A ball is dropped from rest out of a high window in a tall building for 5 seconds. Assuming the we ignore air resistance and assume upwards to be positive. A) What will be the final velocity of the ball B) What is the height of the building if it hits the ground after those 5 seconds. *​

Answer:

I think it might be 8kg grams because it is bigger

Answer:

Explanation:

Equivalent resistance of parallel combination of two resistors of 3 ohms and 4 ohms .

R₁ = 3 x 4 / ( 3 + 4 )

= 12 / 7 = 1.7 ohms .

This equivalent resistance is connected with 4 ohms in series

Total resistance = 1.7 ohms + 4 ohms

= 5.7 ohms.

Hence effective resistance of the circuit = 5.7 ohms .

Answer:

3.504 MeV

Explanation:

Given that;[tex]\frac{A}{Z} X ---->\frac{A-4}{Z-2} Y +\alpha + Q[/tex]

Also;

[tex]Q= KE_{\alpha } (M_{Y} + M_{\alpha } /M_{Y[/tex])

Mass number of X = 229

Mass number of Y = 225

Mass number of alpha particles = 4

Kinetic energy of alpha particles = 3.443 MeV

Q = 3.443 MeV (225 + 4/225)

Q= 3.504 MeV

Answer: A roller coaster does not have an engine to generate energy. The climb up the first hill is accomplished by a lift or cable that pulls the train up. This builds up a supply of potential energy that will be used to go down the hill as the train is pulled by gravity

Hope this helps! Good luck with future homework and exams!

Answer:

The uniy which is accepted all over the world is called SI unit.

Explanation:

The system of measurement that is agreed by the international convention if scientists that is held in paris of France to adopt an international unit is called SI unit unit.

Answer:

=>8x+9x+5+3

=>8x+9x+5+3=>17x+8

Answer:

60 °C

Explanation:

From the question given above, the following data were obtained:

Mass of 1st sample (M₁) = 200 g

Temperature of 1st sample (T₁) = 80 °C

Mass of 2nd sample (M₂) = 100 g

Temperature of 2nd sample (T₂) = 20 °C

Equilibrium temperature (Tₑ) =?

NOTE: Since the sample are the same, the specific heat capacity is constant.

We can obtain the equilibrium temperature as follow:

Heat lost by 1st = heat gained by the 2nd

M₁C(T₁ – Tₑ) = M₂C(Tₑ – T₂)

Cancel out C

M₁(T₁ – Tₑ) = M₂(Tₑ – T₂)

200 (80 – Tₑ) = 100 (Tₑ – 20)

Clear bracket

16000 – 200Tₑ = 100Tₑ – 2000

Collect like terms

16000 + 2000 = 100Tₑ + 200Tₑ

18000 = 300Tₑ

Divide both side by 300

Tₑ = 18000 / 300

Tₑ = 60 °C

Therefore, the equilibrium temperature (i.e the final temperature) of the mixture is 60 °C.

Answer:

28Nm

Explanation:

Torque is expressed as the prduct of force and radius

Given

Force = 140N

radius  = 0.20m

Torque = 140 * 0.2

Torque = 140 * 1/5

Torque = 28Nm

Hence the amouunt of torque applied is 28Nm

Answer:

The larger the amplitude of the waves, the louder the sound. Pitch (frequency) – shown by the spacing of the waves displayed. The closer together the waves are, the higher the pitch of the sound.

Explanation:

Answer:

Explanation:

That is a fun question!

Without a hot radiating core like Earth, Mars will have very different geological and seismic events. The Mars core will be relatively cold and there will not be any molten magma. So Mars will not have earthquakes or volcano activities. Both only occur when there is magma flowing or tectonic plate motion and they will not occur with a cold core.

Answer:

Waves are disorders that travel through a fluid medium. Several simple wave features include frequency, period, wavelength, and amplitude.

Explanation:

Answer:

   i = 1.09 10⁴ A

Explanation:

For this exercise we will look for the resistance of the wire

          R = ρ L / a

the area of ​​the wire is

          a =ππ r² = π πd² / 4

we substitute

         R = ρ L 4 / π d²

          R = 1.72 10⁻⁸   20  4/π 8 10⁻⁴

         R = 5.47 10⁻⁴ Ω

to calculate the current we use ohm's law

           V = R i

            i = V / R

            i = 6 / 5.47 10⁻⁴

            i = 1.09 10⁴ A

Answer:

a.  9.99625 cm b. 68 °C

Explanation:

(a) What is the length of the rod at the freezing point of water (0 0C)?

Before we find the length of the rod, we need to find the coefficient of linear expansion, α = (L - L₀)/[L₀(T - T₀)] where L₀ = length of rod at temperature T₀ = 10.0 cm, T₀ = 20 °C, L = length of rod at temperature T = 10.015 cm and T = 100 °C

Substituting the values of the variables into the equation, we have

α = (L - L₀)/[L₀(T - T₀)]

α = (10.015 cm - 10.0 cm)/[10.0 cm(100 °C - 20 °C)]

α = 0.015 cm/[10.0 cm × 80 °C]

α = 0.015 cm/[800.0 cm °C]

α = 0.00001875 /°C

We now find the length L₁ at T₁ = 0 °C from

L₁ = L₀(1 + α(T₁ - T₀))

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

L₁ = L₀(1 + α(T₁ - T₀))

L₁ = 10.0 cm[1 +  0.00001875 /°C(0° C - 20 °C)]

L₁ = 10.0 cm[1 +  0.00001875 /°C × -20° C]

L₁ = 10.0 cm[1 - 0.000375]

L₁ = 10.0 cm[0.999625]

L₁ = 9.99625 cm

(b) What is the temperature if the length of the rod is 10.009 cm?

With length L₃ = 10.009 cm at temperature T₃, using

L₃ = L₀(1 + α(T₃ - T₀))

making T₃ subject of the formula, we have

L₃/L₀ = 1 + α(T₃ - T₀)

L₃/L₀ - 1 = α(T₃ - T₀)

T₃ - T₀ = (L₃/L₀ - 1)/α

T₃ = T₀ + (L₃/L₀ - 1)/α

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

T₃ = 20 °C + (10.009 cm/10.0 cm - 1)/0.00001875 /°C

T₃ = 20 °C + (1.0009 - 1)/0.00001875 /°C

T₃ = 20 °C + 0.0009/0.00001875 /°C

T₃ = 20 °C + 48 °C

T₃ = 68 °C