Ask question

Ask question

All categories

2 years ago

14

Estimate how many electrons there are in the human's body. asssume the mass is 70 kg. (hint: most of the atoms in your body have

equal numbers of electrons, protons, and neutrons.)

1 answer:

Karolina [17]2 years ago

8 0

How about 8 X 10 to the 100th power.

You might be interested in

Right or left? Summer and Winter or day and night?

evablogger [386]

Answer:

The answer is left

Explanation:

Sorry if im worng

6 0

2 years ago

Read 2 more answers

(25 pts) Estimate how much collector area and storage capacity would be required for an active solar hot-water system designed t

Aneli [31]

Answer:

The required total area is 1.48 m²

Explanation:

Given that,

Latitude = 44+° N

New Mexico,

Latitude= 35+° N

Heat capacity = 4200 J/Kg°C

Temperature = 60°C

Let us assume the input temperature 22°C

Estimate volume of water 100 ltr for 4 person.

We need to calculate the heat

Using formula of heat

$H=mc_{p}\Delta T$

$H=mc_{p}(T_{f}-T_{i})$

Put the value into the formula

$H=100\times4200\times(60-22)$

$H=15960\ KJ$ ...(I)

Let solar radiation for 6 hours/day.

We need to calculate the total energy per unit area

Using formula of energy

$E=1000\times6\times3600\ J/m^2$

$E=21600\ KJ/m^2$

Let the efficiency of collector is 50 %

Then, the total energy per unit area will be

$E=21600\times\dfrac{50}{100}$

$E=10800\ KJ/m^2$ ....(II)

We need to calculate the required total area

Using equation (I) and (II)

$A=\dfrac{H}{E}$

Where, H = heat

E = total energy

Put the value into the formula

$A=\dfrac{15960}{10800}$

$A=1.48\ m^2$

Hence, The required total area is 1.48 m²

6 0

3 years ago

A chinook salmon needs to jump a waterfall that is 1.5 m high. If the fish starts from a distance of 1.00 m from the base of the

____ [38]

<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal

it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
--------------------------------------...
let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
=====================
t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s

so it will able to cross</span>

8 0

2 years ago

The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x

Diano4ka-milaya [45]

Answer:

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Explanation:

Given that

At X=0 V=Vo

At X=X1 V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

$V^2=U^2-2aS$

$0=V_o^2-2a X_1$

$a=\dfrac{V_o^2}{2X_1}$

So the friction force on the box

Ff= m x a

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Where m is the mass of the box.

4 0

3 years ago

Given: Q1 = +10 uc = 1.0 x 10-5C

ser-zykov [4K]

Answer:

-0.038 N

Explanation:

F=K Q1 Q2/r^2 by COULOMB'S LAW

F= 9×10^9×1×10^-5×-1.5×10^-5/(6)^2

F= -0.038 N

5 0

2 years ago