Answer: The pressure that one experiences on the Mount Everest will be different from the one, in a classroom. It is because pressure and height are inversely proportional to each other. This means that as we move up, the height keeps on increasing but the pressure will keep on decreasing. This is the case that will be observed when one stands on the Mount Everest as the pressure is comparatively much lower there.
It is because as we move up, the amount of air molecules keeps on decreasing but all of the air molecules are concentrated on the lower part of the atmosphere or on the earth's surface.
Thus a person in a low altitude inside a classroom will experience high pressure and a person standing on the Mount Everest will experience low pressure.
Answer:
The least uncertainty in the momentum component px is 1 × 10⁻²³ kg.m.s⁻¹.
Explanation:
According to Heisenberg's uncertainty principle, the uncertainty in the position of an electron (σx) and the uncertainty in its linear momentum (σpx) are complementary variables and are related through the following expression.
σx . σpx ≥ h/4π
where,
h is the Planck´s constant
If σx = 5 × 10⁻¹²m,
5 × 10⁻¹²m . σpx ≥ 6.63 × 10⁻³⁴ kg.m².s⁻¹/4π
σpx ≥ 1 × 10⁻²³ kg.m.s⁻¹
Answer:
15.065ft
Explanation:
To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.
By definition the drag force is expressed as:
Where
is the density of the flow
V = Velocity
= Drag coefficient
A = Area
For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3
For second Newton's Law the Force is also defined as,
Equating both equations we have:
Integrating
Here,
Replacing:
Answer: 0.006in/s
Explanation:
Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s
Also let the rate at which the radius of the balloon is increasing be dr/dt
Given r = 4.7in and Π = 3.14
Applying the chain rule method
dV/dt = dV/dr × dr/dt
If the volume of the sphere is 4/3Πr³
V = 4/3Πr³
dV/dr = 4Πr²
If r = 4.7in
dV/dr = 4Π(4.7)²
dV/dr = 277.45in²
Therefore;
1.68 = 277.45 × dr/dt
dr/dt = 1.68/277.45
dr/dt = 0.006in/s
Answer:Bruce is knocked backwards at
14
m
s
.
Explanation:
This is a problem of momentum (
→
p
) conservation, where
→
p
=
m
→
v
and because momentum is always conserved, in a collision:
→
p
f
=
→
p
i
We are given that
m
1
=
45
k
g
,
v
1
=
2
m
s
,
m
2
=
90
k
g
, and
v
2
=
7
m
s
The momentum of Bruce (
m
1
) before the collision is given by
→
p
1
=
m
1
v
1
→
p
1
=
(
45
k
g
)
(
2
m
s
)
→
p
1
=
90
k
g
m
s
Similarly, the momentum of Biff (
m
2
) before the collision is given by
→
p
2
=
(
90
k
g
)
(
7
m
s
)
=
630
k
g
m
s
The total linear momentum before the collision is the sum of the momentums of each of the football players.
→
P
=
→
p
t
o
t
=
∑
→
p
→
P
i
=
→
p
1
+
→
p
2
→
P
i
=
90
k
g
m
s
+
630
k
g
m
s
=
720
k
g
m
s
Because momentum is conserved, we know that given a momentum of
720
k
g
m
s
before the collision, the momentum after the collision will also be
720
k
g
m
s
. We are given the final velocity of Biff (
v
2
=
1
m
s
) and asked to find the final velocity of Bruce.
→
P
f
=
→
p
1
f
+
→
p
2
f
→
P
f
=
m
1
v
1
f
+
m
2
v
2
f
Solve for
v
1
:
v
1
f
=
→
P
f
−
m
2
v
2
f
m
1
Using our known values:
v
1
f
=
720
k
g
m
s
−
(
90
k
g
)
(
1
m
s
)
45
k
g
v
1
f
=
14
m
s
∴
Bruce is knocked backwards at
14
m
s
.
Explanation:
