Which layer of the atmosphere is the least dense?

Question 5 options:

exosphere


thermosphere


mesosphere


troposphere

Answer:

[tex]5.76\times 10^{-18}\ \text{N}[/tex] perpendicular to the velocity and magnetic field

Explanation:

B = Magnetic field = [tex]4\times 10^{-5}\ \text{T}[/tex]

[tex]\theta[/tex] = Angle the magnetic field makes with the horizontal = [tex]30^{\circ}[/tex]

v = Velocity of electron = [tex]9\times 10^5\ \text{m/s}[/tex]

q = Charge of electron = [tex]1.6\times 10^{-19}\ \text{C}[/tex]

Magnetic force is given by

[tex]F=qvB\sin\theta\\\Rightarrow F=1.6\times 10^{-19}\times 9\times 10^5\times 4\times 10^{-5}\sin30^{\circ}\\\Rightarrow F=2.88\times 10^{-18}\ \text{N}[/tex]

The magnitude of the magnetic force is [tex]2.88\times 10^{-18}\ \text{N}[/tex] and the direction is perpendicular to the velocity and magnetic field.

Answer:

c. Double Replacement

Explanation:

As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.

Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2

and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.

Answer:

25N/s or 25kg*m/s^2

Explanation:

It is written wrong because the unit of momentum is kgm/s^2 or N/s

Answer:

a) y = 127.6 m, b) 11.9s,  c)  v = 103.6  m / s, θ’= 326.7º

Explanation:

This is a missile throwing exercise

let's start by breaking down the initial velocity

         sin 30 = [tex]\frac{v_{oy} }{v_o}[/tex]

         cos 30 = v₀ₓ / v₀

         v_{oy} = v₀ go sin 30

         v₀ₓ = v₀ cos 30

         v_{oy} = 100 sin 30 = 50 m / s

         v₀ₓ = 100 cos 30 = 86.6 m / s

a) the maximum height is requested.

At this point the vertical velocity is zero (v_y = 0)

         v_y² = [tex]v_{oy}^2[/tex] - 2 g y

         0 = v_{oy}^2 - 2g y

         y = [tex]\frac{v_{oy}^2 }{2g}[/tex]

         y = 50² / (2  9.8)

         y = 127.6 m

b) Flight time

this is the time it takes to reach the ground, the reference system for this movement is taken on the ground this is a height of y = 0 m and the body is at an initial height of i = 100m

         y = y₀ + v₀ t - ½ g t²

       0 = 100 + 50 t - ½ 9.8 t²

we solve the quadratic equation

        4.9 t² - 50 t - 100 = 0

         t = [tex]\frac{50 \pm \sqrt{50^2 + 4 \ 4.9 \ 100} }{2 \ 4.9}[/tex]

         t = [tex]\frac{50 \ \pm \ 66.8}{9.8}[/tex]

         t₁ = 11.9 s

         t₂ = -8.4 s

flight time is 11.9s

c) The time 1 s before hitting the ground is

        t1 = 11.9 -1

        t1 = 10.9 s

let's find the vertical speed

       v_y =[tex]v_{oy}[/tex] - g t

       v_y = 50 - 9.8 10.9

       v_y = -56.8 m / s

the negative sign indicates that the direction of the velocity is downward.

On the x-axis there is no acceleration therefore the speed is constant.

Let's use the Pythagorean theorem

        v = [tex]\sqrt{v_x^2+v_y^2}[/tex]

        v = [tex]\sqrt { 86.6^2 + 56.8^2}[/tex]

        v = 103.6  m / s

let's use trigonometry

        tan θ = [tex]\frac{v_y}{v_x}[/tex]

        θ = tan⁻¹  \frac{v_y}{v_x}

        θ = tan⁻¹  (-56.8 / 86.60)

        θ = -33.3º

the negative sign indicates that it is measured clockwise from the x-axis

for a counterclockwise measurement

          θ’= 360 - θ

          θ' = 360 - 33.3

          θ’= 326.7º

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A  = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]

Therefore, the length of the wire is 134.62 m.

Answer:

I think c I dont know sorry if I'm wrong

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:  

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t[tex]_{min[/tex] = λ/2n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.20

we substitute

t[tex]_{min[/tex] = 750 / 2(1.20)

t[tex]_{min[/tex] = 750 / 2.4

t[tex]_{min[/tex] = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

b)

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t[tex]_{min[/tex] = λ/4n

given that; wavelength λ = 750 nm and  index of refraction of the oil n = 1.50

we substitute

t[tex]_{min[/tex] = 750 / 4(1.50)

t[tex]_{min[/tex] = 750 / 6

t[tex]_{min[/tex] = 125 nm

Therefore, the minimum thickness be now will be 125 nm

Answer:

Newton's first law of motion concerns any object that has no force applied to it.

Explanation:

three laws of motion and the law of universal gravitation.

Answer: i believe  its B Find the velocity of each block after they have moved apart sorry

Explanation: have a nice day buddy

Answera black cube or dark colors cause dark colors suck in heat

Answer:

I'd say D

Explanation:

because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)

Answer:

The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".

Explanation:

The given values are:

Potential difference,

Δv = 400 v

Radius,

r = 0.5580 cm

As we know,

⇒  [tex]W=e \Delta v[/tex]

and,

⇒  [tex]\frac{1}{2}mv^2=e \Delta v[/tex]

then,

⇒  [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]

On substituting the values, we get

⇒     [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]

⇒     [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]

⇒     [tex]=1.185\times 10^7 \ m/sec[/tex]

Answer:

3.35

Explanation:

Got it on Acellus

The light of wavelength 485 nm passes through a single slit. The single between the first (m=1) and second (m=2) interference minima is 3.36°.

Diffraction is the phenomenon of bending of waves through obstacles.  

Given is the wavelength λ= 485 nm, silt width d =  8.32 *10⁻⁶ m, then the angle θ will be

d sinθ =mλ

for m=1, sin θ₁ = λ/d

for m=2, sin θ₂ = 2λ/d

Substitute the values into both expressions to find the angles,

sin θ₁ = 485 x 10⁻⁹ /  8.32 *10⁻⁶

θ₁ = 3.34°

and sin θ₂ = (2 x  485 x 10⁻⁹ )/  8.32 *10⁻⁶

θ₂ = 6.7°

The angle between m =1 and m=2 will be

θ₂ -θ₁ = 6.7° - 3.34° =3.36°

Thus, angle between the first (m=1) and second (m=2) interference minima is 3.36°.

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Answer:

The mechanical energy of the ball is 112.667 J

Explanation:

Mechanical energy = kinetic energy + potential energy

Where:

kinetic energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]

potential energy = mgh

So that for the given ball, its mechanical energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] + mgh

                                = m([tex]\frac{1}{2}[/tex][tex]v^{2}[/tex] + gh)

But, m = 400 g = 0.4 kg, v = 23.0 m/s, g = 9.81 m/[tex]s^{2}[/tex] and h = 1.75 m.

Mechanical energy of the ball = 0.4([tex]\frac{1}{2}[/tex]([tex]23^{2}[/tex]) + 1.75*9.81)

                                                  = 0.4(264.5 + 17.1675)

                                                  = 0.4 x 281.6675

                                                  = 112.667

The mechanical energy of the ball is 112.667 J.

A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall, the maximum speed of the block when the spring is compressed to one-half of the maximum distance is 4.33 m/s

From the conservation of energy; the kinetic energy of the mass is equal to the work done on the spring.

i.e.

[tex]\mathbf{\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2_{max}}[/tex]

Given that:

∴

From the equation above, multiplying both sides with 2, we have:

[tex]\mathbf{mv^2 =kx^2_{max}}[/tex]

Making (k) the subject of the formula;

[tex]\mathbf{k = \dfrac{mv^2}{x^2_{max}}}[/tex]

[tex]\mathbf{k = \dfrac{4 \times 5^2}{0.68^2}}[/tex]

k = 216.26 N/m

However, when compressed to one-half of the maximum distance; the speed is computed as follows:

x = 0.68/2 = 0.34 m

∴

[tex]\mathbf{\dfrac{1}{2}mv_o^2 - \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2}[/tex]

[tex]\mathbf{m(v_o^2 -v^2) =kx^2}[/tex]

[tex]\mathbf{(v_o^2 -v^2) =\dfrac{kx^2}{m}}[/tex]

[tex]\mathbf{(5^2 -v^2) =\dfrac{216.26 \times 0.34^2}{4.0}}[/tex]

25 - v² = 6.25

25 -6.25 = v²

v² = 18.75

[tex]\mathbf{ v= \sqrt{18.75 }}[/tex]

v = 4.33 m/s

Therefore, we can conclude that the speed of the block when the spring is compressed to only one-half of the maximum distance is 4.33 m/s

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Answer:

please give me brainlist and follow

Explanation:

Using locally available resources for art help in the preservation of environment. A significant and practical aspects of art is material significance. The items used by artists while making an art piece affects both the form and the material. Every material delivers something special in the creative process.

Answer:

Explanation:

mass per unit length ρ = .100 / 1.65 = .0606 . kg /m

length of wire L = 1.65 m

For fundamental frequency , the expression is as follows

n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]

L = 1.65 , T = 16 n and m = .0606

n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]

= 4.9 /s .

This is fundamental frequency .

other mode of vibration ( first three ) will be as follows

4.9 x 2 = 9.8 /s ,

4.9 x 3 = 14.7 /s .

Answer:

a) v = 1.1 m/s

b) A = 0.315 m

c) v = 1.1 m/s  A= 0.15 m

Explanation:

a)

        [tex]v = \lambda * f (1)[/tex]

       [tex]f = \frac{1}{T} = \frac{1}{5.20s} = 0.19 Hz (3)[/tex]

      [tex]v = \lambda * f = 5.70 m * 0.19 Hz = 1.10 m/s (4)[/tex]

b)

      [tex]A = \frac{0.630m}{2} = 0.315 m (5)[/tex]

c)  

       v = 1.10 m/s (6)

       [tex]A = \frac{0.30m}{2} = 0.15 m (7)[/tex]

Explanation:

As pressure increases, with temperature constant, density increases. Conversely when temperature increases, with pressure constant, density decreases. Air density will decrease by about 1% for a decrease of 10 hPa in pressure or 3 °C increase in temperature.

Answer:

T = 2.5 s

Explanation:

Given that,

Number of complete orbits = 10

Time, t = 25 seconds

We need to find the period of the orbiting ball. Let it is T. We know that number of oscillations per unit time is called frequency and the reciprocal of frequency is called period of the ball.

So,

[tex]T=\dfrac{t}{n}\\\\T=\dfrac{25}{10}\\\\T=2.5\ s[/tex]

So, the period of the orbiting ball is equal to 2.5 seconds.

Answer:

  x = 0.455 L

Explanation:

For this exercise we must use the rotational equilibrium condition

        Σ τ = 0

it has two forces, the first is perpendicular to the rod, so its stub is

         τ₁ = F₁ L

the second force is applied with an angle, so we can use trigonometry to find its components

          sin θ = F_parallel / F₂

          cos θ = F_perpendicular / F₂

         F_parallel = F₂ sin θ

         F _perpendicular = F₂ cos θ

torque is

         τ₂ = F_perpendicular x + F_parallel 0

the parallel force is on the rod therefore its distance is zero

we apply the equilibrium equation

          τ₁  - τ₂ = 0

          F₁ L = F₂ cos θ  x

          x = [tex]\frac{L}{cos \theta} \ \frac{F_1}{F_2}[/tex]

let's calculate

          x = [tex]\frac{L}{cos \ 42.9} \ \frac{2.00}{6.00}[/tex]

          x = 0.455 L

The normal force exerted by the floor of the elevator on the student during her brief acceleration is 1357.2 N.

To determine the normal force exerted by the floor of the elevator on the student during the brief acceleration, we need to analyze the forces acting on the student.

The force of gravity acting on the student can be calculated using the equation F_gravity = m * g, where m is the mass of the student and g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, F_gravity = 87.0 kg * 9.8 m/s² = 852.6 N.

Since the elevator and the student are accelerating upward, there is an additional upward force acting on the student. This force is equal to the mass of the student multiplied by the acceleration of the elevator, which is given as 5.80 m/s². So, the upward force is F_upward = m * a = 87.0 kg * 5.80 m/s² = 504.6 N.

Now, the normal force exerted by the floor of the elevator on the student is equal to the sum of the gravitational force and the upward force. In this case, the normal force is given by F_normal = F_gravity + F_upward = 852.6 N + 504.6 N = 1357.2 N.

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Answer:

Explanation:

proton

Answer:

245 divided by 5.14=47.6653696 or 47.66

Explanation:

Explanation:

it's true

Kiss me if I'm wrong. But dinosaurs still exist right?

Answer:

True

Definetely true

Answer:

the mass of the object multiplied by the acceleration of the object

Explanation:

N2L states that F = ma (force equals mass times acceleration).

Answer:

The rocket will appear larger than it actually is

The tension in the strings are 31.47 and 19.25 N respectively.

Mass of the block, m = 3 kg

From the figure, consider the vertical components,

T₁ sin45° + T₂ sin30° = mg

(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4

Also, consider the horizontal components,

T₁ cos45° = T₂ cos30°

T₁/√2 = T₂ x√3/2

T₁ = T₂ x √3/2 x √2

So,

T₁ = 0.612T₂

Applying in the first equation,

(T₁/√2) + (T₂/2) = 29.4

(0.612T₂/1.414) + 0.5T₂ = 29.4

0.434 T₂ + 0.5 T₂ = 29.4

0.934 T₂ = 29.4

Therefore, the tension,

T₂ = 29.4/0.934

T₂ = 31.47 N

So, the tension,

T₁ = 0.612 T₂

T₁ = 0.612 x 31.47

T₁ = 19.25 N

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