Answer:
a) A = 500 N/s²
b) I = 5812.50 N-s
c) Δv = 2.7034 m/s
Explanation:
Given info
m = 2150 kg
F(t) = At²
F(1.25 s) = 781.25 N
a) A = ?
We use the equation
F(t) = At² ⇒ 781.25 N = A*(1.25 s)²
⇒ A = F(t) / t² = (781.25 N) / (1.25 s)²
⇒ A = 500 N/s²
b) I = ? if 2.00 s ≤ t ≤ 3.50 s
we apply the equation
I = ∫F(t) dt = ∫At² dt = A ∫t² dt = (500/3)*t³ + C
Since the limits of integration are 2 and 3.5, we obtain
I = (500/3)*((3.5)³-(2)³) = 5812.50 N-s
c) Δv = ?
we can apply the equation
I = m*Δv ⇒ Δv = I / m
⇒ Δv = 5812.50 N-s / 2150 kg
⇒ Δv = 2.7034 m/s
Answer:
Propagation of an Electromagnetic Wave. Electromagnetic waves are waves which can travel through the vacuum of outer space. Mechanical waves, unlike electromagnetic waves, require the presence of a material medium in order to transport their energy from one location to another.
Answer: C = Q/4πR
Explanation:
Volume(V) of a sphere = 4πr^3
Charge within a small volume 'dV' is given by:
dq = ρ(r)dV
ρ(r) = C/r^2
Volume(V) of a sphere = 4/3(πr^3)
dV/dr = (4/3)×3πr^2
dV = 4πr^2dr
Therefore,
dq = ρ(r)dV ; dq =ρ(r)4πr^2dr
dq = C/r^2[4πr^2dr]
dq = 4Cπdr
FOR TOTAL CHANGE 'Q', we integrate dq
∫dq = ∫4Cπdr at r = R and r = 0
∫4Cπdr = 4Cπr
Q = 4Cπ(R - 0)
Q = 4CπR - 0
Q = 4CπR
C = Q/4πR
The value of C in terms of Q and R is [Q/4πR]
A uniform metal rod with of length 80cm and a mass of 3.2kg is supported horizontally by two vertical spring balances C and D. Balance C is 20cm from one end while D is 30cm from the other end would show the reading of 1.06 Kg and 2.13 kg respectively
<h3>What is gravity?</h3>
It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another, The gravity varies according to the mass and size of the body for example the force of gravity on the moon is the 1/6th times of the force of gravity on the earth.
As given in the problem, A uniform metal rod of the length of 80cm and mass of 3.2kg is supported horizontally by two vertical springs balance C and D. Balance C is 20cm from one end while D is 30cm from the other end
The weight of the rod acting downward is from the center of the rod at 40 cm
Let us suppose the reading on the spring balance C and D are P and Q respectively
By using the equilibrium for the vertical force
Fv=0
P + C = 3.2
By using the equilibrium for the moment around the left corner
20×P+ 50×Q= 40 ×3.2
By solving for both P and Q from the above two equations we would get
P =1.06 and Q = 2.13
Thus, the reading on the spring balance C and D would be 1.06 Kg and 2.13 kg respectively
Learn more about gravity from here
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