Answer:
The velocity of the truck after the collision is 20.93 m/s
Explanation:
It is given that,
Mass of car, m₁ = 1200 kg
Initial velocity of the car, 
Mass of truck, m₂ = 9000 kg
Initial velocity of the truck, 
After the collision, velocity of the car, 
Let
is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.




So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.
Answer:
The type of light and the material of lenz.
Explanation:
1) As the investigation is based on how the thickness of a lens effect the other variable. Thickness of the lenz is independent variable. So Lidia has to experiment with the different thicknesses in order to find the effect on dependent variable.
2) As the investigation is based to find the point where the beam of light is focused. It's a dependent variable and Lidia has no control over it. So the only thing she can do is to measure and observe how it respond to the changes in independent variable.
3) For conclusion, she has to make sure that the other variables are not effecting the output or results that is the beam point where the light is focused. So she must have to kept constant the type of light and material of lenz otherwise she won't be able to discriminate the effect of thickness of lenz from other causes.
Answer:
a) f = 615.2 Hz b) f = 307.6 Hz
Explanation:
The speed in a wave on a string is
v = √ T / μ
also the speed a wave must meet the relationship
v = λ f
Let's use these expressions in our problem, for the initial conditions
v = √ T₀ /μ
√ (T₀/ μ) = λ₀ f₀
now it indicates that the tension is doubled
T = 2T₀
√ (T /μ) = λ f
√( 2To /μ) = λ f
√2 √ T₀ /μ = λ f
we substitute
√2 (λ₀ f₀) = λ f
if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change
λ₀ = λ
f = f₀ √2
f = 435 √ 2
f = 615.2 Hz
b) The tension is cut in half
T = T₀ / 2
√ (T₀ / 2muy) = f = λ f
√ (T₀ / μ) 1 /√2 = λ f
fo / √2 = f
f = 435 / √2
f = 307.6 Hz
Traslate
La velocidad en una onda en una cuerda es
v = √ T/μ
ademas la velocidad una onda debe cumplir la relación
v= λ f
Usemos estas expresión en nuestro problema, para las condiciones iniciales
v= √ To/μ
√ ( T₀/μ) = λ₀ f₀
ahora nos indica que la tensión se duplica
T = 2T₀
√ ( T/μ) = λf
√ ) 2T₀/μ = λ f
√ 2 √ T₀/μ = λ f
substituimos
√2 ( λ₀ f₀) = λ f
si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia
λ₀ = λ
f = f₀ √2
f = 435 √2
f = 615,2 Hz
b) La tension se reduce a la mitad
T = T₀/2
RA ( T₀/2μ) = λ f
Ra(T₀/μ) 1/ra 2 = λ f
fo /√ 2 = f
f = 435/√2
f = 307,6 Hz
Answer:
The frequency of the signal is 2 GHz
Explanation:
Given;
period of the clock signal, T = 500 ps = 500 x 10⁻¹² s
the frequency of the signal is given by;

F = 2 GHz
Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz
The volume of the room is the product of its dimensions:

Now, from the equation

where d is the density, m is the mass and V is the volume, we deduce

So, multiply the density and the volume to get the mass of air in the room.