All categories

2 years ago

Which is the smallest unit of life?

Physics

1 answer:

emmasim [6.3K]2 years ago

7 0

Answer:

the cell is the smallest unit

You might be interested in

A wire with a weight per unit length of 0.082 N/m is suspended directly above a second wire. The top wire carries a current of 3

FromTheMoon [43]

Answer:

d = 4.5079mm

Explanation:

from the question we are given that

$\frac{F}{L}$ = 0.082N/m

I₁= 30.6 A , I₂ = 60.4 A , and μ₀ = 4 π × 10⁻⁷ T · m / A

In order for the system to be in equilibrium, the repulsive magnetic force

per unit length on the top wire must equal the weight per unit length of the wire. Then we have

F/L = μ₀I₁I₂ / 2πd , making d as the subject of formular

we have that d = Lμ₀I₁I₂ / 2πF

= 4 π × 10⁻⁷ × 30.6 ×60.4 / 2π × 0.082

=45079.02× 10⁻⁷

in mm .. 45079.02× 10⁻⁷ × 10³

45079.02 ×10⁻⁴

d = 4.5079mm

4 0

3 years ago

The temperature of a sample of silver increased by 24.0 °C when 269 J of heat was applied. What is the mass of the sample?

statuscvo [17]

Answer:

Mass of the silver will be equal to 46.70 gram

Explanation:

We have given heat required to raise the temperature of silver by 24°C is 269 J , so $\Delta T=24^{\circ}C$

Specific heat of silver = 0.240 J/gram°C

We have to find the mass of silver

We know that heat required is given by

$Q=mc\Delta T$ , here m is mass, c is specific heat of silver and $\Delta T$ is rise in temperature

So $269=m\times 0.240\times 24$

m = 46.70 gram

So mass of the silver will be equal to 46.70 gram

3 0

3 years ago

True or false. The super continent Pangea started to move 225 million years ago.

sammy [17]

TRUE 225 million years ago

7 0

3 years ago

Read 2 more answers

As more and more bulbs are connected in series to a flashlight battery, what happens to the brightness of each bulb? Assuming th

balu736 [363]

Answer:

$P_1 = P_2 = \frac{P}{2}$

so each bulb brightness becomes half of its given or indicated power

$P_1 = P_2 = \frac{V^2}{R}$

so both bulb will glow same power as indicated

Explanation:

Let the indicated power on the bulbs is given as P and its rated voltage is V

so here resistance of each bulb is given as

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

now if the two bulbs are connected in series so we will have

$R_{eq} = R_1 + R_2$

$R_{eq} = 2\frac{V^2}{P}$

now the current in the circuit is given as

$i = \frac{V}{R_{eq}$

$i = \frac{P}{2V}$

now brightness of each bulb is given as

$P_1 = P_2 = i^2 R$

$P_1 = P_2 = \frac{P}{2}$

so each bulb brightness becomes half of its given or indicated power

Now if the two bulbs are connected in parallel

then the net voltage across each bulb is "V"

so we will have

$P_1 = P_2 = \frac{V^2}{R}$

so both bulb will glow same power as indicated

6 0

4 years ago

Kate is working on a project in her tech education class. She plans to assemble a fan motor. Which form of energy does the motor

andreyandreev [35.5K]

Motion Energy

I am writing this so it can be more than 20 letters

7 0

4 years ago