This photo shows a beam of light entering and exiting a piece of glass.
What happens when the light enters the glass?
A. The light is absorbed.
B. The light is reflected.
C. The light is scattered.
D. The light is refracted.

Answer:

the pressure the balloon exerts on the table is 1,730.77 N/m²

Explanation:

Given;

weight of the water balloon, F = 4.5 N

area of the balloon, A = 2.6 x 10⁻³ m²

The pressure the balloon exerts on the table is calculated as follows;

[tex]P = \frac{F}{A}[/tex]

substitute the given values and solve for pressure, P;

[tex]P = \frac{4.5}{2.6 \times 10^{-3}} \\\\P = 1,730.77 \ N/m^2[/tex]

Therefore, the pressure the balloon exerts on the table is 1,730.77 N/m²

Answer:

im sorry i would help but thats too much

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:

The cart mark (a) has the most potential energy and the cart marked (b) has the most kinetic energy

Answer:

1992.008J

Explanation:

Answer:

A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").

A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).

a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.

b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.

Answer:

A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").

A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).

a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.

b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.

Explanation:

Explanation:

When you get closer to the mirror than the focal point a virtual image is formed behind the mirror and this image is not inverted. That's why the image flips as you get closer. ... With a virtual image the light rays never come to a focus so there is no place you can put a piece of paper to see the image.

Answer:

a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

Explanation:

a. Who reaches the bottom first

The kinetic energy of the objects is given by

K = 1/2mv² + 1/2Iω² where m = mass of object, v = velocity of object, I = moment of inertia and ω = angular velocity = v/r where r = radius of object

For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder

So, its kinetic energy, K = 1/2mv² + 1/2(mr²/2)(v/r)²

K = 1/2mv² + 1/4mv²

K = 3mv²/4

For the steel hoop, I' = mr'² where m' = mass of steel hoop and r' = radius of steel hoop and v' = velocity of steel hoop

So, its kinetic energy, K' = 1/2m'v'² + 1/2(m'r'²)(v'/r')²

K' = 1/2m'v'² + 1/2m'v'²

K' = m'v'²

Since both kinetic energies are the same, since the drop from the same height,

K = K'

3mv²/4 = m'v'²

v²/v'² = 4m/3m'

v²/v'² = 4/3(m/m')

v/v' = √[4/3(m/m')]

Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.

Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.

b. Why

Since the kinetic energy, K = translational + rotational

We find the translational kinetic energy of each object.

For the wooden cylinder,

K = K₀ + 1/2Iω² where K₀ = translational kinetic energy of wooden cylinder

K - 1/2Iω² = K₀

3/4mv² - 1/2(mr²/2)(v/r)² = K₀

3/4mv² - 1/4mv² = K₀

K₀ = 1/2mv²

For the steel hoop,

K' = K₁ + 1/2I'ω'² where K₁ = translational kinetic energy of steel hoop

K' - 1/2I'ω'² = K₁

m'v'² - 1/2(m'r'²)(v'/r')² = K₁

m'v'² - 1/2m'v'² = K₁

K₁ = 1/2m'v'²

So, K₀/K₁ =  1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².

Since (m/m') < 1 ⇒  (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁  < 1.33 ⇒ K₀ > K₁

So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.

So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

The energy of the body due to its movement in a particular direction under the influence of a force like a free-falling body due to gravitaional force is called  Kinetic energy.

The kinetic energy of the objects is given by

[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}Iw^2[/tex]

where

m = mass of object,

v = velocity of object,

I = moment of inertia and

ω = angular velocity = v/r where r = radius of object

For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder

So, its kinetic energy,

[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{mr^2}{2})\dfrac{v}{r}^2[/tex]

[tex]K = \dfrac{3mv^2}{4}[/tex]

For the steel hoop,

I' = mr'²

where

m' = mass of steel hoop and

r' = radius of steel hoop and

v' = velocity of steel hoop

So, its kinetic energy,

[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}(m'r'^2)\dfrac{v'}{r'}^2[/tex]

[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}m'v'^2[/tex]

K' = m'v'²

Since both kinetic energies are the same, since the drop from the same height,

K = K'

[tex]\dfrac{3mv^2}{4 }= m'v'^2[/tex]

[tex]\dfrac{v^2}{v'^2} =\dfrac{ 4m}{3m'}[/tex]

[tex]\dfrac{v^2}{v'^2} = \dfrac{4}{3}(\dfrac{m}{m'})[/tex]

[tex]\dfrac{v}{v'} = \sqrt{[\dfrac{4}{3}(\dfrac{m}{m'})][/tex]

Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.

Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.

(b) Since the kinetic energy, K = translational + rotational

We find the translational kinetic energy of each object.

For the wooden cylinder,

[tex]K = K_o + \dfrac{1}{2}Iw^2[/tex]

where

K₀ = translational kinetic energy of wooden cylinder

[tex]K - \dfrac{1}{2}Iw^2 = K_o[/tex]

[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{2}(\dfrac{mr^2}{2})(\dfrac{v}{r})^2 = K_a[/tex]

[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{4}mv^2 = K_o[/tex]

[tex]K_o = \dfrac{1}{2}mv^2[/tex]

For the steel hoop,

[tex]K' = K_1 + \dfrac{1}{2}I'w'^2[/tex]

where

K₁ = translational kinetic energy of steel hoop

[tex]K' - \dfrac{1}{2}I'w'^2 = K_1[/tex]

[tex]m'v'^2 - \dfrac{1}{2}(m'r'^2)(\dfrac{v'}{r'})^2 = K_1[/tex]

[tex]m'v'^2 - \dfrac{1}{2}m'v'^2 = K_1[/tex]

[tex]K_1= \dfrac{1}{2}m'v'^2[/tex]

So, K₀/K₁ =  1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².

Since (m/m') < 1 ⇒  (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁  < 1.33 ⇒ K₀ > K₁

So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.

So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.

To know more about Kinetic energy follow

https://brainly.com/question/25959744

Answer:

[tex]24.9\ \text{m/s}[/tex]

[tex]206.7\ \text{m}[/tex]

Explanation:

m = Mass of skydiver = 80 kg

[tex]x_1[/tex] = Height for which the parachute is closed = 1000-200 = 800 m

[tex]x_2[/tex] = Height for which the parachute is open = 200 m

[tex]f_1[/tex] = Resistive force when parachute is closed = 50 N

[tex]f_2[/tex] = Resistive force when parachute is open = 3600 N

v = Velocity of skydiver on the ground

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

h = Height from which the skydiver jumps = 1000 m

The energy balance of the system will be

[tex]mgh-f_1x_1-f_2x_2=\dfrac{1}{2}mv^2\\\Rightarrow 80\times 9.81\times 1000-50\times 800-3600\times 200=\dfrac{1}{2}\times 80\times v^2\\\Rightarrow v=\sqrt{\dfrac{2(80\times 9.81\times 1000-50\times 800-3600\times 200)}{80}}\\\Rightarrow v=24.9\ \text{m/s}[/tex]

The velocity fo the skydiver when he lands will be [tex]24.9\ \text{m/s}[/tex]

x = Height where the person opens the parachute

v = 5 m/s

[tex]mgh-f_1x_1-f_2x_2=\dfrac{1}{2}mv^2\\\Rightarrow 80\times 9.81\times 1000-50\times (1000-x)-3600\times x=\dfrac{1}{2}\times 80\times 5^2\\\Rightarrow 80\times 9.81\times 1000-50000+50x-3600x=\dfrac{1}{2}\times 80\times 5^2\\\Rightarrow x=\dfrac{80\times 9.81\times 1000-50000-\dfrac{1}{2}\times 80\times 5^2}{3550}\\\Rightarrow x=206.7\ \text{m}[/tex]

The height at which the parachute is to be opened is [tex]206.7\ \text{m}[/tex]

Answer:

Your answer is B

because it's on sneel's law.

that is sin of incident ray / sin of refracted ray is refractive index

Answer:

como  ías

Explanation:

Explanation:

mass, m = 5kg

initial velocity, u = 16m/s

final velocuty, v = -22m/s

change in momentum, ∆p = ?

