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4 years ago

7

Why can't you stand with your heels and back to a wall and then bend over to touch your toes and return to your stand-up positio

n?

2 answers:

snow_tiger [21]4 years ago

7 0

Answer:

Due to the rotating effect created due to opposite acting offset forces of the body weight and the normal reaction from the heels.

Explanation:

We cannot stand on heels with a wall at the back support and then bend over to touch the toes and regain the position back by standing up because we have the center of mass of our body somewhere at the abdominal portion of our body.
So, while standing on our heels we are giving our body only one point to support and balance itself.
On bending the center of mass of our body shifts forward and it is now not vertically in line to the direction of the support reaction from the heels because of the wall at the back we cannot bend our knees and move our weight over our heels for gaining the balance. Due to a difference between the these two opposite forces we have a turning effect called moment on the center of mass of our body. This tries to rotate our body in the forward direction about the center of mass of our body.

igor_vitrenko [27]4 years ago

7 0

Explanation:

We cannot stand with our heels and back to a wall and then bend over to touch your toes and return to our stand-up position because,

Our centre of gravity above our base of support is too high. A vertical line through our body's center of gravity extends behind our feet, beyond our base of support.

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In which arrangement of magnets will all the magnets attract?

Orlov [11]

Answer:

B

Explanation:

opposites attract ;)

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4 years ago

A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane

Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, $\phi=8.6\times 10^{-3}\ T/m^2$

The magnetic flux linked through the loop is :

$\phi=B.A$

$\phi=BA\ cos\theta$

Here, $\theta=0$

$B=\dfrac{\phi}{A}$

or

$B=\dfrac{\phi}{\pi r^2}$

$B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}$

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0

3 years ago

Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at

bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

a)

The electric field produced by a single-point charge is given by

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

$q=9\mu C = 9\cdot 10^{-6} C$

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as $q_0$ , and the charge at x = 8 m as $q_8$ .

For a point located at x = -2 m, both the fields $E_0$ and $E_8$ produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

$E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C$

b)

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C$

And since the sign is positive, the direction is to the right.

c)

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C$

And the negative sign indicates that the electric field in this case is towards the left.

d)

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

$E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C$

And the positive sign means the field is to the right.

e)

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0$

This means that we have to solve the equation

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0$

Re-arranging it,

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0\\\frac{(8-x)^2-x^2}{x^2(8-x)^2}=0$

So

$(8-x)^2-x^2=0\\64+x^2-16x-x^2=0\\64-16x=0\\64=16x\\x=4 m$

So, the electric field is zero at x = 4 m, exactly halfway between the two charges (which is reasonable, because the two charges have same magnitude)

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago

Read 2 more answers

What is the purpose of mucilage in the spaces around the cells of hornworts? Why is this important?

Bumek [7]

New Zealand hornworts and general information on them. ... in the top of the plant and remain attached to it and continue to grow throughout its life. ... are that the spaces between the cells are filled with mucilage rather then air chambers

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3 years ago

what is the efficiency of a machine that can carry a load of 100 kg with an effort of 20N when its velicity ratio is 10 ASAP

Salsk061 [2.6K]

Answer:

Efficiency of machine=4.9

Explanation:

We are given that

Mass of load=100 kg

Effort=20 N

Velocity ratio=10

We have to find the efficiency of machine.

Load= $mg=100\times 9.8=980 N$

Efficiency= $\frac{Load}{effort}\times \frac{1}{velocity\;ratio}$

Using the formula

Efficiency of machine= $\frac{980}{20}\times \frac{1}{10}$

Efficiency of machine= $\frac{49}{10}$

Efficiency of machine=4.9

7 0

3 years ago