the car travels 34 mi in one hour.
then, in 6 hours car travels
34 x 6 mi
= 204 mi
Answer:
The weight of the girl = 1045.86 kg/m³
Explanation:
Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³.
From Archimedes principle,
R.d = Density of the person/Density of water = Weight of the person in air/Upthrust.
⇒ D₁/D₂ = W/U............................... Equation 1.
Where D₁ = Density of the person, D₂ = Density of water, W = Weight of the person in air, U = Upthrust in water.
Making D₁ the subject of the equation,
D₁ = D₂(W/U)................................... Equation 2
<em>Given: D₂ = 1000 kg/m³ , W = 509.45 N, U = lost in weight = weight in air - weight in water = 509.45 - 22.34 = 487.11 N</em>
<em>Substituting these values into equation 2</em>
D₁ = 1000(509.45/487.11)
D₁ = 1045.86 kg/m³
Thus the weight of the girl = 1045.86 kg/m³
<em></em>
Answer:
The speed of the block is 8.2 m/s
Explanation:
Given;
mass of block, m = 2.1 kg
height above the top of the spring, h = 5.5 m
First, we determine the spring constant based on the principle of conservation of potential energy
¹/₂Kx² = mg(h +x)
¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)
0.03125K = 118.335
K = 118.335 / 0.03125
K = 3786.72 N/m
Total energy stored in the block at rest is only potential energy given as:
E = U = mgh
U = 2.1 x 9.8 x 5.5 = 113.19 J
Work done in compressing the spring to 15.0 cm:
W = ¹/₂Kx² = ¹/₂ (3786.72)(0.15)² = 42.6 J
This is equal to elastic potential energy stored in the spring,
Then, kinetic energy of the spring is given as:
K.E = E - W
K.E = 113.19 J - 42.6 J
K.E = 70.59 J
To determine the speed of the block due to this energy:
KE = ¹/₂mv²
70.59 = ¹/₂ x 2.1 x v²
70.59 = 1.05v²
v² = 70.59 / 1.05
v² = 67.229
v = √67.229
v = 8.2 m/s
Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,
Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀ =A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values
b)
, where 
is time period of damped SHM
⇒
let 
be number of oscillations made
then, 
⇒
