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3 years ago

What do you mean by pressure

2 answers:

8 0

Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed.
P=F/A

bagirrra123 [75]3 years ago

6 0

Pressure is when forced is applied to an object

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Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce

Marat540 [252]

Answer:

15.32°

Explanation:

We have given the wavelength $\lambda =591nm=591\times 10^{-9}m$

Diffraction grating is 1460 lines per cm

So $d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m$ (as 1 m=100 cm )

For maximum diffraction

$dsin\Theta =m\lambda$ here m is order of diffraction

So $6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}$

$sin\Theta =0.264$

$\Theta =15.32^{\circ}$

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3 years ago

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6. What is the resultant force on each car below? Remember TWO pieces of

tamaranim1 [39]

Explanation:

In first case, the forces on LHS and on RHS is the same i.e. 3 N. The force acting on the car is balanced force. As a result, the car will not move at all.

In second case,

Force on RHS = 2000 N

Force on LHS = -6000 N

Net force acting on it is given by :

F = 2000+(-6000)

= -4000 N

Hence, this is the required solution.

4 0

2 years ago

Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the pos

deff fn [24]

Answer:

E = (0, 0.299) N

Explanation:

Given,

Charge $q_1\ =\ 8.0\ nC$
Charge $q_2\ =\ 6.0\ nC$
Distance of the first charge from the origin = (16m, 0)
Distance of the second charge from the origin = (-9, 0)
Point where the electric field required = (0, 12m)

Let $\theta_1\ and\ theta_2$ be the angle of the electric fields by first and second charge at the point A.

$\therefore sin\theta_1\ =\ \dfrac{12}{20}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{20}\ \right )\\\Rightarrow \theta_1\ =\ 36.87^o\\\\\therefore sin\theta_1\ =\ \dfrac{12}{9}\\\Rightarrow \theta_1\ =\ sin^{-1}\left (\dfrac{12}{9}\ \right )\\\Rightarrow \theta_1\ =\ 53.13^o\\$

Electric field by charge $q_1$ at point A,

$F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 8\times 10^{-9}}{20^2}\\\Rightarrow F_1\ =\ 0.18\ N/C$

Electric field by the charge $q_2$ at point A,

$F_1\ =\ \dfrac{kq_1}{r_1^2}\\\Rightarrow F_1\ =\ \dfrac{9\times 10^9\times 6.0\times 10^{-9}}{16^2}\\\Rightarrow F_1\ =\ 0.24\ N/C$

Now,

Net electric field in horizontal direction at point A $F_x\ =\ F_{1x}\ +\ F_{2x}\\\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\\\Rightarrow F_x\ =\ 0.18\times( -cos36.87^o)\ +\ 0.24\times cos53.13^o\\\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\\\Rightarrow F_x\ =\ 0\ N/C$

Net electric field in vertical direction at point A.

$F_y\ =\ F_{1y}\ +\ F_{2y}\\\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\\\Rightarrow F_y\ =\ 0.18\times sin36.87^o\ +\ 0.24\times sin53.13^o\\\Rightarrow F_y\ =\ 0.180\ +\ 0.192\\\Rightarrow F_y\ =\ 0.299\ N/C$

Hence, the net electric field at point A,

$F\ =\ ( 0, 0.299 )\ N/C.$

5 0

3 years ago

What do you mean by pressure​

What do you mean by pressure