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Maksim231197 [3]

2 years ago

5

Why can you plant plants, but not water water?

2 answers:

SpyIntel [72]2 years ago

7 0

Answer:

have a great day

Explanation:

okay you cant grow water frist of all and u can water water but there no poin cuz it don't need to and ti wounled be a wast of time and plants cant plant plants but plant can have plans grow on then

Grace [21]2 years ago

3 0

Answer:

Because water is water and its a liquid meanwhile plants are plants living things that you can move around.

Explanation:

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A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

3 0

2 years ago

Electricity is the flow of electrons. The questions relate to how electricity is quantified. Electrons are charged particles. Th

xxTIMURxx [149]

The amount of charge that passes per unit time is called <em>electric current</em> .

Current has dimensions of [Charge] / [Time] .

It's measured and described in units of ' Ampere ' .

1 Ampere means 1 Coulomb of charge passing a point every second.

7 0

2 years ago

the area of 1 end of U-tube is 0.0 metre square and that of the other end is 1 metre square when you force was applied on the li

Lapatulllka [165]

Answer:

The question is wrong Since if you apply Force on 0.0m²It would mean That the pressure exerted=F/A=F/0

An since we can't divide a number by 0, the question is wrong

6 0

3 years ago

he product side of a chemical reaction is shown. → 7Ti2(SO4)3 Which number represents a coefficient? 2 3 4 7

valkas [14]

I think it would be 7

3 0

2 years ago

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid ra

BabaBlast [244]

Explanation:

It is given that,

Length of the helicopter, l = 3.1 m

The helicopter rotates, the length of helicopter will become the radius of circular path, r = 3.1 m

Angular speed of the helicopter, $\omega=280\ rev/min=29.32\ rad/s$

(a) The centripetal acceleration in terms of angular velocity is given by :

$a_c=r\times \omega^2$

$a_c=3.1\times (29.32)^2$

$a_c=2664.95\ m/s^2$

(b) Let v is the linear speed of the tip. The relation between the linear and angular speed is given by :

$v=r\times \omega$

$v=3.1\times 29.32$

v = 90.89 m/s

$\dfrac{v_{tip}}{v_{sound}}=\dfrac{90.89}{340}=0.267$

Hence, this is the required solution.

4 0

2 years ago