What is the mass of 2.5 mol of Ca, which has a molar mass of 40 g/mol?

With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a

Aleks04 [339]

Answer:

The correct option is (b).

Explanation:

Given that,

Electric field, $E=2\times 10^2\ N/C$

We need to find the magnitude of the force on the electron as a result of the electric field.

We know that, the electric force is given by :

$F=qE\\\\=1.6\times 10^{-19}\times 2\times 10^2\\\\F=3.2\times 10^{-17}\ N$

So, the required force on the electron is equal to $3.2\times 10^{-17}\ N$ .

5 0

3 years ago

Given a 45 45 90 prism with index of 1.5, immersed in air. The hypotenuse acts as the reflecting face by TIR. A ray of light ent

Genrish500 [490]

Answer:

83.6°

Explanation:

For the ray to be totally internally reflected, at the boundary, the angle of refraction is 90. Using the law of refraction where

n₁sinθ₁ = n₂sinθ₂ where n₁ = refractive index of prism = 1.5, θ₁ = critical angle in prism, n₂ = refractive index of air = 1 and θ₂ = refractive angle = 90°.

So, substituting these values into the equation,

n₁sinθ₁ = n₂sinθ₂

1.5 × sinθ₁ = 1 × sin90

1.5 × sinθ₁ = 1

sinθ₁ = 1/1.5

sinθ₁ = 0.6667

θ₁ = sin*(0.6667)

θ₁ = 41.8°

So, for total internal reflection, an incidence angle of 41.8° is required. So, a full convergence angle of 2 × 41.8° = 83.6° is required for the whole bundle of rays.

5 0

3 years ago

1. Set frequency of wave generator to 5Hz.

ivann1987 [24]

Separate the barriers so they have a 2cm gap between them.

3 0

3 years ago

(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg

valina [46]

Answer:

Explanation:

From the given information:

Let the first weight be $m_ 1$ = 80 kg

The weight of the buddy be $m_2$ = 120 kg

The weight of Bubba be $m_3$ = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

$x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2$

The length of the boat be $x_2$ = 4 m

∴

We can find the center of mass of the system by using the formula:

$X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}$