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2 years ago

1-Find the gravitational force

Physics

1 answer:

butalik [34]2 years ago

5 0

$g = \frac{GMm}{r^2}$

where $G$ is the gravitational constant and is about $6.67 * 10^{-11}$

$g = \frac{6.67 * 10^{-11} * 25 * 0.55}{35^2}$

$g = \frac{6.67 * 10^{-11} * 25 * 0.55}{1225} = 0.000000000000749 N$ :D

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Can I PLEASE get some help?

Rus_ich [418]

1.) C
2.) B
3.) D
4.) B
Good luck with your work!

3 0

3 years ago

The total blood volume of a human is 75 ml/kg. If the volumetric flow rate is the same in adults and children, how much more blo

Tema [17]

The amount of blood that flows through the venae cavea of the adult is 3750 ml more than that of the child.

This is the volume of fluid flowing in a vessel per unit time.

First, we need to get the volume of blood flowing through the venae cavae of the adult and the child in 1 hour, then subtract the volume of blood for the child from that of the adult.

Using,

V = F'm................ Equation 1

Where:

V = Volume of the blood
F' = Flow rate
m = mass of blood.

For the Adult,

Given:

m = 70 kg
F = 75 ml/kg

Substitute these values into equation 1

V = 70(75)
V = 5250 ml.

For the child,

Given:

m = 20 kg
F = 75 ml/kg

Substitute these values into equation 1 also

V = 75(20)
V = 1500 ml.

The amount of blood that flows more through the adult than the child is

A = 5250-1500
A = 3750 ml.

Hence, the amount of blood that flows through the venae cavea of the adult is 3750 ml more than that of the child.

Learn more about flow rate here: brainly.com/question/21630019

5 0

2 years ago

A charged particle A exerts a force of 2.45 μN to the right on charged particle B when the particles are 12.2 mm apart. Particle

Brilliant_brown [7]

Answer:

$F_2 = 1.10 \mu N$

Explanation:

As we know that the electrostatic force is a based upon inverse square law

so we have

$F = \frac{kq_1q_2}{r^2}$

now since it depends inverse on the square of the distance so we can say

$\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$

now we know that

$r_2 = 18.2 mm$

$r_1 = 12.2 mm$

also we know that

$F_1 = 2.45 \mu N$

now from above equation we have

$F_2 = \frac{r_1^2}{r_2^2} F_1$

$F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)$

$F_2 = 1.10 \mu N$

5 0

3 years ago

What makes output energy less than input energy?

AleksAgata [21]

Answer:

part of input energy is wasted

it is used to increase the entropy of the surrounding

therefore the useful energy output is necessarily smaller than the energy input

in other words the efficiency of heat engine is always less than hundred percent

6 0

2 years ago

Read 2 more answers

A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance,

Lisa [10]

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

$F = \frac{Kq_{1}q_{2}}{r^{2}}$

So, the force between the charges q and Q - q is given by

$F = \frac{K\left ( Q-q \right )q}}{r^{2}}$

For maxima and minima, differentiate the force with respect to q.

$\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )$

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now $\frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}$

the double derivate is negative, so the force is maxima when q = Q / 2 .

6 0

3 years ago