Avg velocity definition | Problem Average velocity vectors…
1 answer:
Answer:
point in the same direction as the change in position.
require knowing the initial and final position, along with the elapsed time.
Explanation:
As we know that average velocity is defined as the ratio of change in position and the time interval of the object
It is given as
so we will have
= displacement
= time interval
the direction is this average velocity is always along the direction of the displacement
so correct answer will be
point in the same direction as the change in position.
require knowing the initial and final position, along with the elapsed time.
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Answer:
a) I = 3.63 W / m² , b) I = 0.750 W / m²
Explanation:
The intensity of a sound wave is given by the relation
I = P / A = ½ ρ v (2π f )²
I = (½ ρ v 4π² s_{max}²) f²
a) with the initial condition let's call the intensity Io
cte = (½ ρ v 4π² s_{max}²)
I₀ = cte s² f₀²
I₀ = cte 10 6
If frequency is increase f = 2.20 10³ Hz
I = constant (2.20 10³) 2
I = cte 4.84 10⁶
let's find the relationship of the two quantities
I / Io = 4.84
I = 4.84 Io
I = 4.84 0.750
I = 3.63 W / m²
b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or
I = cte (f s)²
I = constant (0.250 10³ 4)²
I = cte 1 10⁶
the relationship
I / Io = 1
I = Io
I = 0.750 W / m²
Answer:
the block reaches higher than the sphere
\frac{y_{sphere}} {y_block} = 5/7
Explanation:
We are going to solve this interesting problem
A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp
Let's use the concept of conservation of energy
starting point. At the top of the ramp
Em₀ = U = m g y₁
final point. At the exit of the ramp
Em_f = K + U = ½ m v² + ½ I w² + m g y₂
notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂
energy is conserved
Em₀ = Em_f
m g y₁ = ½ m v² + ½ I w² + m g y₂
angular and linear velocity are related
v = w r
the moment of inertia of a sphere is
I = m r²
we substitute
m g (y₁ - y₂) = ½ m v² + ½ ( m r²) (
)²
m g h = ½ m v² (1 + )
where h is the difference in height between the two sides of the ramp
h = y₂ -y₁
mg h = (
m v²)
v = √5/7 √2gh
This is the exit velocity of the vertical movement of the sphere
v_sphere = 0.845 √2gh
B) is the same case, but for a box without friction
starting point
Em₀ = U = mg y₁
final point
Em_f = K + U = ½ m v² + m g y₂
Em₀ = Em_f
mg y₁ = ½ m v² + m g y₂
m g (y₁ -y₂) = ½ m v²
v = √2gh
this is the speed of the box
v_box = √2gh
to know which body reaches higher in the air we can use the kinematic relations
v² = v₀² - 2 g y
at the highest point v = 0
y = vo₀²/ 2g
for the sphere
y_sphere = 5/7 2gh / 2g
y_esfera = 5/7 h
for the block
y_block = 2gh / 2g
y_block = h
therefore the block reaches higher than the sphere
Answer:
distance is 13 m for 100 dB
distance is 409 km for 10 dB
Explanation:
Given data
distance r = 2.30 m
source β = 115 dB
to find out
distance at sound level 100 dB and 10 dB
solution
first we calculate here power and intensity and with this power and intensity we will find distance
we know sound level β = 10 log(I/) ......................a
put here value (I/) = 10^−12 W/m² and β = 115
115 = 10 log(I/10^−12)
so
I = 0.316228 W/m²
and we know power = intensity × 4π r² ...............b
power = 0.316228 × 4π (2.30)²
power = 21.021604 W
we know at 100 dB intensity is 0.01 W/m²
so by equation b
power = intensity × 4π r²
21.021604 = 0.01 × 4π r²
so by solving r
r = 12.933855 m = 13 m
distance is 13 m
and
at 10 dB intensity is 1 × 10^–11 W/m²
so by equation b
power = intensity × 4π r²
21.021604 = 1 × 10^–11 × 4π r²
by solving r we get
r = 409004.412465 m = 409 km
Answer:
α = 0.0135 rad/s²
Explanation:
given,
t = 133 min = 133 x 60 = 7980 s
angular speed varies from 570 rpm to 1600 rpm
now,
570 rpm =
= 59.69 rad/s
1600 rpm = =
= 167.6 rad/s
using equation of rotational motion
ωf = ωi + αt
167.6 = 59.7 + α x 7980
α x 7980 = 107.9
α = 0.0135 rad/s²