A negative point charge - is at the center of a hollow insulating spherical shell, which has an inner radius R1 and an outer radius R2. There is a total charge of +7Q spread uniformly throughout the volume of the insulating shell, not just on its surface. Determine the electric field for the following. (Use any variable or symbol stated above along with the following as necessary: r and E
(a) r < R1
(b) R1 (c) R2

Answer:

dasd

Explanation:

dsdas

Answer:

mass, weight, strength of gravity increases, weight decreases

Explanation:

got it on edge

Answer:

An object’s

✔ mass

will remain constant throughout the universe, but its

✔ weight

can change from planet to planet.

If you increase the mass of a planet, what happens to its gravity?

✔ strength of gravity increases

If the gravity on a planet decreases, what happens to the weight of an object on that planet?

✔ weight decreases

Explanation:

right on edge 22

they're parallelograms

Answer:

R ≈ 15 ohms

Explanation:

Using ohm's law equation,

I = V/R, to solve for the resistance of the heating coil.

R = V/I

Known:

V = 220 v = 220 kgm^2s^-3A^-1

I = 15 A

Unknown:

R =?

Solution:

R = (220 kgm^2s^-3A^-1)/ 15.0 A

R = 14.6 kgm^2s^-3A^-2

R ≈ 15 kgm^2s^-3A^-2

R ≈ 15 ohms

Answer:

[tex]0.0018833\ \text{m/s}[/tex]

Explanation:

[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]

[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]

[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]

We have the relation

[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]

[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]

The required answer is [tex]0.0018833\ \text{m/s}[/tex].

Answer:

violet region (UV) is wavelength less than 400 nm, the most common is between 200 and 400 nm,

The infrared (IR) range wavelengths greater than 700 nm

Explanation:

The spectral region near the visible is what humans can see with our eyes.

The range of the ultra violet region (UV) is wavelength less than 400 nm, the most common is between 200 and 400 nm, this radiation is responsible for tanning and skin burns.

The infrared (IR) range is a very wide range that begins at wavelengths greater than 700 nm and continues up to approximately 25,000 nm (1012 Hz).

Answer:

1.04 s

Explanation:

The computation is shown below:

As we know that

t = t' × 1 ÷ (√(1 - (v/c)^2)

here

v = 0.5c

t = 1.20 -s

So,

1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)

1.20 = t' × 1 ÷ (√(1 - (0.5)^2)

1.20 = t' ÷ √0.75

1.20 = t' ÷ 0.866

t' = 0.866 × 1.20

= 1.04 s

The above formula should be applied

im confused hold on imma send you a link to the answerExplanation:

Answer:

The correct solution is "1.2 m".

Explanation:

The given values are:

Wavelength of waves,

λ = 1.5 m

Speed of waves on surface,

V = 2 m/sec

Speed of waves in water,

V₁ = 1.6 m/sec

As we know,

⇒  [tex]V=f\times \lambda[/tex]

or,

⇒  [tex]f=\frac{V}{\lambda}[/tex]

On substituting the values, we get

⇒      [tex]=\frac{2}{1.5}[/tex]

⇒      [tex]=1.33 \ Hz[/tex]

hence,

⇒  [tex]\lambda_1=\frac{V_1}{f}[/tex]

On substituting the values, we get

⇒       [tex]=\frac{1.6}{1.33}[/tex]

⇒       [tex]=1.2 \ m[/tex]

Answer:

Frictional Force

Explanation:

i think so

Answer:

A force is a push or pull upon an object resulting from the object's interaction with another object.

Momentum is force or speed of movement.

Velocity defines the path of the motion of the frame or the object

Answer:

W' = 1.66 x 10¹⁴ N

Explanation:

First, we will calculate the mass:

[tex]W = mg[/tex]

where,

W = weight on earth = 690 N

m = mass = ?

g = acceleration due to gravity on earth = 9.8 m/s²

Therefore,

[tex]m = \frac{W}{g} = \frac{690\ N}{9.8\ m/s^2}\\\\m = 70.4\ kg[/tex]

Now, we will calculate the value of g on the neutron star:

[tex]g' = \frac{GM}{R^2}[/tex]

where,

g' = acceleration due to gravity on the surface of the neutron star = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of the Neutron Star = 1.99 x 10³⁰ kg

R = Radius of the Neutron Star = 15 km/2 = 7.5 km = 7500 m

Therefore,

[tex]g' = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1.99\ x\ 10^{30}\ kg)}{(7500\ m)^2}\\\\g' = 2.36\ x\ 10^{12}\ m/s^2[/tex]

Therefore, the weight on the surface of the neutron star will be:

[tex]W' = mg'\\W' = (70.4\ kg)(2.36\ x\ 10^{12}\ m/s^2)[/tex]

W' = 1.66 x 10¹⁴ N

Explanation:

The answer is In the picture. Thanks.

Answer:

a) v₂ = 30 m/s

b) m₁ = 12600 kg

c) m₂ = 12600 kg

Explanation:

a)

Using the continuity equation:

[tex]A_1v_1 = A_2v_2[/tex]

where,

A₁ = Area of inlet = π(0.15 m)² = 0.07 m²

A₂ = Area of outlet = π(0.05 m)² = 0.007 m²

v₁ = speed at inlet = 3 m/s

v₂ = speed at outlet = ?

