True or False
Microscopic organisms grow on the rocks near a volcano if the rocks are cooled to 120 degrees Celsius or less.

Answer:

less than one lightyear=d

Explanation:

I took the test.:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D::):):):):):):):):):):):):):):):):):):):):):)

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:

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:

not enought info

Explanation:

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

Answer:

B. 5 A

Explanation:

10/2= 5

Educere

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

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:

Answer:

im sorry i would help but thats too much

Answer: The half-life is the amount of time it takes for a given isotope to lose half of its radioactivity. If a radioisotope has a half-life of 14 days, half of its atoms will have decayed within 14 days. In 14 more days, half of that remaining half will decay, and so on.

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²  

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²  

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

Answer:

ultraviolet light

plz mark me as brainliest.

Answer:

Ultra violet

Explanation:

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:

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.

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!

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:

the electric field strength inside the resistor is 2.57 V/m

Explanation:

Given;

current flowing through the wire, I = 1.10 A

resistance of the wire, R = 7.00 Ω

length of the wire, L = 3.00 m

The emf created inside the resistor is calculated as;

V = IR

V = 1.10 x 7

V = 7.7 V

The electric field strength inside the resistor is calculated as;

E = V/L

E = 7.7 / 3

E = 2.57 V/m

Therefore, the electric field strength inside the resistor is 2.57 V/m

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 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: [tex]240\ rad/s^2[/tex]

Explanation:

Given

Length of beam [tex]l=2\ m[/tex]

mass of beam [tex]m=5\ kg[/tex]

Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude

[tex]\tau =F\times l=200\times 2=400\ N.m[/tex]

Also, the beam starts rotating about its center

So, the moment of inertia of the beam is

[tex]I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2[/tex]

Torque is the product of moment of inertia and angular acceleration

[tex]\Rightarrow \tau=I\alpha\\\\\Rightarrow 400=\dfrac{5}{3}\times \alpha\\\\\Rightarrow \alpha =240\ rad/s^2[/tex]

Answer:

The team must have a vast knowledge of thermodynamics

Explanation:

Just took the test!!!

Answer:

C. Thermodynamics

Explanation:

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: 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:

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 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:

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:

1.72 m

Explanation:

Potential energy = mgh, where m is mass, g is acceleration due to gravity (9.8), and h is height

76 = (3.5)(9.8)h

76=44.1h

h=1.72335600907 ≈1.72 m

Answer:

:r

Explanation:r

Tobnbv346468this Ishmael

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

Answer:

  Λ = 5.14 10⁻⁴ m

Explanation:

This is a double slit experiment, which for the case of constructive interference

          d sin θ = m λ

let's use trigonometry

         tan θ = y / L

as the angles are very small

         tan θ = [tex]\frac{sin \theta}{cos \theta}[/tex] = sin θ

         sin θ = y / L

we substitute

         d y / L = m λ

        y = m λ L / d

we calculate for the interference of order m = 5

         y = 5  527 10⁻⁹  1.54/0.102 10⁻³

         y = 3.978 10⁻² m

Now we can find the difference in length between the two rays, that of the central maximum and this

let's use the Pythagorean theorem

           L’= [tex]\sqrt{L^2 +y^2}[/tex]

           L ’= [tex]\sqrt{1.54^2 +(3.978 \ 10^{-2})^2 }[/tex]

           L ’= 1.54051 m

optical path difference

          Λ = L’- L

          Λ = 1.54051 - 1.54

          Λ = 5.14 10⁻⁴ m

Answer:

it is safe to stand at the end of the table

Explanation:

For this exercise we use the rotational equilibrium condition

         Στ = 0

         W x₁ - w x₂ - w_table x₃ = 0

         M x₁ - m x₂ - m_table x₃ = 0

where the mass of the large rock is M = 380 kg and its distance to the pivot point x₁ = 850 cm = 0.85m

the mass of the man is 62 kg and the distance

            x₂ = 4.5 - 0.85

            x₂ = 3.65 m

the mass of the table (m_table = 22 kg) is at its geometric center

            x_{cm} = L/2 = 2.25 m

            x₃ = 2.25 -0.85

            x₃ = 1.4 m

let's look for the maximum mass of man

            m_{maximum} = [tex]\frac{ M x_1 -m_{table} x_3}{ x_2}[/tex]

let's calculate

             m_{maximum} = [tex]\frac{ 380 \ 0.85 - 22 \ 1.4}{3.65}[/tex](380 0.85 - 22 1.4) / 3.65

             m_{maximum} = 80 kg

we can see that the maximum mass that the board supports without turning is greater than the mass of man

             m_{maximum}> m

consequently it is safe to stand at the end of the table