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Andreyy89
2 years ago
4

Antonio wants to replace the lightbulbs in his house. How much money would he save in one year by replacing 10 incandescent bulb

s with 10 CFL bulbs?
incandescent$1.47 per bulb CFL $3.99
annual operating cost incandescent $32.86 CFL $7.67
Physics
2 answers:
ycow [4]2 years ago
7 0

Answer:

226.70

Explanation:

Soloha48 [4]2 years ago
5 0

Answer: $226.70 on edg

Explanation:

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Consider a boat heading due east at 15 miles/hour. The water's current is moving at 7.1 miles/hour at 45º south of east. Drag ve
givi [52]

If a boat is going East at 15mph and there is a water current going southeast at 45° then the boat is being drifted southward.  So since the current is going at an angle then it has a x and y component.  So Rx refers to the x-component force of the current and Ry refers to the y-component of the current, and |R| refers to the magnitude of these forces.

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2 years ago
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Gold and silicon are mutually insoluble in the solid state and form a eutectic system with a eutectic temperature of 636 k and a
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Yupp its c because my dad farted 
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2 years ago
A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv
Juli2301 [7.4K]

Answer:

<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>

Explanation:

The initial charge on 4 mF capacitor  = 4 mf  x 50 V = 200 mC

The initial Charge on 6 mF capacitor  = 6 mf x 30 V =180 mC

Since the negative ends are joined together  the total charge on both capacity would be;

q = q_{1} -q_{2}

q = 200 - 180

q = 20 mC

In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

V = 2 V

For the final charge on 6.0 mF;

q = CV

q = 6.0 mF x 2 V

q =  12 mC

Therefore the final charge on the 6.0 mF capacitor would be 12 mC

5 0
2 years ago
Read 2 more answers
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
DENIUS [597]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle \theta with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

Where y_o is the initial height above the ground level, v_{oy} is the vertical component of the initial velocity and t is the time

The y-component of the speed is

v_y=v_{oy}-gt

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of v_o

The object will reach the maximum height when v_y=0. It allows us to compute the time to reach that point

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum heigh is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Solving for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Replacing the known values

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
2 years ago
A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?
Fed [463]
I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
W_x =mg \sin \theta
where
m is the mass of the car
g is the acceleration of gravity
\theta is the angle of the hill

Substituting numbers into the formula, we find
W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N
6 0
2 years ago
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