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miv72 [106K]
3 years ago
15

The technological challenges of installing cellphones across the globe

Physics
1 answer:
Leona [35]3 years ago
3 0

Answer:

I don't know sorry For this question

You might be interested in
2.
horrorfan [7]

Answer:

B. 6HgO → 6Hg + 3O_{2}

Explanation:

A decomposition reaction is a reaction in which a single reactant is broken down into 2 or more products.

6 0
3 years ago
9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o
Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).

Now the force of friction is F_s=\mu*N, where N is the normal force and from the diagram it is F_y=mg*cos(\theta).

Thus F_s=\mu*N=\mu*mg*cos(\theta).

Therefore,

F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).

Substituting the value for F_H,m,\mu, and \:\theta we get:

F_{tot}= -38.63N.

Now acceleration is simply

a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.

The negative sign indicates that the acceleration is directed up the incline.

Part B:

d=\frac{1}{2} at^2

Which can be rearranged to solve for t:

t=\sqrt{\frac{2d}{a} }

Substitute the value of d=0.50m and a=7.7m/s and we get:

t=0.36s.

which is our answer.

Notice that in using the formula to calculate time we used the positive value of a, because for this formula absolute value is needed.

5 0
3 years ago
A child (approximately 4 years old) takes her metal “slinky Toy” (a flexible coiled metal spring) and does various tests to dete
anzhelika [568]

Answer:

248

Explanation:

L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H

l = length of the slinky = 3 m

N = number of turns in the slinky

r = radius of slinky = 4 cm = 0.04 m

Area of slinky is given as

A = πr²

A = (3.14) (0.04)²

A = 0.005024 m²

Inductance is given as

L = \frac{\mu _{o}N^{2}A}{l}

130\times 10^{-6} = \frac{(12.56\times 10^{-7})N^{2}(0.005024)}{3}

N = 248

8 0
3 years ago
Using a crowbar, a person can remove a nail by exerting little force, whereas pulling directly on the nail requires a large forc
algol [13]
Here we deal with a lever law. It states that product of force and distance from a fixed point on a lever is equal on both sides.

F₁*d₁ = F₂*d₂

By analysing this formula we can see that applying small force on a great length equals great force on a small length.
To remove nail we need to apply certain force. If we use F₁ for this required force we can see that on other side we need to apply certain force. If we have greater arm length we need smaller force. In a crowbar arm length along which we apply force is greater than length of our arm. This leads to a conclusion that we need smaller force when using crowbar. Depending on the length of a nail it is possible that we need to apply force that is greater than force required to remove nail.
7 0
3 years ago
When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2.
VMariaS [17]

Answer:

A)d=\dfrac{1}{2F}mv^2

B)\Delta KE'=2\times \dfrac{1}{2}mv^2

Explanation:

Given that

Force  = F

Increase in Kinetic energy = \dfrac{1}{2}mv^2

\Delta KE=\dfrac{1}{2}mv^2

we know that

Work done by all the forces =change in the kinetic energy

a)

Lets distance = d

We know work done by force F

W= F .d

F.d=ΔKE

F.d=\dfrac{1}{2}mv^2

d=\dfrac{1}{2F}mv^2

b)

If the force become twice

F' = 2 F

F'.d=ΔKE'

2 F .d = ΔKE'                          ( F.d =Δ KE)

2ΔKE = ΔKE'

\Delta KE'=2\times \dfrac{1}{2}mv^2

Therefore the final kinetic energy will become the twice if the force become twice.

8 0
3 years ago
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