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3 years ago

a student have difficulty in reading books what should be the defect student suffering from? how can it be corrected?

Physics

1 answer:

MatroZZZ [7]3 years ago

8 0

Answer: concave lens, and myopia.

Explanation:A student has difficulty It shows that he is unable to see distant objects clearly. He is suffering from myopia. This defect can be corrected by using a concave lens.

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Every atom of an element has the same _____.

kirill [66]

The same number of protons. brainliest?(:

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3 years ago

Read 2 more answers

Blood flows through the major artery at 1 m/s for 0.5 m then at a 0.6 m/s over a distance of another 0.5 m through the small art

topjm [15]

Blood flows through the major artery at 1 m/s for 0.5 m then at a 0.6 m/s over a distance of another 0.5 m through the small artery the average speed of blood is 0.4 m/s.

We know that average speed = $\frac{0.5\times1+0.6\times0.5}{2}$ =0.4 m/s

Average speed is an important component in determining how long it takes to finish a journey. Average speed is simply a technique that assists us in calculating trip time and distance. It is obvious that the speed changes throughout the travel, making determining the average speed even more critical.

There are various methods for determining an object's or vehicle's average speed.

It is most desired when the speed of the object remains constant during the voyage, i.e. does not rise or decrease.

The approach for determining the average Speed is to divide. Divide the distance the vehicle travels by the time it travels to get the result.

Learn more about average speed brainly.com/question/12322912

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2 years ago

How to get a + b on a graph

Nataly [62]

The first positively essential requirement is that
you absolutely have to know what 'a' and 'b' are.

I have no clue, so this is as far as I can go.

3 0

4 years ago

In a RLC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect o

Marrrta [24]

Answer:

<h2>Case i) if $\omega L > \frac{1}{\omega c}$ </h2><h2>So initially if the circuit is inductive in nature then its net impedance will decrease after this</h2><h2>Case ii) if $\omega L < \frac{1}{\omega c}$ </h2><h2>So initially if the circuit is capacitive in nature then its net impedance will increase after this</h2>

Explanation:

As we know that the impedance of the circuit is given as

$z = \sqrt{(\omega L - \frac{1}{\omega c})^2 + R^2}$

when we join another identical capacitor in parallel with previous capacitor in the circuit then we will have for parallel combination

$c_{eq} = c_1 + c_2$

so it is

$c_{eq} = 2c$

now we have

$z = \sqrt{(\omega L - \frac{1}{2\omega c})^2 + R^2}$

Case i) if $\omega L > \frac{1}{\omega c}$

So initially if the circuit is inductive in nature then its net impedance will decrease after this

Case ii) if $\omega L < \frac{1}{\omega c}$

So initially if the circuit is capacitive in nature then its net impedance will increase after this

7 0

4 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:

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

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

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

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

3 years ago