Urgent!!!!! A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes. The data are shown in the table. Estimate the average temperature of the air in the room at 20 min. explain your answer.

Answer:

4 seconds

Explanation:

Assuming that the object started from rest,

v = at

--> t = v/a = (12 m/s) / (3 m/s^2)

= 4 seconds

Answer: C

Explanation:

USAtestprep

Answer:

b

Explanation:

f1=-f2 that could be thank u

Answer:

  K_{total} = 8.91 J

Explanation:

In this exercise you are asked to find the kinetic energy of the can of coca-cola

         K_total = K_ {Translation} + K_ {rotation}

the translational kinetic energy is

         K_ {translation} = ½ m v²

the kinetic energy of rotation is

         K_ {rotation} = ½ I w²

The moment of inertia of a cylinder is

           I = ½ m r²

we substitute

          K_ {total} = ½ m v² + ½ (½ m r²) w²

angular and linear velocity are related

          v = w r

we substitute

          K_ {total} = ½ m v² + ¼ m r² v² / r²

          K_ {total} = m v² (½ + ¼)

          K_ {total} = ¾ m v²

let's calculate

          K_ {total} = ¾ 0.33 6.00²

           K_{total} = 8.91 J

The answer is A. velocity .

Velocity is a vector quantity because it need magnitude as well as direction to be defined.

Answer: hetero i think i dont know

Explanation:

Answer:

When two substances that cannot dissolve each other are mixed, a mixture is formed.

i hope this helps a little bit.

Answer:

D

Explanation:

Answer:

 θ = 32.4º

Explanation:

For this exercise let's use Malus's law

         I = Io cos² θ

in this case it indicates that the incident intensity is 370 W/m², when the first polarization passes, only the radiation with the same polarization of the polarizer emerges, that is, vertical

         I₀ = 370/2 = 185 W / m²

this is the radiation that affects the second polarizer, let's apply the expression of Maluz

         θ = cos⁻¹ ([tex]\sqrt{\frac{I}{I_o} }[/tex])

         θ = cos⁻¹ ([tex]\sqrt{132/185}[/tex])

         θ = cos⁻¹ (0.844697)

         θ = 32.4º

Answer:

24 J

Explanation:

[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]

Question: Two people stand facing each other at a roller-skating rink then push off each other. If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity of the boy skater if his mass is 50 kg?

Answer:

3 m/s

Explanation:

Applying,

The Law of conservation of momentum

Momentum of the girl skater = momentum of the boy skater

MV = mv...................... Equation 1

Where M = mass of the girl skater, V = velocity of the girl skater, m = mass of the boy skater, v = velocity of the boy skater

From the question, we were asked to calculate v

v = MV/m.................. Equation 1

Given: M = 30 kg, V = 5 m/s, m = 50 kg

Substitute these values into equation 1

v = (30×5)/50

v = 3 m/s

Hence the velocity of the the boy skater is 3m/s

Answer:

Thrust developed = 212.3373 kN

Explanation:

Assuming the ship is stationary

Determine the Thrust developed

power supplied to the propeller ( Punit ) = 1900 KW

Duct distance ( diameter ; D  ) = 2.6 m

first step : calculate the area of the duct

A = π/4 * D^2

   =  π/4 * ( 2.6)^2  = 5.3092 m^2

next : calculate the velocity of propeller

Punit = (A*v*β ) / 2  * V^2     ( assuming β = 999 kg/m^3 ) also given V1 = 0

∴V^3 = Punit * 2 / A*β

         = ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )

hence V2 = 8.9480 m/s

Finally determine the thrust developed

F = Punit / V2

  = (1900 * 10^3) / ( 8.9480)

  = 212.3373 kN

Answer:

D th

Explanation:

D B. The force applied to the ball is a balanced force.

Answer:

Ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California

Explanation:

If increasing the Diameter of a Telescope beyond a given value will increase the ability of the telescope to capture more light and also capture astronomical objects located in a very distant position without improving resolution.

Hence the superiority of Keck telescope atop Mauna Kea over Hale Telescope atop Palomar mountain in California is the ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California

Answer:

magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

Explanation:

Given the data in the question;

Speed of carousel N = 24 rpm

From the diagram below, selected path direction defines the Axis of slip.

Hence, The Coriolis is acting along the axis of transmission

Now, we determine the angular speed ω of the carousel.

ω = 2πN / 60

we substitute in the value of N

ω = (2π × 24) / 60

ω = 2.5133 rad/s

Next, we convert the given velocity from mph to ft/s

we know that; 1 mph = 1.4667 ft/s

so

[tex]V_{slip[/tex] = 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s

Now, we determine the magnitude of the Coriolis acceleration

[tex]a_c[/tex] = 2( [tex]V_{slip[/tex] × ω )

we substitute

[tex]a_c[/tex] = 2( 8.8002 ft/s × 2.5133 rad/s )

[tex]a_c[/tex] = 44.235 ft/s²

Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

Answer:

i dont know but i should know try g o o g l e

Explanation:

Answer:

the answer is false.

Explanation:

i took the test and it is false trust me!!!!!!!!!

Answer: I think B.)

Explanation:

Answer:

Surface tension is,  the surface where the water meets the air, water molecules cling even more tightly to each other.

