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Artyom0805 [142]
3 years ago
7

Anyone willing to help?

Physics
2 answers:
grigory [225]3 years ago
5 0

Answer:

I think it is "A" because the wind systems are created by uneven heating of Earth's surface.It also may help form large global wind patterns.

Explanation:

Ierofanga [76]3 years ago
3 0
D is the answer reeeeeeeeeeetaaaaaaardddddddddddd
You might be interested in
Which force below acts at 90 degrees and is applied by a surface/wall to an object.
Molodets [167]

Answer:

B. normal

please mark as brainliest

4 0
3 years ago
The eye of the Atlantic giant squid has a diameter of 3.50 × 10^2 mm. If the eye
Lunna [17]

Answer:

   q = 224 mm,   h ’= - 98 mm, real imagen

Explanation:

For this exercise let's use the constructor equation

        \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

       

where f is the focal length, p and q are the distance to the object and the image respectively.

In a mirror the focal length is

        f = R / 2

indicate us radius of curvature is equal to the diameter of the eye

       R = 3,50  10² mm

       f = 3.50 10² /2 = 1.75 10² mm

they also say that the distance to the object is p = 0.800 10³ mm

        1 / q = 1 / f - 1 / p

        1 / q = 1 / 175 - 1 /800

        1 / q = 0.004464

         q = 224 mm

to calculate the size let's use the magnification ratio

          m = \frac{h'}{h} = - \frac{q}{p}

          h '= - \frac{q}{p} \ h

          h ’= - 224 350 / 800

          h ’= - 98 mm

in concave mirrors the image is real.

3 0
3 years ago
I need help with this
mel-nik [20]
It’s 4. Definitely 4. 99% sure it’s 4.
4 0
3 years ago
A network of streams that drains an area of land
cupoosta [38]

Answer:

a water bottle.

Explanation:

a water bottle drains an area of land, just pour a water bottle on a small land on the surface of the earth it'll be called drained.

4 0
2 years ago
A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
rjkz [21]

Answer:

a) F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}

b) \mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

\Sigma F_{y}=0

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

-F_{y}-W+N=0

When solving for the normal force we get:

N=F_{y}+W

we know that

W=mg

and

F_{y}=Fsin \theta

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

f_{k}=\mu_{k}N

so we can find the kinetic friction by substituting for N, so we get:

f_{k}=\mu_{k}(F sin \theta +mg)

Now we can find the sum of forces in x:

\Sigma F_{x}=0

so after analyzing the diagram we can build our sum of forces to be:

-f+F_{x}=0

we know that:

F_{x}=Fcos \theta

so we can substitute the equations we already have in the sum of forces on x so we get:

-\mu_{k}(F sin \theta +mg)+Fcos \theta=0

so now we can solve for the force, we start by distributing \mu_{k} so we get:

-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0

we add \mu_{k}mg to both sides so we get:

-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg

Nos we factor F so we get:

F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg

and now we divide both sides of the equation into (cos \theta-\mu_{k} sin \theta) so we get:

F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

f_{s}=\mu_{s}(F sin \theta +mg)

and

F cos θ = f

when substituting one into the other we get:

F cos \theta=\mu_{s}(F sin \theta +mg)

which can be solved for the static friction coefficient so we get:

\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}

which is the answer to part b.

3 0
3 years ago
Read 2 more answers
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