Therefore,

[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]

v₂ = 30 m/s

b)

[tex]m_1 = \rho A_1v_1t[/tex]

where,

m₁ = mass of water flowing in = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]

m₁ = 12600 kg

c)

[tex]m_1 = \rho A_1v_1t[/tex]

where,

m₂ = mass of water flowing out = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]

m₂ = 12600 kg

Answer:

Distance is 300 and displacement is 100

Explanation:

Distance= 100+100+100=300

Displacement=100

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Answer:

   [tex]r_{cm}[/tex]= 1/5 L

Explanation:

To find the center of mass of the system let's use

        [tex]r_{cm} = \frac{1}{M}[/tex] ∑ r_i x_i

where m is the total mass of the system

let's apply this expression to our case

Let's set the reference frame on the star

         [tex]r_{cm} = \frac{1}{M +4M} ( 4M 0 + M L)[/tex]

         r_{cm} = [tex]\frac{1}{5}[/tex] L

         [tex]r_{cm}[/tex]= 1/5 L

Hey, the breast center is 1/5 of the distance between the star and the planet.

Answer:98.6 degrees Fahrenheit

Explanation:

Answer:

a) Battery-operated devices have: small led lights, flashlights, wireless keyboards and mice, watches, electronic weights

b) ed lights and flashlights transform into light energy and thermal energy

c)  Em₀ = U = m gh,  Em_f = K = ½ m v²

Explanation:

In this exercise ask to complete the sentences

a) Battery-operated devices have: small led lights, flashlights, wireless keyboards and mice, watches, electronic weights

b) These devices transform the chemical energy stored in the batteries into other forms of energy.

Led lights and flashlights transform into light energy and thermal energy

Keyboards transform into electromagnetic energy that is emitted

clocks transform to mechanical energy from the movement of the needles

Electronic weights transforms the chemical energy of the baria into gravitational potential energy that prevents the movement of the plate and this translates into the reading of the body weight

c) The total energy of the cylinder mechanical energy when sustained is

          Em₀ = U = m gh

it is transformed as it descends into kinetic energy, at any point

          Emₙ = K + U = 1/2 m v² + m g y

at the lowest point of the trajectory all energy is transformed

           Em_f = K = ½ m v²

Answer:

36 times less.

Explanation:

The distance from you to the sun is 6AU's, and from the sun to the earth is 1 AU.

Therefore,

At Earth sunlight received per unit cm² is:

[tex]I_{earth} = \dfrac{I_o}{4 \pi \times (1)^2}[/tex]

[tex]I_{earth} = \dfrac{I_o}{4 \pi}[/tex]

[tex]I_{me} =\dfrac{I}{4 \pi (6)^2}[/tex]

[tex]I_{me} = \dfrac{I_o}{36(4 \pi)}[/tex]

Thus, [tex]I_{earth} = 36 \times I_{me}[/tex]

Thus, the right answer is 36 times less.

Answer:

0.02 kg/cm³

Explanation:

Answer:

 w = 0.319 rad / s

Explanation:

This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.

initial instant. Before the squirrel jumps

           L₀ = m v r

final instant. After the trough and the squirrel are together

          L_f = (I_fetter + I_ardilla) w

angular momentum is conserved

          L₀ = L_f

          m v r = (I_fetter + I_ardilla) w

          w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]

the moment inercial ofbody is

         I_thed = 2.00 kg m²

We approach the squirrel to a specific mass

          I_ardilla = m r²

we substitute

            w = m v r / ( I_[feefer  + m r²)

let's calculate

              w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )

              w = 0.6426 / 2.0119

               w = 0.319 rad / s

Answer:

C

Explanation:

a=f/m

10Kgm/s2/4kg

2.5m/s2

When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity, or g. It is a vector quantity whose strength or magnitude is determined by the norm and whose direction correlates with a plumb bob.

Given in the question, force 10 N and mass 4.0 Kg the acceleration is,

a = 10/4 = 2.5 m/sec²

When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².

To learn more about acceleration refer to the link:

brainly.com/question/12550364

#SPJ2

Answer:

uv radiation cause cancer

uv radiation effect our eyes

Answer:

you can get

1:skin cancer

2:eye damage

3:skin damage

4:immune system suppression

choose which two u want

hope this helped

:)

Explanation:

Answer:

This is because of friction and heat lost.

Explanation:

Answer:

Time taken by Mr. smith = 4.80 hour (Approx.)

Explanation:

Given:

Distance travel by Mr. smith = 523 kilometer

Average speed of Mr. smith = 109 km/hr

Find;

Time taken by Mr. smith

Computation:

Time taken = Distance cover / Speed

Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.

smith

Time taken by Mr. smith = 523 / 109

Time taken by Mr. smith = 4.798 hr

Time taken by Mr. smith = 4.80 hour (Approx.)

Answer:

Hello! Your answer would be, O False

Explanation:

Hope I helped! Brainiest plz!♥ Have a nice morning! Hope you make a 100%! -Abby

Answer:

False is the correct answer.

Explanation:

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Answer:

the smallest separation between two objects is 0.8067 m

Explanation:

Given the data in the question;

Altitude h = 5.75 km = 5750 m

Diameter D = 4.0 mm = 0.004 m

λ = 460 nm = 4.6 × 10⁻⁷ m

Now, Using Rayleigh criterion for Airy disks resolution.

we know that, Minimum angular separation for resolving two points is;

θ = 1.22λ / D

so we substitute

θ = (1.22 × 4.6 × 10⁻⁷)  / 0.004

θ = 5.612 × 10⁻⁷ / 0.004

θ = 1.403 × 10⁻⁴ rad  

so minimum separation [tex]d_{min[/tex] =  θh

so we substitute

[tex]d_{min[/tex] = (1.403 × 10⁻⁴) × 5750 m

[tex]d_{min[/tex] = 0.8067 m

Therefore, the smallest separation between two objects is 0.8067 m