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:
24 cm/s
Explanation:
Applying
Pythagoras theorem,
a² = b²+c²............. Equation 1
Where a = resultant, b = vertical component, c = horizontal component
From the question,
Given: a = 26 cm/s, c = 10 cm/s
Substitute these values into equation 1
26² = b²+10²
676 = b²+100
b² = 676-100
b² = 576
b = √576
b = 24 cm/s
Answer:
The center of the Milky Way most likely contains a supermassive black hole.
Explanation:
Because it is an eleptical galaxy, it has a little rotation to it but not enough to flatten out so the center will contain a supermassive black hole.
The answer is shown below:
"The diurnal motion" of the stars is defined as the apparent daily motion of stars around the Earth. The cause for this "apparent motion of stars" was proposed by Heraclides (who was a Greek philosopher and astronomer). He afirmed that this was created by the Earth's rotation on its axis, which is completed in 23 hours, 56 minutes and 4.09 seconds.
Answer:
The quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.
Explanation:
Given;
emf of the battery, V = 12 V
resistance of the resistor, R = 100-Ω
time of current flow, t = 1 min
charge of 1 electron = 1.602 x 10¹⁹ C
The current through this circuit is given by;
I = V / R
I = (12) / (100)
I = 0.12 A
The quantity of charge or electron flowing the wire in the given time is calculated as;
Q =It
where;
I is the current flowing through the wire
t is the time of current flow = 1 x 60s = 60 s
Q = 0.12 x 60
Q = 7.2 C
1.602 x 10⁻¹⁹ C --------------- 1 electron
7.2 C -----------------------------? electron
Therefore, the quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.