Accelerating objects are changing their velocity. Velocity is often thought of as an object's speed with a direction. Thus, objects which are accelerating are either changing their speed or changing their direction. They are either speeding up, slowing down or changing directions. Changing the velocity in any one of these three ways would be an example of an accelerated motion.

Answer:

The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].

Explanation:

The work done is given as in terms of

[tex]W=\Delta TE[/tex]

Where ΔTE is the change in total energy.

This is given as

[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]

Rearranging it in  terms of K_x gives

[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]

Answer:

You get Day and Night

It takes 24 hour

Answer:

Explanation:

The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.

This question is incomplete, the complete question is;

A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.

Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.

Hint : ΔV = Ed

Answer:

the required potential difference, between the capacitor plates is 600 V

Explanation:

Given the data in the question;

B = 0.60 T

d = 2.0 mm = 0.002 m

v = 5.0 × 10⁵ m/s.

since particle pass straight through without deflection.

F[tex]_{net[/tex] = 0

so, F[tex]_E[/tex] = F[tex]_B[/tex]

qE = qvB

divide both sides by q

E = vB

we substitute

E = (5.0 × 10⁵) × 0.6

E = 300000 N/C

given that; potential difference ΔV = Ed

we substitute

ΔV = 300000 × 0.002

ΔV = 600 V

Therefore, the required potential difference, between the capacitor plates is 600 V

Answer:

a)  P = 807.85 N,  b)  P = 392.15 N,  c)  P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

         sin θ = Wₓ / W

         cos θ = W_y / W

         Wₓ = W sin θ

         W_y = W cos θ

         Wₓ = 1200 sin 30 = 600 N

          W_y = 1200 cos 30 = 1039.23 N

Y axis  

      N- W_y = 0

      N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

       P -Wₓ -fr = 0

       P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

      fr = μ N

      fr = 0.20 1039.23

      fr = 207.85 N

we substitute

       P = 600+ 207.85

       P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

        P + fr -Wx = 0

       fr = μ N

       fr = 0.20 1039.23

        fr = 207.85 N

we substitute

        P =  Wₓ -fr

        P = 600 - 207,846

        P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

         P - Wₓ + fr = 0

         P = Wₓ - fr

         fr = 0.15 1039.23

         fr = 155.88 N

         P = 600 - 155.88

         P = 444.12 N

Answer:

It will go up to 93.75 m before it is moving at 20 m/s

Explanation:

As we know that

[tex]v^2 - u^2 = 2aS[/tex]

here v is the final speed i.e 20 m/s

u is the initial speed i.e 5 m/s

a is the acceleration due to gravity i.e 2 m/s^2

Substituting the given values in above equation, we get -

[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters

Answer:

[tex]523269.9\ \text{N/m}[/tex]

Explanation:

q = Charge

r = Distance

[tex]q_1=25\ \text{C}[/tex]

[tex]r_1=3000\ \text{m}[/tex]

[tex]q_2=40\ \text{C}[/tex]

[tex]r_2=850\ \text{m}[/tex]

The electric field is given by

[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]

The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]

Answer:

Truck [tex]\dfrac{g}{10}[/tex]

Road [tex]-\dfrac{g}{10}[/tex]

Explanation:

[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]

[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]

Frictional force is given by

[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]

Net acceleration is given by

[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]

The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.

Answer:

[tex]10.54\ \text{rad/s}[/tex]

Explanation:

[tex]\omega_i[/tex] = Initial angular velocity = 500 rpm = [tex]500\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]

[tex]\omega_f[/tex] = Final angular velocity

t = Time = 34 s

[tex]\theta[/tex] = Angular displacement = 170 revs = [tex]170\times 2\pi\ \text{rad}[/tex]

[tex]\alpha[/tex] = Angulr acceleration

From the kinematic equations of angular motion we have

[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \alpha=\dfrac{\theta-\omega_it}{\dfrac{1}{2}t^2}\\\Rightarrow \alpha=\dfrac{170\times 2\pi-500\times \dfrac{2\pi}{60}\times 34}{\dfrac{1}{2}\times 34^2}\\\Rightarrow \alpha=-1.23\ \text{rad/s}^2[/tex]

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=500\times \dfrac{2\pi}{60}+(-1.23)\times 34\\\Rightarrow \omega_f=10.54\ \text{rad/s}[/tex]

The rate at which the wheel is spinning when the power comes back on is [tex]10.54\ \text{rad/s}[/tex].

Answer:

the decay of uranium ending in lead,  of potassium (40K) that becomes argon,  the decay of rubidium

Explanation:

For the radioactive dating process, a material is needed that has a known average life time and that we can know the amount of material at a given moment,

In the case of carbon 14 (14C), living beings stop capturing it from the air and plants when they die, so knowing the amount they currently have, it is possible to calculate the time in which they stopped absorbing, but the life time average is 5730 years, the maximum time that can be used is up to about 10 average visa times

To analyze extra samples have high half-life times

* the decay of uranium ending in lead

* the decay of potassium (40K) that becomes argon T1 / 2 = 1,251 10⁹ years

* the decay of rubidium (87Ru) which becomes strontium T1 / 2 = 4.92 10¹⁰ years