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iren [92.7K]
3 years ago
5

Which of the following is NOT an example of a medium.

Physics
1 answer:
andriy [413]3 years ago
8 0

Answer:

D.) Metal

Explanation:

light can not pass through metal.

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WHY light travel in straight line?reason?
Dafna1 [17]

Light travels in a straight line because it doesn't usually interact with its medium. When it does it is called refraction.

8 0
4 years ago
Read 2 more answers
What is the speed of a beam of electrons when under the simultaneous influence of E = 1.64×104 V/m B = 4.60×10−3 T Both fields a
andrezito [222]

Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s

Explanation:

Magnetic force(B) = 4.60×10^-3 T

Electric force(E) = 1.64×10^4 V/m

Both forces having equal magnitude ;

Magnetic force = electric force

qvB = qE

vB = E

v = (1.64×10^4) ÷ (4.60×10^-3)

v = 3.57×10^6 m/s

2.) Assume no electric field

qvB = ma

Where a = v^2 ÷ r

R = radius

a = acceleration

v = velocity

qvB = m(v^2 ÷ R)

R = (m×v) ÷ (|q|×B)

q=1.6×10^-19C

m = 9.11×10^-31kg

R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)

R = 32.5227×10^-25 ÷ 7.36×10^-22

R = 4.42×10^-3m

3.) period(T)

T = (2*pi*R) ÷ v

T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)

T = (27.775×10^-3) ÷(3.57×10^6)

T = 7.78×10^-9 s

6 0
4 years ago
An electron has a charge of - 1.6x10 ^ - 19 C and a proton has a charge of 1.6x 10 ^ - 19 C . In a hydrogen atom, the distance b
ra1l [238]

Answer:, , 8.2 . 10^-6, ATTRACTIVE

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law

4 0
3 years ago
1. Calcular la masa de mercurio que pasó de 30 °C hasta 120 °C y absorbió 4400 cal. Calor específico del
timofeeve [1]

Answer:

Masa, m = 0.088 kg

Explanation:

Given the following data;

Temperatura inicial = 30°C

Temperatura final = 120°C

Capacidad calorífica específica = 138J/kg.K

Calor absorbido, Q = 4400 cal.

Para encontrar la masa;

La capacidad calorífica viene dada por la fórmula;

Q = mct

Dónde;

Q representa la capacidad calorífica o la cantidad de calor.

m representa la masa de un objeto.

c representa la capacidad calorífica específica del agua.

dt representa el cambio de temperatura.

dt = T2 - T1

dt = 120 - 30

dt = 90°C to kelvin = 273 + 90 = 363K

Sustituyendo en la fórmula, tenemos;

4400 = m*138*363

4400 = 50094m

m = \frac {4400}{50094}

Masa, m = 0.088 kg

7 0
3 years ago
A block of mass m = 2.20 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 6.80
Bogdan [553]

Answer:

(a) The speed of the lighter block is v_{2x} = 3.7~m/s.

    The speed of the heavier block is v_{3x} = 3.55~m/s.

(b) The smaller block goes up to 0.69 m.

Explanation:

We will divide this question into three parts: Part A is for the smaller mass from the top of the incline to the collision. Part B is the collision. And Part C is for the smaller mass from the bottom to the highest point it can achieve.

In order to solve this question, I will assume that the smaller mass is initially at rest.

Part A:

We will use conservation of energy.

K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}mv_1^2  + 0\\(2.2)(9.8)(3.6) = \frac{1}{2}(2.2)v_1^2\\v_1 = 8.4 ~m/s

This is the speed of the smaller mass just before the collision. The velocity of the mass is directed 30° above horizontal, since the mass is sliding down the incline.

Part B:

Momentum is a vector identity, so the x- and y-components of momentum are to be investigated separately. Since the collision occurs at the horizontal surface, only the x-component of momentum is conserved.

P_1 = P_2\\mv_{1x} = mv_{2x} + Mv_{3x}\\(2.2)(8.4\cos(30^\circ)) = (2.2)v_{2x} + (6.8)v_{3x}\\16 = 2.2v_{2x} + 6.8v_{3x}

During the collision kinetic energy is also conserved. Since kinetic energy is a scalar quantity, we don't have to separate its components.

K_{initial} = K_{final}\\\frac{1}{2}mv_1^2 = \frac{1}{2}mv_{2}^2 + \frac{1}{2}Mv_{3x}^2\\(2.2)(8.4)^2 = (2.2)v_{2}^2 + (6.8)v_{3x}^2\\155.23 = 2.2v_{2}^2 + 6.8v_{3x}^2

The following relation will be used when combining the two equations:

v_{2x} = v_2\cos{30^\circ}

The following equation is useful for combining the two equations:

v_{3x} = \frac{2m}{(m+M)}v_{1x} = \frac{2(2.2)}{(2.2 + 6.8)}(8.4\cos(30^\circ)) = 3.55~m/s

Therefore from the first equation,

16 = 2.2v_{2x} + 6.8v_{3x} = 2.2v_{2x} + 6.8(3.55) \\v_{2x} = -3.7~m/s

Part C:

We will again use the conservation of energy to find the highest point that the mass can go:

K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv_{2x}^2 + 0 = 0 + mgH\\\frac{1}{2}(2.2)(-3.7)^2 = (2.2)(9.8)H\\H = 0.69 ~m

4 0
4 years ago
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