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3 years ago

Cells & Cellular Processes

Physics

1 answer:

never [62]3 years ago

7 0

Answer:

one or more cells

Explanation:

I need to have 20 characters to submit lol

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If you connected 10 of these 12V (10 W) lamps in parallel, to the 12V source, how much current would the source have to supply

Flauer [41]

Answer:

$T=8.33A$

Explanation:

From the question we are told that:

Number of battery $n=10$

Voltage source $E=12V$

Lamp Power $P=10W$

Generally the equation for Resistance is mathematically given by

$R=\frac{V^2}{P}$

$R=\frac{12^2}{10}$

$R=14.4ohms$

Therefore

$R_{eq}=\frac{14.4}{10}$

$R_{eq}=1.44$

Generally the equation for Current is mathematically given by

$T=\ffrac{V}{Req}$

$T=\frac{12}{1.44}$

$T=8.33A$

6 0

2 years ago

A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20\, \text N20N20, start text, N, end text

meriva

Answer:

300 Nm ; 300 J

Explanation:

Given that:

Force (F) = 20 N

Distance (d) = 15 m

The kinetic energy (Workdone) = Force * Distance

Kinetic Energy = 20N * 15m

Kinetic Energy = 300Nm

K. E = 1/2

4 0

2 years ago

Read 2 more answers

A 0.0663 kg ingot of metal is heated to 241◦C

Westkost [7]

Answer:280.216j/kg°C

Explanation:

Mass of metal=0.0663kg

mass of water=0.395kg

Final temperature=27.4°C

Temperature of metal=241°C

Temperature of water=25°C

specific heat capacity of water=4186j/kg°C

0.0663xax(241-27.4)=0.395x4186x(27.4-25)

0.0663xax213.6=0.395x4186x2.4

14.16168a=3968.328

a=3968.328 ➗ 14.16168

a=280.216j/kg°C

4 0

2 years ago

What is the atomic mass of Jupiter?

lana [24]

The mass of Jupiter is 1.9 x 1027 kg.

4 0

3 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago