An object is thrown 16m/s straight up from a 7m tall cliff. How much time does it take to hit the ground below

A solid uniformly charged insulating sphere has uniform volume charge density p and radius R. Apply Gauss's law to determine an

RUDIKE [14]

Answer:

electric field E = (1 /3 e₀) ρ r

Explanation:

For the application of the law of Gauss we must build a surface with a simple symmetry, in this case we build a spherical surface within the charged sphere and analyze the amount of charge by this surface.

The charge within our surface is

ρ = Q / V

Q ’= ρ V '

The volume of the sphere is V = 4/3 π r³

Q ’= ρ 4/3 π r³

The symmetry of the sphere gives us which field is perpendicular to the surface, so the integral is reduced to the value of the electric field by the area

I E da = Q ’/ ε₀

E A = E 4 πi r² = Q ’/ ε₀

E = (1/4 π ε₀) Q ’/ r²

Now you relate the fraction of load Q ’with the total load, for this we use that the density is constant

R = Q ’/ V’ = Q / V

How you want the solution depending on the density (ρ) and the inner radius (r)

Q ’= R V’

Q ’= ρ 4/3 π r³

E = (1 /4π ε₀) (1 /r²) ρ 4/3 π r³

E = (1 /3 e₀) ρ r

4 0

3 years ago

A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Negle

Artist 52 [7]

Answer:

θ₀ = 84.78° (OR) 5.22°

Explanation:

This situation can be treated as projectile motion. The parameters of this projectile motion are:

R = Range of Projectile = 150 m

V₀ = Launch Speed of Projectile = 90 m/s

g = 9.8 m/s²

θ₀ = Launch angle (OR) Angle of Elevation = ?

The formula for range of a projectile is given as:

R = V₀² Sin 2θ₀/g

Sin 2θ₀ = Rg/V₀²

Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²

2θ₀ = Sin⁻¹ (0.18)

θ₀ = 10.45°/2

Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:

θ₀ = 90° - 5.22°

7 0

3 years ago

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$

$F_{net} = 300 - 150$

$F_[net} = 150 N$

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0

3 years ago

In one type of solar energy system, sunlight heats the air within solar panels, which heats copper tubes filled with water. What

Anna71 [15]

The type of energy that is produced by the system that is described is heat energy. The correct answer is D.

3 0

3 years ago

Read 2 more answers

Light propagate faster through medium “a” than medium “b”

dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

1)

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

$n=\frac{c}{v}$

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

2)

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1, n_2$ are the index of refraction of the two mediums

$\theta_1, \theta_2$ are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)