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!
A bird in flight pushes itself upward with
a 7.28 N force. If the bird is climbing at a
constant rate of 1 m/s (no acceleration),
what is the weight of the bird?
[?] N
Answer:
The weight of the bird is equal to 7.28 N.
Explanation:
The upward force acting on the bird = 7.28 N
The bird is climbing at a constant rate of 1 m/s.
We need to find the weight of the bird.
We know that the weight of an object is the force of gravity acting on it. It can be calculated as follows :
W = mg
In this case, 7.28 N of force is acting on the object. Hence, the weight of the bird is equal to 7.28 N.
8× +5+9×+3 how can I solve this
Answer:
collect like terms then add=>8x+9x+5+3
=>8x+9x+5+3=>17x+8
La resistividad de cierto alambre es 1.72x10 -8 Ωm a 20 oC .Una batería de 6v se conecta a una bobina de 20 m de este alambre, con un diámetro de 0.8 mm. ¿Cuál es la corriente en el alambre?
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
Which object will take the most force
to accelerate? *
4 kg
6 kg
8 kg
02 kg
Answer:
I think it might be 8kg grams because it is bigger
A nucleus with mass number 229 emits a 3.443 MeV alpha particle. Calculate the disintegration energy Q for this process, taking the recoil energy of the residual nucleus into account.
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
Explain how frequency and amplitude affect sound vibrations?
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:
An aluminum-alloy rod has a length of 10.0 cm at 20°C and a length of 10.015 cm at the boiling point of water (1000C). (a) What is the length of the rod at the freezing point of water (0 0C)? (b) What is the temperature if the length of the rod is 10.009 cm? /4mks
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
What are the si units
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.
A (200 g) of water at (80 °C) is mixed with (100 g)of water at (20 °C). What is the final temperature of the water?
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.
You have just learned that the planet core of Mars is no longer radiating heat like the core of Earth. With this information what would you communicate to building designers as the lead geologist in regards to suitable building locations on the surface of Mars? Remember your informing builders about locations that would best avoid natural disasters like marsquakes and volcanoes? What evidence or theories do you have to support your statements?
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.