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3 years ago

How much heat in joules would you need to raise the temperature of 1 kg of water by 5 °C?

1 answer:

7 0

First we need to know the specific heat capacity of water. You can look it up online because it is uniform. The specific heat capacity of water is 4184 kg^-1 °C^-1.

To calculate the energy required to heat a specific thing up, we can use the formula
E = mc△T
Where E is the total energy required, m as mass, c as specific heat capacity and △T

Substitute the numbers in
E = 1 x 4184 x 5
E = 20920J

Therefore the answer is 20920 joules

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De manera individual subraya la respuesta correcta: A) El Método Científico de la física experimental y su búsqueda de respuesta

alexira [117]

Answer:

a. La ciencia.

Explicación:

El método científico de la física experimental y su búsqueda de respuestas tiene un principio básico que es la ciencia. La ciencia es la aplicación del conocimiento del mundo natural y social siguiendo una metodología sistemática basada en evidencias. Sin evidencias no hay aceptación de la observación ni se debe hacer teoría porque la evidencia es el principal criterio para aceptar o rechazar una hipótesis.

6 0

3 years ago

Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.

KIM [24]

Answer:

The velocity is $v = 4.76 \ m/s$

Explanation:

From the question we are told that

The first distance is $d_1 = 4.0 \ km = 4000 \ m$

The first speed is $v_1 = 5.0 \ m/s$

The second distance is $d_2 = 1.0 \ km = 1000 \ m$

The second speed is $v_2 = 4.0 \ m/s$

Generally the time taken for first distance is

$t_1 = \frac{d_1 }{v_1 }$

$t_1 = \frac{4000}{5}$

$t_1 = 800 \ s$

The time taken for second distance is

$t_1 = \frac{d_2 }{v_2 }$

$t_1 = \frac{1000}{4}$

$t_1 = 250 \ s$

The total time is mathematically represented as

$t = t_1 + t_2$

=> $t = 800 + 250$

=> $t = 1050 \ s$

Generally the constant velocity that would let her finish at the same time is mathematically represented as

$v = \frac{d_1 + d_2}{t }$

=> $v = \frac{4000 + 1000}{1050 }$

=> $v = 4.76 \ m/s$

7 0

4 years ago

a ball is dropped from rest at a height of 89m above the ground. (a)what is it's speed just before it hits the ground? (b) how l

kykrilka [37]

Answer:

(a) 41.75m/s

(b) 4.26s

Explanation:

Let:

Distance, D = 89m

Gravity, $g$ = 9.8 m/ $s^{2}$

Initial Velocity, $u$ = 0m/s

Final Velocity, $v$ = ?

Time Taken, $t$ = ?

With the distance formula, which is

D = $ut$ + $\frac{1}{2} gt^2$

and by substituting what we already know, we have:

89 = $\frac{1}{2}$ ×9.8× $t^{2}$

With the equation above, we can solve for $t$ :

$t=\sqrt{\frac{89(2)}{9.8}} \\t=\sqrt{\frac{178}{9.8} } \\t=\sqrt{18.16} \\t=4.26 seconds$

Now that we have solved $t$ , we can use the following velocity formula to solve for $v$ :

$v = u + at$ , where $a$ is also equals to $g$ , so we have

$v = u + gt$

By substituting $u = 0$ , $g = 9.8$ , and $t = 4.26$ ,

We have:

$v = 0 + 9.8(4.26)\\v = 41.75m/s$

4 0

3 years ago

F each fission reaction in a nuclear reactor emits three neutrons, how many neutrons are produced from three fission reactions?

IrinaK [193]

The Three fission reactions will produce nine neutrons

<h3>Meaning of Fission</h3>

Fission can be defined as a nuclear process that involves the breaking of a whole nuclear matter into smaller bits.

In a fission reaction, the matter is broken down into simpler bits.

In conclusion, The Three fission reactions will produce nine neutrons

Learn more about fission:brainly.com/question/3992688

#SPj1

8 0

2 years ago

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

dimulka [17.4K]

Answer:

$-1.7908787542\times 10^{-9}\ C$

$3.4260289211\times 10^{-9}\ C$

$1.7908787542\times 10^{-9}\ C$

$1.6351501669\times 10^{-9}\ C$

Explanation:

r = Radius

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field is given by

$E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C$

The charge is $-1.7908787542\times 10^{-9}\ C$

$Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C$

The charge is $3.4260289211\times 10^{-9}\ C$

The charge inside will have the polarity changed

$q=+1.7908787542\times 10^{-9}\ C$

Outside the charge will be

$3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C$

3 0

3 years ago