Answer:
4
Explanation:
We know that intensity I = P/A where P = power and A = area through which the power passes through.
Now, let the initial intensity of the speaker be I₀ and its initial power be P₀. Since the intensity is increased by a factor of 4, the new intensity be I and new power be P.
So, I = P/A and I₀ = P₀/A
Now, if I = 4I₀,
P/A = 4P₀/A
P = 4P₀
Now, energy E = Pt, where t = time. So, P = E/t and P₀ = E₀/t
Substituting P and P₀ into the equation, we have
P = 4P₀
E/t = 4E₀/t
E = 4E₀
Since the energy is four times the initial energy, the energy output increases by a factor of 4.
Answer:
P = 4000 [Pa]
Explanation:
Pressure is defined as the relationship between Force and the area where the body rests.
The support area is equal to:
![A=50*20=1000[cm^{2} ]](https://tex.z-dn.net/?f=A%3D50%2A20%3D1000%5Bcm%5E%7B2%7D%20%5D)
But we must convert from square centimeters to square meters.
![1000[cm^{2}]*\frac{1^{2}m^{2} }{100^{2}m^{2} }=0.1[m^{2} ]](https://tex.z-dn.net/?f=1000%5Bcm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7Dm%5E%7B2%7D%20%20%7D%7B100%5E%7B2%7Dm%5E%7B2%7D%20%20%7D%3D0.1%5Bm%5E%7B2%7D%20%5D)
And the pressure is:
![P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BF%7D%7BA%7D%20%5C%5CP%3D400%2F0.1%5C%5CP%3D4000%5BN%2Fm%5E%7B2%7D%20%5Dor%204000%5BPa%5D)
Answer:
Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.
Explanation:
From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.
Answer:
a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)
b) The acceleration of point A is 3.25 m/s²
The acceleration of point E is 0.75 m/s²
Explanation:
a) The relative acceleration of B with respect to D is equal:

Where
aB = absolute acceleration of point B = 2.5 j (m/s²)
aD = absolute acceleration of point D = 1.5 j (m/s²)
(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam
(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)


We have that
(aB/D)t = BDα
Where α = acceleration of the beam
BDα = 1 m/s²
Where
BD = 2

b) The acceleration of point A is:

(aA/D)t = ADαj

The acceleration of point E is:
(aE/D)t = -EDαj
