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sdas [7]
3 years ago
5

Battery life is what distinguishes one type of mobile computer from another.

Physics
1 answer:
BlackZzzverrR [31]3 years ago
8 0
I think the statement is false. It is not only the battery life that would distinguish one type of mobile computer from another. There are other aspects that would characterize every mobile computer. It could be the brand, the size, the operating system, the hardware specifications like the RAM, processor and the video card. Every mobile computer would vary depending on the brand, model and intended use.
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I believe that the answer is B
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3 years ago
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A car slid off an icy 10m bridge and landed 12m away from the bridge. How much time was the car in the air? (Hunt: Projectile)
Contact [7]

You will use the height of the bridge from the ground.

Solution:

Formula to be used is y=Viy(t)+g(t^2)/2

Where:

Vi=initial velocity which is 0 m/s

 y=10 m

Gravitational acceleration or g =9.8m/s^2

T= time you need

Substitute all the given to the formula

10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2

10mx2=9.8m/s^2(t^2)

Now isolate the variable you want to find which is T or time

10mx2/9.8m/s^2=t^2

20m/9.8m/s^2=t^2

Square root of 2.04= square root of t^2

T=1.43 secs

The answer is 1.43 seconds

8 0
3 years ago
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Develop expression for equation for uniform motion in a straight line
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3 years ago
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}

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3 years ago
Three point charges are fixed in place in a right triangle, as shown in the figure.
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