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Masteriza [31]
3 years ago
11

Identify the type of electrification of the road

Physics
1 answer:
Arte-miy333 [17]3 years ago
6 0
Electrification is widely considered as a viable strategy for reducing the oil dependency and environmental impacts of road transportation. In pursuit of this strategy, most attention has been paid to electric cars. However, substantial, yet untapped, potentials could be realized in urban areas through the large-scale introduction of electric two-wheelers.
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Light travels _____________ in a material with a higher index of refraction.
Aleks [24]
<span>Light travels Slower  in a material with a higher index of refraction.
because </span><span>Water has a higher index of refraction than air and it travels slower (bends) in water
high index mean slow travel
hope it helps</span>
4 0
2 years ago
Read 2 more answers
En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
2 years ago
PLEASE HELP!!! 20 PIONTS!!!
timama [110]

Answer:

Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves.

Explanation:

The movement of the medium is different. In the longitudinal wave, the medium moves left to right, while in thee transverse wave, the medium moves vertically up and down. Longitudinal waves have a compression and rarefaction, while the transverse wave has a crest and a trough. Longitudinal waves have a pressure variation, transverse waves don't have pressure variation. Longitudinal waves can be propagated in solids, liquids and gases, transverse waves can only be propagated in solids and on the surfaces of liquids. Longitudinal waves have a change in density throughout the medium, transverse waves don't.

4 0
2 years ago
During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t
tia_tia [17]

1.A) 4.9 m  

AL2006 Ace

The instant it was dropped, the ball had zero speed.


After falling for 1 second, its speed was 9.8 m/s straight down (gravity).


Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.


Falling for 1 second at an average speed of 4.9 m/s, is covered 4.9 meters.


ANYTHING you drop does that, if air resistance doesn't hold it back.


Read more on Brainly.com - brainly.com/question/11776597#readmore

2 idk sorry

5 0
2 years ago
Read 2 more answers
A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot
denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

P_r = 345 Watt

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

P_r = 345 Watt

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

\Delta V = E + i r

here we have

E = 12 V

i = 56 A

r = 0.11

now we have

\Delta V = 12 + (0.11)(56) = 18.16 V

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

Part c)

Rate of energy conversion into EMF is given as

P_{emf} = i E

P_{emf} = (56)(12)

P_{emf} = 672 Watt

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

V = E - i r

V = 12 - (56)(0.11)

V = 12 - 6.16 = 5.84 V

Part e)

now the rate of energy dissipation is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

7 0
3 years ago
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