Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)
• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)
•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
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:
See explanation
Explanation:
First, in order for you to understand, remember the basic concept of meniscus in graduated cylinder.
<em>"The meniscus is the curve seen at the top of a liquid in response to its container. The meniscus can be either concave or convex, depending on the surface tension of the liquid and its adhesion to the wall of the container".</em>
Now, according to this definition, and for water, the reading of the volume must be donde at the bottom of the curve of the meniscus. This is because the water gives a concave curve.
If you read it and matches the height of water, you are getting two results:
One, get an accurate value or volume, because it's been done at eye level.
The second fact is that when you do the reading this way, The total pressure is made equal to the atmospheric pressure by adjusting the height of the cylinder until the water level is equal.
1) Current in the wire: 0.0875 A
The current in the wire is given by:

where
Q is the charge passing a given point in the conductor
t is the time elapsed
In this problem, we have
Q = 420 C is the total charge passing through a given point in a time of
t = 80 min = 4800 s
So, the current is

2) Drift velocity of the electrons: 
The drift velocity of the electrons in the wire is given by:

where
I = 0.0875 A is the current
is the number of free electrons per cubic meter
A is the cross-sectional area
is the charge of one electron
The radius of the wire is

So the cross-sectional area is

So, the drift velocity is

Answer:

Explanation:
The free body diagram of the block on the slide is shown in the below figure
Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces
N is the reaction force between the block and the slide
For equilibrium along x-axis we have

Using value of N from equation β in α we get value of force as

Applying values we get
