Answer:
The speed and direction of the two players immediately after the tackle are 3.3 m/s and 53.4° South of West
Explanation:
given information:
mass of fullback,
= 92 kg
speed of full back,
= 5.8 to south
mass of lineman,
=110 kg
speed of lineman,
= 3.6
according to conservation energy,
assume that the collision is perfectly inelastic, thus
initial momentum = final momentum
=
'
m₁v₁ = (m₁+m₂)
'
' = m₁v₁/(m₁+m₂)
= (92) (5.8)/(92+110)
= 2.64 m/s
=
'
m₂v₂ = (m₁+m₂)
'
' = m₁v₁/(m₁+m₂)
= (110) (3.6)/(92+110)
= 1.96 m/s
thus,
' = √
'²+
'²
= 3.3 m/s
then, the direction of the two players is
θ = 90 - tan⁻¹(
'/
')
= 90 - tan⁻¹(1.96/2.64)
= 53.4° South of West
Answer:
the cat is 0.4238 m in front of the dog as it leaps through the window
Explanation:
Given that;
acceleration a = 0.85 m/s²
speed v = 1.40 m/s
the cat is at rest, so initial velocity u = 0
we know that, since the cat is sleeping on the floor in the middle of a 2.8-m-wide room, it needs to cover (2.8 m / 2 ) distance to get to the window;
using the second equation equation of motion;
s = ut + 1/2 at²
we substitute
2.8/2 = 0×t + 1/2 × 0.85 × t²
1.4 = 0.425t²
t = √( 1.4 / 0.425 )
t = 1.81497 sec
now, at acceleration 0.10 m/s²
the dog has to cover the distance;
s = ut + 1/2 at²
s = ( 1.4 × 1.81497) - 1/2 × 0.10 × 1.81497²
s = 2.540958 - 0.1647
s = 2.3762 m
The cant in front of the dog as it leaps through the window;
distance = 2.8 m - 2.3762 m
distance = 0.4238 m
Therefore, the cat is 0.4238 m in front of the dog as it leaps through the window
For the rest 3 km
Bholu need to run in 5 minutes so
He need to 3 × 1000 / 5 × 60 m/s
3000/ 300 = 10 m/s
So , his speed should be 1 m/s to reach in 5 minute
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I think it’s D sorry If I’m wrong