V = u + at where u is initial velocity (15 m/s), a is acceleration (2m/s^2) and t is time (15 seconds)
V = 15 + 2 X 15
V = 45 m/s
We'll look at two properties:
1. The variation in temperature
2. The material's heat transfer coefficient
By taking an example;
Use a circular rod made of a certain material (for example, steel) that is insulated all the way around.
One end of the rod is immersed in a huge reservoir of 100°C water, while the other is immersed in water at 40°C. The cold water is kept in an insulated cylinder on both sides. The temp of the chilly water is measured using a meter as a time - dependent.
Conclusion of experiment;
- Heat is transferred from a hot location to a cooler region.
- Whenever heat is applied to a body, its thermal power rises, and its temperature rises.
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Answer:
12m/s
Explanation:
Given parameters:
Power = 6.5 x 10⁴W
Force = 5.5 x 10³N
Unknown:
The resulting velocity = ?
Solution:
The velocity of a body is related to force and power using the expression below;
Power = Force x velocity
Insert the parameters and solve for velocity
6.5 x 10⁴ = 5.5 x 10³ x velocity
velocity = = 12m/s
λ=v/f
λ-wavelength
v-speed
f-frequency
we have the wavelength(6.2 x 10^-6meters) and we use the speed of light which is equal to 3*10^8m/s
6.2*10^-6m=3*10^8m/s/f
f=(3*10^8m/s)/(6.2*10^-6)≈0.48*10^14Hz
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 × m
dynamic viscosity = 1.75 × Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ
so
= µ ............1
put here value
= 1.75× ×
= 0.175 v
and
area between air and puck is given by
Area =
area =
area = 7.85 × m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 ×
force = 1.374 × v
and now apply newton second law
force = mass × acceleration
- force =
- 1.374 × v =
t =
time = 2.18
so time required after impact for a puck is 2.18 seconds