Answer:
depth of well is 163.30 m
Explanation:
Given data
speed of sound = 343 m/s
timer = 6.25 s
to find out
depth of well
solution
let us consider depth d
so equation will be
depth = 1/2 ×g ×t² ..............1
and
depth = velocity of sound × time .................2
here we have given time 6.25 that is sum of 2 time
when stone reach at bottom that time
another is sound reach us after stone strike on bottom
so time 1 + time 2 = 6.25 s
so from equation 1 and 2 we get
1/2 ×g ×t² = velocity of sound × time
1/2 ×9.8 × t1² = 343 × (6.25 - t1 )
t1 = 5.77376 sec
so height = 1/2 ×g ×t²
height = 1/2 ×9.8 × (5.773)²
height = 163.30 m
Answer:
a) θ₁ = 23.14 °
, b) θ₂ = 51.81 °
Explanation:
An address network is described by the expression
d sin θ = m λ
Where is the distance between lines, λ is the wavelength and m is the order of the spectrum
The distance between one lines, we can find used a rule of proportions
d = 1/600
d = 1.67 10⁻³ mm
d = 1-67 10⁻³ m
Let's calculate the angle
sin θ = m λ / d
θ = sin⁻¹ (m λ / d)
First order
θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)
θ₁ = sin⁻¹ (3.93 10⁻¹)
θ₁ = 23.14 °
Second order
θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)
θ₂ = sin⁻¹ (0.786)
θ₂ = 51.81 °
Answer:
A. Remove everything in the refrigerator to lighten the load.
B. Put a lubricant between the surface of the object and the floor
C. Use round objects, like pencils , to decrease the friction and push the refrigerator over the pencils more easily
Explanation:
Force of friction is a resistance force which acts between two surfaces which are in relative motion. Friction is both boon and bane. Due to friction, we are able to sit, walk etc but also, due to friction there is dissipation of energy. Friction can be reduced by applying lubricants, reducing contact area, reducing the load.
F = μN where N is the normal force which depends on the mass.
Thus, by reducing the load, force of friction can be reduced. Round objects like wheels can also be used. By this the contact area reduces.
Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
Larger molecules will move slower and smaller molecules will move faster. Did this answer your question?
