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inn [45]
3 years ago
9

A graph of angular position v. time has the following equation:

Physics
1 answer:
Zielflug [23.3K]3 years ago
3 0

Answer:

3.5x^2 +25x +250-y=0 derivate this eqn you get 7x +25-1 =0 7x =-24 divided the eqn and you get it then inter the value what you get in first eqn

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If an X-ray beam of wavelength 1.4 × 10-10 m makes an angle of 20° with a set of planes in a crystal causing first order constru
Flura [38]

Answer:

43.16°

Explanation:

λ = Wavelength = 1.4×10⁻¹⁰ m

θ₁ = 20°

n can be any integer

d = distance between the two slits

Since for the first bright fringe, n₁ = 1

n₂ = 2 for second order line

The relation between the distance of the slits and the angle through which it is passed is:

dsinθ=nλ

As d and λ are constant

\frac{n_1\lambda}{sin \theta_1}=\frac{n_2\lambda}{sin \theta_2}\\\Rightarrow \frac{1}{sin20}=\frac{2}{sin\theta_2}\\\Rightarrow sin\theta_2=\frac{2}{\frac{1}{sin20}}\\\Rightarrow \theta_2=sin^{-1}{\frac{2}{\frac{1}{sin20}}}\\\Rightarrow \theta_2=43.16^{\circ}

∴ Angle by which the second order line appear is 43.16°

5 0
3 years ago
Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a
butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0
3 years ago
If the different parts of an electric circuit are found on separate branches of the circuit, the circuit is called a (an)
Vladimir [108]

Answer:

C parallel circuit

Explanation:

4 0
3 years ago
In an elastic collision, ________. A. an individual molecule in the collision never loses energy B. the molecules involved in th
Aleks04 [339]

Answer:

Im pretty sure its B im very sorry if its wrong.

3 0
3 years ago
A 2.0-kg block is held at rest against a spring of constant 2700 N/m, compressing it 5.5 cm. The block and the spring are locate
KonstantinChe [14]

Answer:

a) 4.1 J

b) -14 J

c) 4.8 m/s

Explanation:

The energy stored in the spring is given by:

U_e=\frac{1}{2}*K*x^2\\\\U_e=\frac{1}{2}*2700N/m*(5.5*10^{-2}m)^2\\\\U_e=4.1J

The mechanical energy loss is because of the work done by the friction force.

The friction force (only presented on the inclined surface) is given by:

F_f=\µ*F_N

F_N=2.0kg*9.8m/s^2*cos(35)\\F_N=16N\\F_f=0.29*16N=4.6N

We need to calculate the length of the ramp in order to calculate the work, the length of the ramp is the hypotenuse:

sin(\theta)=\frac{OC}{h}\\h=\frac{OC}{sin(\theta)}\\\\h=\frac{1.7m}{sin(35)}\\\\h=3.0m

So the work done by the friction force is:

W_f=F_f*d*cos(\alpha)\\W_f=4.6N*3.0*cos(180)\\W_f=-14J

the angle is 180 degrees because the force is opposite to the motion.

In order the know the final velocity we need to apply the Energy Conservation Theorem:

K_1+U_{g1}+U_{e}+W_f=K_2+U_{g2}\\\\0+m*g*h_1+4.1J-14J=\frac{1}{2}*m*v^2+m*g*h_2\\\\2.0kg*9.8m/s^2*1.7m+4.1J-14J=\frac{1}{2}*2.0kg*v^2+2.0kg*9.8m/s^2*(0)\\\\33J+4.1J-14J=v^2\\\\v=\sqrt{23.1}\\\\v=4.8m/s

5 0
3 years ago
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