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Alenkasestr [34]
3 years ago
9

A garage door opener has a power rating of 350 watts.

Physics
2 answers:
Lelechka [254]3 years ago
6 0
We Know, Power = Energy/Time
Substitute the known value. which is P = 350 watt, & T = 30 sec
350 = E/30 
E = 350 * 30
E = 1050 Joules.

So, your answer is 1050 Joules.
Hope this helps! Keep studying! 
34kurt3 years ago
5 0

Answer:

10,500 J

Explanation:

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Ann is driving down a street at 61 km/h.
Wittaler [7]

Answer:

12.267 seconds approximately.

Explanation:

The units can be simplified into m/s, in which case you would have 61000/3600. Simplify that to 16 and 17/18. This is your meters per second, so multiply that by .724 to get the answer.

4 0
3 years ago
2in+3in+4in+4in+6in+7in calculate the area and perimeter of each shape
ololo11 [35]
Perimeter=26in

Area would be to multiply the units together
4 0
2 years ago
Read 2 more answers
How long does it take a microwave of power 0.2kW to sue 10000 J of energy
olasank [31]

Answer:

50s .

Explanation:

\frak{\pink{Given}}\begin{cases}\textsf{ The power of microvave is 0.2kW .}\\\textsf{ Amount of energy is 10000 J .}\end{cases}

Here the power of the microwave is 0.2kW . And as we know that ,

  • Rate of doing work is called power .

So from the definition , we have ;

\sf\longrightarrow Power =\dfrac{Work}{time}

  • Here the work done is equal to the energy consumed by the microwave i.e. 10000 J .So we can write it as ,

\sf\longrightarrow Power =\dfrac{Energy}{time}

\sf\longrightarrow 0.2kW = \dfrac{10^4 J }{t} \\

\sf\longrightarrow 0.2 * 1000 W =  \dfrac{10^4 J }{t}

Cross multiply ,

\sf\longrightarrow t = \dfrac{ 10^4 }{ 0.2 * 10^3}s=\dfrac{10^4}{2*10^2} s

Simplify ,

\sf\longrightarrow \boxed{\bf t = 50s}

<h3>Hence the time taken is 50s .</h3>
5 0
1 year ago
The Bugatti Chiron is currently the fastest production car in the world reaching a top speed of 261 mph
VMariaS [17]

Answer:304 not 261

Explanation:

That is, in prototype form anyway – the custom Chiron Longtail hit 304 mph (490 km/h) last year, but the production version dubbed the Chiron Super Sport 300+ will be electronically limited to only 273 mph (440 km/h).

...

7 0
2 years ago
Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge
Serhud [2]

Answer:

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}=23.495N

Vector direction FC

\beta=35.91 degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

\alpha = tan^{-1}(\frac{3}{4}) = 36.86 degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2} Equation (1): Magnitude of the electric force of the charge qA over the charge qC

F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2} Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

k=8.99*10^9 \frac{N*m^2}{C^2}

q_{A}=2.70*10^{-4} C

q_{B}=-6.36*10^{-4} C

q_{C}=1.04*10^{-4} C

r_{AC} =3

r_{BC}=\sqrt{4^2+3^2} = 5

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N

F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N

Components of the FBC force at x and y:

F_{BCx}=23.785 *Cos(36.86)=19.03N

F_{BCy}=23.785 *Sin(36.86)=14.27N

Components of the resulting force acting on qC:

F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N

F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}= \sqrt{19.03^2+13.78^2} =23.495N

Vector direction FC

\beta = tan^{-1} (\frac{13.78}{19.03})=35.91 degrees: angle that forms FC with the horizontal

7 0
3 years ago
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