∆p = m (v-u)

5(-22-16)

5(38)

∆p = 190kgm/s

check the calculations!

Answer:

B. it increases

Explanation:

As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).

Answer:

B is the correct answer.

Explanation:

I hope this will help you. Here is the link  http://web.physics.ucsb.edu/~phys6c/summer2006/hw4solutions

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]

Answer:

Explanation:

angle of incidence.

Answer:

ultraviolet light

plz mark me as brainliest.

Answer:

Ultra violet

Explanation:

Answer: 2.67

Explanation: it said he went from 0 to 8 in 3 seconds so if we divide eight By three we get 2.67 rounded to the nearest hundredth so you accelerated that 2.67 m/s

Answer:

3.54

Explanation:

some nerd thing I found it on Yahoo answers

Answer:

3.54º

Explanation:

Find the blue θ first

sin⁻¹(540x10⁻⁹/2.88x10⁻⁶)=8.99°

Then find the orange θ

sin⁻¹(625x10⁻⁹/2.88x10⁻⁶)=12.53°

Take the differences and subtract

12.53°－8.99°=3.54°

Answer:

The team must have a vast knowledge of thermodynamics

Explanation:

Just took the test!!!

Answer:

C. Thermodynamics

Explanation:

Answer:

the distance is 6.46 cm.

Explanation:

Given

current in the wire, I = 4.2 A

magnitude of the magnetic field, B = 1.3 x 10⁻⁵ T

The distance from the wire is determined by using Biot-Savart Law;

[tex]B = \frac{\mu_o I}{2\pi r} \\\\r = \frac{\mu_o I}{2\pi B}[/tex]

Where;

r is the distance from the wire where the magnetic field is experienced

[tex]r = \frac{\mu_o I}{2\pi B}\\\\r = \frac{4\pi \times 10^{-7} \times 4.2 }{2\pi \times 1.3 \times 10^{-5}}\\\\r = 0.0646 \ m\\\\r = 6.46 \ cm[/tex]

Therefore, the distance is 6.46 cm.

Answer:

The correct answer is "6666.67 N".

Explanation:

The given values are:

Mass,

m = 0.100

Relative speed,

v = 4.00 x 10³

time,

t = 6.00 x 10⁻⁸

As we know,

⇒  [tex]F=m(\frac{\Delta v}{\Delta t} )[/tex]

On substituting the given values, we get

⇒      [tex]=0.100\times 10^{-6}(\frac{4\times 10^3}{6\times 10^{-8}} )[/tex]

⇒      [tex]=6666.67 \ N[/tex]

Answer:

1) It seems that he would need the central gravitational force

   (the sun)

2) Also the planets would need to be included (orbits around the sun)

    Mercury, Venus, Earth, Mars, Jupiter, Saturn, etc.

3. Then, many of the planets have significant objects (moons) rotating about them.

Those would seem to be objects to be included in a model of the solar system.

1) He would need the central gravitational force (the sun)

2) The planets would need to be included: Mercury, Venus, Earth, Mars, Jupiter, Saturn, etc.

3) Many of the planets have specific moons rotating about them.

Larry should put the Earth between the planets Venus, and Mars.

Answer:

the magnitude of the magnetic force on the wire is 0.2298 N

Explanation:

Given the data in the question;

we know that, the magnitude of magnetic force is given as;

|F[tex]_{mg}^>[/tex] | = I([tex]B^>[/tex] × [tex]L^>[/tex] )

given that

I = 2.6 A

[tex]B^>[/tex] = 0.17

[tex]L^>[/tex] = 0.52

so we substitute

|F[tex]_{mg}^>[/tex] | = 2.6( 0.17i" × 0.52j" )

|F[tex]_{mg}^>[/tex] | = 0.2298 N

Therefore, the magnitude of the magnetic force on the wire is 0.2298 N

Answer:

You can breathe in too much carbon monoxide, which will eliminate the flow of oxygen to your bloodstream and can kill you.

Explanation:

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

I believe the answer is a

Answer:

not enought info

Explanation:

tbh I just know it's not 5 10 or 1

Answer:

B. 5 A

Explanation:

10/2= 5

Educere