Answer:
buoyant force on the block due to the water= 10 N
Explanation:
We know that
buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.
Given:
A block of metal weighs 40 N in air and 30 N in water.
F_B = 40-30= 10 N
therefore, buoyant force on the block due to the water= 10 N
Answer:
F₁ = F₂ = F₃ = 0 N
Explanation:
given,
Arrow 1 mass = 80 g speed = 10 m/s
Arrow 2 mass = 80 g speed = 9 m/s
Arrow 3 mass = 90 g speed = 9 m/s
Horizontal Force:- F₁ , F₂ and F₃
There is no air resistance.
If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.
We know,
According to newton's second law
F = m a
If Acceleration is equal to zero
Then Force is also equal to zero.
Hence, F₁ = F₂ = F₃ = 0 N
Answer:
There is 148.35 Joules of heat is released in the process.
Explanation:
Given that,
Heat capacity of the object, 
Initial temperature, 
Final temperature, 
We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :
Let the mass of the object is 10 g or 0.01 kg
So,
Q = 148.35 Joules
So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.
Answer:
Explanation:
As we know by force equation that force along the inclined planed due to gravity is given as
so the acceleration due to gravity along the plane is given as
now we have
now we know that
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
