Answer: Anthony will be warmer after the game.
Explanation :
Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing a white shirt. Anthony will be warmer after the game. The black color is a good absorber of radiation and a bad reflector.
The black color absorbs heat until a thermal equilibrium is attained. So, it is advisable to wear cotton clothes in summers not dark colored clothes.
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
Answer:
The acceleration of the cheetahs is 10.1 m/s²
Explanation:
Hi there!
The equation of velocity of an object moving along a straight line with constant acceleration is the following:
v = v0 + a · t
Where:
v = velocity of the object at time t.
v0 = initial velocity.
a = acceleration.
t = time
We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.
Let's convert mi/h into m/s:
50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s
Then, using the equation:
v = v0 + a · t
22.4 m/s = 0 m/s + a · 2.22 s
Solving for a:
22.4 m/s / 2.22 s = a
a = 10.1 m/s²
The acceleration of the cheetahs is 10.1 m/s²
Answer:
The frequency of the photon decreases upon scattering
Explanation:
Here we note that when a photon is scattered by a charged particle, it is referred to as Compton scattering.
Compton scattering results in a reduction of the energy of the photon and hence an increase in the wavelength (from λ to λ') of the photon known as Compton effect.
Therefore, since the wavelength increases, we have from
λf = λ'f' = c
f = c/λ
Where:
f and f' = The frequency of the motion of the photon before and after the scattering
c = Speed of light (constant)
We have that the frequency, f, is inversely proportional to the wavelength, λ as follows;
f = c/λ
As λ = increases, and c is constant, f decreases, therefore, the frequency of the photon decreases upon scattering.
<span>As seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.
Let's assume that both Barbara and Neil start out at coordinate (0,0) and skate for exactly 1 second. Where do they end up?
Barbara is going due south at 5.9 m/s, so she's at (0,-5.9)
Neil is going due west at 1.4 m/s, so he's at (-1.4,0)
Now to see Neil's relative motion to Barbara, compute a translation that will place Barbara back at (0,0) and apply that same translation to Neil. Adding (0,5.9) to their coordinates will do this.
So the translated coordinates for Neil is now (-1.4, 5.9) and Barbara is at (0,0).
The magnitude of Neil's velocity as seen by Barbara is
sqrt((-1.4)^2 + 5.9^2) = sqrt(1.96 + 34.81) = sqrt(36.77) = 6.1 m/s
The angle of his vector relative to due west will be
atan(5.9/1.4) = atan(4.214285714) = 76.7 degrees
So as seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.</span>
