V_y = 7.32125957526$

$V_y = 7.32\ m/s$ --- Approximated

Calculating Horizontal Components:

$V_x = Vcos\theta$

$V_x = 18 * cos24$

$V_x = 18 * 0.91354545764$

$V_x = 16.4438182375$

$V_x = 16.44\ m/s$ --- Approximated

4 0

3 years ago

A resistor and an inductor are connected in series to a battery. The time constant for the circuit represents the time required

sattari [20]

E, 63% of the maximum current

5 0

4 years ago

Two spherical shells have a common center. A -1.50 × 10-6 C charge is spread uniformly over the inner shell, which has a radius

Murrr4er [49]

Answer:

a) At 0.20 m, the magnitude of the field is 675.0 kV

The direction of the field is acting outwards from the center of the charged spheres

b) At 0.10 m, the magnitude of the field is 135 kV

The direction is acting outwards from the center of the charged spheres

c) At 0.025 m

The magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres

Explanation:

The charged spherical shell parameters are;

The charge on the inner sphere, q₁ = -1.50 × 10⁻⁶ C

The radius of the inner shell, R₁ = 0.050 m

The charge on the outer sphere, q₂ = +4.50 × 10⁻⁶ C

The radius of the outer shell, R₂ = 0.15 m

Let 'r', represent the distance at which the electric field is measured, the following relationships can be obtained;

When r < R₁ < R₂,

$V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )$

When R₁ < r < R₂,

$V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )$

When R₁ < R₂ < r,

$V = k \cdot \left( \dfrac{q_1 + q_2}{r^2} \right )$

a) When r = 0.20 m, we have;

R₁ < R₂ < r, therefore

$V = k \cdot \left( \dfrac{q_1 + q_2}{r^2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} + 4.50\times 10^{-6} }{0.20^2} \right ) = 675.0 \ kV$

Therefore, the magnitude of the field, V = 675.0 kV

The direction of the field is outwards

b) When r = 0.10 m, we have;

When R₁ < r < R₂, therefore;

$V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} }{0.10} + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = 135 \ kV$

Therefore, the magnitude of the field, V = 135 kV

The direction of the field is outwards from the center

c) When r = 0.025 m, we have;

When r < R₁ < R₂, therefore;

$V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} }{0.05} + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = -270 \ kV$

Therefore, the magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres.

4 0

3 years ago