Hi there!
We can use the rotational equivalent of Newton's Second Law:
Στ = Net Torque (Nm)
I = Moment of inertia (kgm²)
α = Angular acceleration (rad/sec²)
We can plug in the given values to solve.
Answer:
2 meters towards the mirror.
Explanation:
In a plane mirror the image distance is equal to the object distance. Therefore, by moving 2 meters towards the mirror, the boy reduces the distance between him and the mirror to two meters which is the object distance. The image distance is also 2 meters. add the two distances you will get four meters.
Answer:
Explanation:
As we know that volume of cylinder is
Where v=volume , h= height or thickness and r= radius
Here,
Putting these values in the previous equation , we get
Therefore thickness is 27.5 m
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02
Alternative solution:
c = sqrt (2) * 58 = 1.414 * 58 = 82.02
Therefore, they have to fly 82.02 mph
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:
