Answer:
First of all the formula is F= uR,( force= static friction× reaction)
mass= 5+25=30
F= 50
R= mg(30×10)=300
u= ?
F=UR
u= F/R
u= 50/300=0.17N
You can tell a lot about an object that's not moving,
and also a lot about the forces acting on it:
==> If the box is at rest on the table, then it is not accelerating.
==> Since it is not accelerating, I can say that the forces on it are balanced.
==> That means that the sum of all forces acting on the box is zero,
and the effect of all the forces acting on it is the same as if there were
no forces acting on it at all.
==> This in turn means that all of the horizontal forces are balanced,
AND all of the vertical forces are balanced.
Horizontal forces:
sliding friction, somebody pushing the box
All of the forces on this list must add up to zero. So ...
(sliding friction force) = (pushing force), in the opposite direction.
If nobody pushing the box, then sliding friction force = zero.
Vertical forces:
gravitational force (weight of the box, pulling it down)
normal force (table pushing the box up)
All of the forces on this list must add up to zero, so ...
(Gravitational force down) + (normal force up) = zero
(Gravitational force down) = -(normal force up) .
Explanation:
Solar technologies convert sunlight into electrical energy either through photovoltaic (PV) panels or through mirrors that concentrate solar radiation. This energy can be used to generate electricity or be stored in batteries or thermal storage.
Answer:
47.8 °C
Explanation:
Use the heat equation:
q = mCΔT
where q is the heat absorbed/lost,
m is the mass of water,
C is the specific heat capacity,
and ΔT is the change in temperature.
Here, q = 100 kJ, m = 0.5 kg, and C = 4.184 kJ/kg/°C.
100 kJ = (0.5 kg) (4.184 kJ/kg/°C) ΔT
ΔT = 47.8 °C
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:
