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8 days ago

The ultimate source of energy that powers the Sun is__________.

Physics

2 answers:

Maru [2.3K]8 days ago

5 0

The primary source of energy powering the Sun is__________.

kicyunya [2.2K]8 days ago

3 0

the answer is D the kinetic energy from the Sun's orbital motion

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An object with charge q=−6.00×10−9C is placed in a region of uniform electric field and is released from rest at point A. After

kicyunya [2264]

Answer:

it’s a

Explanation:

utilized Google

8 0

21 day ago

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Softa [2029]

Answer:

The plane's speed in relation to the ground is 300.79 km/h.

Explanation:

Provided details include:

Wind speed = 75.0 km/hr

Plane's airspeed = 310 km/hr

Next, we must find the ground speed of the plane

Calculating the angle

Using the angle formula

$\sin\theta=\dfrac{v'}{v}$

Where v' represents the wind speed

v represents the plane's speed

We will substitute the values into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

Now, we must find the resultant speed

Using the resultant speed formula

$\cos\theta=\dfrac{v''}{v}$

Insert the values into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Consequently, the plane's speed in relation to the ground equals 300.79 km/h.

6 0

28 days ago

A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate sepa

Keith_Richards [2256]

Answer:2.53*10^-10F

Explanation:

C=£o£r*A/d

Where £ represents the permittivity constant

£o= 8.85*10^-12f/m

£r=6.3

A=150mm^2=0.015m^2

d=3.3mm= 0.0033m

C=8.85*10^-12*6.3*0.015/0.0033

C=8.85*6.3*10^-12*0.015/0.0033

C=55.755*0.015^-12/0.003

C=8.36/3.3*10^-13+3

C=2.53*10^-10F

7 0

22 days ago

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Yuliya22 [2420]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

It is known that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v are constant on both sides. Therefore we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let the final pressure be P and the temperature $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide equation (2) by equation (3)

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, the left side temperature equals 1.48 times the right side temperature

6 0

12 days ago

The flight of a kicked football follows the quadratic function f(x)=−0.02x2+2.2x+2, where f(x) is the vertical distance in feet

Keith_Richards [2256]

The ball covers a horizontal distance of 0.902 meters. The trajectory of a kicked football adheres to a quadratic equation expressed as: f(x), where f(x) indicates the vertical distance in feet, and x signifies how far the ball travels horizontally. To compute the distance the ball will advance before striking the ground, we set the condition f(x) = 0. Upon solving this quadratic equation, we find that the horizontal distance traveled by the ball is: x = -0.902 meters, leading us to conclude that it travels 0.902 meters across the field.

7 0

4 days ago