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3 years ago

The gland which is like a medium for the passage of the sperms is known as

Physics

1 answer:

Inessa [10]3 years ago

4 0

Answer:

seminal glands or vesicular glands

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A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x ax

alexdok [17]

Answer:

Explanation:

We shall apply conservation of momentum law in vector form to solve the problem .

Initial momentum = 0

momentum of 12 g piece

= .012 x 37 i since it moves along x axis .

= .444 i

momentum of 22 g

= .022 x 34 j

= .748 j

Let momentum of third piece = p

total momentum

= p + .444 i + .748 j

applying conservation law of momentum

p + .444 i + .748 j = 0

p = - .444 i - .748 j

magnitude of p

= √ ( .444² + .748² )

= .87 kg m /s

mass of third piece = 58 - ( 12 + 22 )

= 24 g = .024 kg

if v be its velocity

.024 v = .87

v = 36.25 m / s .

6 0

2 years ago

A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

$W=mg$ : weight of the ladder, with m = 20 kg (mass) and $g=9.8 m/s^2$ (acceleration of gravity)

$W_M=Mg$ : weight of the person, with M = 54 kg (mass)

$N_1$ : normal reaction exerted by the wall on the ladder

$N_2$ : normal reaction exerted by the floor on the ladder

$F_f = \mu N_2$ : force of friction between the floor and the ladder, with $\mu$ (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

$W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}$

where

$Wsin 21^{\circ}$ is the component of the weight of the ladder perpendicular to the ladder

$W_M sin 21^{\circ}$ is the component of the weight of the man perpendicular to the ladder

$N_1 sin 69^{\circ}$ is the component of the normal force perpendicular to the ladder

And solving for $N_1$ , we find the force exerted by the wall on the ladder:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N$

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of $N_2$ .

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

$\sum F_y = 0\\N_2 - W - W_M =0$

And substituting and solving for N2, we find:

$N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N$

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

$\sum F_x = 0\\F_f - N_1 = 0$

And re-writing the equation,

$\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257$

So, the minimum value of the coefficient of friction is 0.257.

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

$\mu = \frac{N_1}{N_2}$

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}$

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)$

And substituting, we get

$N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N$

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

$\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332$

Learn more about torques and equilibrium:

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7 0

3 years ago

what happens to light when it falls upon a material that has a natural frequency equal to the frequency of the light?

mixas84 [53]

The energy from the light is transferred to the material, causing it to vibrate and absorb the light.

What is energy?
In physics, energy is the quantitative quality that is transmitted to the a body or a physical system, and is discernible in the work performed as well as in the form of light and heat. The law of conservation states that although energy can change its form, it cannot be created or destroyed. Energy is indeed a conserved quantity. The International System of Units' (SI's) joule is the measurement unit for energy (J). A moving object's kinetic energy, a solid object's elastic energy, chemical energy caused by chemical reactions, and the potential energy that an object stores (for instance because of its position inside a field) are examples of common forms of energy.

When light falls upon a material that has a natural frequency equal to the frequency of the light, the light will be absorbed by the material. This is due to resonance, which occurs when the frequency of the light matches the natural frequency of the material. The energy from the light is transferred to the material, causing it to vibrate and absorb the light.

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1 year ago

By experiment, determine what makes a force attractive or repulsive. Describe your experiments and observations with some exampl

drek231 [11]

The charge present determines a force to be attractive or repulsive.

The charges acquired by two bodies determines the Force as Attractive Or Repulsive.

Electric Force applied due to Electrical charges is same in magnitude but opposite in direction. This corresponds this phenomenon equivalent to the Newton's Third Law.

Examples of the experiments and observations:

On combing hair through a comb and then keeping it close to small pieces of paper shows attraction of paper pieces towards the comb.

This occurs due to the Electric charges present in the comb that induces charge in paper pieces leading to their attraction.

In both Gravitational Force and Coulomb force, the force remains inversely proportional to the square of the distance following the Inverse Square Law being the Central Force system. This only differs by the fact that in Gravitational Force, masses are used and in Coulomb force, charges are used.

The more the distance between the charges, the less is the Electric Force.

The lesser the distance between the charges, the more is the Electric Force.

If both the objects are charged the same i.e. either positive or negative then the Force is Repulsive and if the charges are Oppositely charged then the force is attractive.

Hence, the charge present determines a force to be attractive or repulsive.

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6 0

1 year ago

Calculate the potential difference across a 10 ohm register carrying a current of 2.5 ampere

Bas_tet [7]

Answer:

using ohm's law

V=IR

V= 10 X 0.2

V = 2 Volt.

Explanation:

4 0

2 years ago

The gland which is like a medium for the passage of the sperms is known as ​

The gland which is like a medium for the passage of the sperms is known as