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lawyer [7]
1 year ago
12

The power rating of an electric lawn mower is 2000 watts if lawnmower is used for 120 seconds how many joules of work van it do

Physics
1 answer:
Scorpion4ik [409]1 year ago
4 0
1 watt = 1 joule/sec

2,000 watts = 2,000 joules/sec

                     (2,000 joule/sec) x (120 sec)

                  = (2,000 x 120)  (joule-sec/sec)

                  =   240,000 joules .
        
You might be interested in
An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.
viktelen [127]

Answer:

The  force that the wire exerts on the electron is -4.128\times10^{-20}i-6.88\times10^{-20}j+0k

Explanation:

Given that,

Current = 8.60 A

Velocity of electron v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s

Position of electron = (0,0.200,0)

We need to calculate the magnetic field

Using formula of magnetic field

B=\dfrac{\mu I}{2\pi d}(-k)

Put the value into the formula

B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}

B=0.0000086\ T

B=-8.6\times10^{-6}k\ T

We need to calculate the force that the wire exerts on the electron

Using formula of force

F=q(\vec{v}\times\vec{B}

F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )

F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k

F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k

Hence, The  force that the wire exerts on the electron is -4.128\times10^{-20}i-6.88\times10^{-20}j+0k

5 0
2 years ago
In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t
gladu [14]

Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

8 0
2 years ago
La luz pasa del medio A al medio B formando un ángulo de 35° con la frontera horizontal entre ambos. Si el ángulo de refracción
zaharov [31]

Answer:

Índice de refracción entre los dos medios = 1,43

Refractive index between the two media = 1.43

Explanation:

El índice de refracción entre dos medios se explica mejor entendiendo primero la refracción.

Cuando las olas se mueven de un medio a otro, a menudo experimentan un cambio de dirección con respecto al medio en el que viajan.

Por lo tanto, el índice de refracción se expresa como el seno del ángulo de incidencia dividido por el seno del ángulo de refracción.

El seno del ángulo de incidencia y la refracción utilizados en esta fórmula de índice de refracción se miden respectivamente con respecto a la vertical.

En esta pregunta Ángulo de incidencia = 35° a la horizontal = (90° - 35°) a la vertical = 55° a la vertical.

Ángulo de refracción = 35°

Índice de refracción entre los dos medios

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 a 2 d.p.

¡¡¡Espero que esto ayude!!!

English Translation

The light passes from medium A to medium B at an angle of 35 ° with the horizontal border between the two. If the angle of refraction is also 35 °, what is the relative refractive index between the two media?

Solution

The refractive index between two media is best explained by first understanding refraction.

When waves move from one medium to another, they often experience a change in direction with respect to the medium in which they are travelling.

Hence, refractive index is expressed as the sine of angle of incidence dibided by the sine of angle of refraction.

The sine of angle of incidence and refraction used in this refractive index formula are both respectively measured with respect to the vertical.

In this question,

Angle of incidence = 35° to the horizontal = (90° - 35°) to the vertical = 55° to the vertical.

Angle of refraction = 35°

Refractive index between the two media

= (Sin 55°) ÷ (Sin 35°)

= 0.8192 ÷ 0.5736

= 1.428 = 1.43 to 2 d.p.

Hope this Helps!!!

3 0
1 year ago
A handbag weighing 162N is carried by two students each holding the handle of the bag next to him. If each handle is inclined at
Mashcka [7]

Answer:

162N

Explanation: resolving into components and equating both vertical components equal to 162 as explained in attachment

Download pdf
3 0
1 year ago
You are driving downhill on a rural road with a 3% grade at a speed of 45 mph. While playing on the side of the road, a child ac
Dennis_Churaev [7]

Answer: a) 95.07m b) 81.88 m

Explanation:

a)

For finding the distance when vehicle is going downhill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1  + \frac{(0.28*72.4)^{2} }{2*9.81*(0.31-0.03)}

Stop sight distance= 95.07 m

b)

For finding the distance when vehicle is going uphill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1  + \frac{(0.28*72.4)^{2} }{2*9.81*(0.31+0.03)}

Stop sight distance= 81.88 m

5 0
2 years ago
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