HELP PLS!!!!!!!!WILL MARK BRAINLIEST!

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Answer 1

Answer: I do not know the answer sorry :(

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Convert 7 (gcm^2)/(min^2) into a value in standard S.I. units. Be sure to use scientific notation if necessary. You do not need to answer units.

Answers

The required value is required in SI units.

The required answer is $1.94*10^(-10)\ \text{kg m}^2/\text{s}^2$

SI units

The SI unit of mass, length and time is kg, m and s respectively.

In order to convert one unit into another it has to be multiplied or divided by the conversion factors.

A definite magnitude which has some quantity which is defined by convention or law is called a unit.

The conversion factors are

$1\ \text{g}=10^(-3)\ \text{kg}$

$1\ \text{cm}=10^(-2)\ \text{m}$

$1\ \text{cm}^2=10^(-4)\ \text{m}^2$

1 min = 60 s

$1\ \text{min}^2=60*60\ \text{s}^2$

So,

$7\ \text{g cm}^2/\text{min}^2=7* (10^(-3)* 10^(-4))/(60* 60)\n =1.94*10^(-10)\ \text{kg m}^2/\text{s}^2$

Learn more about SI units:

brainly.com/question/16393390

Hi hi hi hi hi hi hi

an object down, but this is not true. If you place a box of mass 8 kg on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed of the belt, which is 5 m/s. The coefficient of kinetic friction between box and belt is 0.6. (a) How much time does it take for the box to reach this final speed

Answers

Answer:

Explanation:

ASSUMING the belt is horizontal

kinetic friction force is μmg = 0.6(8)(9.8) = 47.04 N

Horizontal acceleration is

a = F/m = 47.04 / 8 = 5.88 m/s²

t = v/a = 5.0 / 5.88 = 0.85034...

t = 0.85 s

Which of the following is correct? *PLEASE HELP MEEEE
1 cm = 100 m
1 mm = 100 cm
100 mm = 1 cm
1 m = 100 cm

Answers

Answer:

The last one

1m = 100 cm

Explanation:

If you do not trust me look it up

What was the main idea of Malthus theory of population

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Answer:

The idea that population growth is potentially exponential while the growth of the food supply or other resources is linear.

Explanation:

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.1) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
2) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, how far downstream is she swept before reaching the opposite bank?
3) For the conditions, how long does it take for her to reach the opposite bank?

Answers

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^(-1)$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^(-1)$

speed of the swimmer with respect to water, $v=4\ ft.s^(-1)$

Now the resultant of the two velocities perpendicular to each other:

$v_r=√(v^2+v_s^2)$

$v_r=√(4^2+8^2)$

$v_r=8.9442\ ft.s^(-1)$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=(v_s)/(v)$

$\tan\beta=(8)/(4)$

$\beta=63.43^(\circ)$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d* \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=√(d^2-w^2)$

$\Delta s=√(1118.034^2-500^2)$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4* \cos37$

$v'=3.1945\ ft.s^(-1)$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^(-1)$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^(-1)$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=(v_n)/(v'_r)$

$\tan\phi=(4.8055)/(2.4073)$

$\phi=63.39^(\circ)$

Now let the distance swam in this direction be d'.

$d'* \cos\phi=w$

$d'=(500)/(\cos63.39)$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=√(d'^2-w^2)$

$\Delta s'=√(1116.344^2-500^2)$

$\Delta s'=998.11\ ft.s^(-1)$

Time taken in crossing the rive in case 1:

$t=(d)/(v_r)$

$t=(1118.034)/(8.9442)$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=(d')/(v'_r)$

$t'=(1116.344)/(2.4073)$

$t'\approx463.733\ s$

A squirrel is running a race where she is on track for her average velocity to be 6.0 m/s. She is distracted by a dummy of an attractive male squirrel and pauses for 3.0 s. As a result her average velocity ends up being 5.0 m/s instead. What is the length of the race? [HINT: construct two equations with the same two unknowns in them and you can solve the system of equations]