Answer:
The value of tension on the cable T = 1065.6 N
Explanation:
Mass = 888 kg
Initial velocity ( u )= 0.8 
Final velocity ( V ) = 0
Distance traveled before come to rest = 0.2667 m
Now use third law of motion 
= 
- 2 a s
Put all the values in above formula we get,
⇒ 0 = 
- 2 × a ×0.2667
⇒ a = 1.2 
This is the deceleration of the box.
Tension in the cable is given by T = F = m × a
Put all the values in above formula we get,
T = 888 × 1.2
T = 1065.6 N
This is the value of tension on the cable.
Answer:
Explanation:
Intensity of light is inversely proportional to distance from source
I ∝ 1 /r² where I is intensity and r is distance from source . If I₁ and I₂ be intensity at distance r₁ and r₂ .
I₁ /I₂ = r₂² /r₁²
If r₂ = 4r₁ ( given )
I₁ / I₂ = (4r₁ )² / r₁²
= 16 r₁² / r₁²
I₁ / I₂ = 16
I₂ = I₁ / 16
So intensity will become 16 times less bright .
"16 times " is the answer .
Considering the total body, there are six elements of fitness: aerobic capacity, body structure, body composition, balance, muscular flexibility and strength
<span>The rate of change in velocity is acceleration.</span>
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:
