We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.
Where,
- F is force
- m is mass
- a is acceleration
In our case,
- F = ?
- m = 2500 kg
- a = 20m/s
<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>
Answer:
3 times louder
Explanation:
The Loudness in decibel Db L = 10㏒(I/I₀) where I = sound intensity level and I₀ = threshold of hearing = 10⁻¹² W/m².
Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹
and I₂ = sound intensity level of Braylee's music = 10⁻³
So, substituting the variables into the equation, we have
L₁ = 10㏒(I₁/I₀)
L₁ = 10㏒(10⁻⁹/10⁻¹²)
L₁ = 10㏒(10³)
L₁ = 3 × 10㏒10
L₁ = 30㏒10
L₁ = 30 dB
Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³
So, substituting the variables into the equation, we have
L₂ = 10㏒(I₁/I₀)
L₂ = 10㏒(10⁻³/10⁻¹²)
L₂ = 10㏒(10⁹)
L₂ = 9 × 10㏒10
L₂ =90㏒10
L₂ = 90 dB
So, the number of times Braylee's music is louder than Jessica's music is L₂/L₁ = 90 dB/30 dB = 3
So, Braylee's music is 3 times louder than Jessica's music
Answer: The result of "the upper bound of the density" does not go on the denominator.
So simplified, no. The answer is no.
Answer:
W = 0.060 J
v_2 = 0.18 m/s
Explanation:
solution:
for the spring:
W = 1/2*k*x_1^2 - 1/2*k*x_2^2
x_1 = -0.025 m and x_2 = 0
W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2
W = 0.060 J
the work-energy theorem,
W_tot = K_2 - K_1 = ΔK
with K = 1/2*m*v^2
v_2 = √2*W/m
v_2 = 0.18 m/s
The change in velocity is +4 m/s to the right (or -4 m/s to the left).
The object's mass is irrelevant.
