A 98-kg fullback is running along at 8.6 m / s when a 76-kg defensive back running in the same direction at 9.8 m / s jumps on h
1 answer:
The total momentum before and after the collision must be conserved.
The total momentum before the collision is:
where m1 and m2 are the masses of the two players, and
and 
their initial velocities. Both are considered with positive sign, because the two players are running toward the same direction.
The final momentum is instead
because now the two players are moving together with a total mass of (m1+m2) and final speed vf.
By requiring that the momentum is conserved
we can calculate vf, the post-collision speed:
and the direction is the same as the direction of the players before the collision.
The total momentum before the collision is:
where m1 and m2 are the masses of the two players, and
The final momentum is instead
because now the two players are moving together with a total mass of (m1+m2) and final speed vf.
By requiring that the momentum is conserved
we can calculate vf, the post-collision speed:
and the direction is the same as the direction of the players before the collision.
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Answer:
the final temperature of the tea is 7.39⁰C.
Explanation:
Given;
mass of the tea, m = 375 g
specific heat capacity of the tea, C = 4.184 JJ/g°C
initial temperature of the tea, t₁ = 95°C
the final temperature of the tea, t₂ = ?
Energy lost by the refrigerator, Q = 137,460 J
The energy lost by the refrigerator is given by the following formula;
-Q = mc(t₂ - t₁)
-137,460 =375 x 4.184(t₂ - 95°C)
-137,460 = 1569(t₂ - 95°C)
Therefore, the final temperature of the tea is 7.39⁰C.
Answer:
The minimum work per unit heat transfer will be 0.15.
Explanation:
We know the for a heat pump the coefficient of performance () is given by
where, is the magnitude of heat transfer between cyclic device and high-temperature medium at temperature
and
is the required input and is given by
,
being magnitude of heat transfer between cyclic device and low-temperature
. Therefore, from above equation we can write,
Given, and
. So, the minimum work per unit heat transfer is given by
Answer:
x = A cos (w \sqrt{2y_{o}/g})
a) maximun Ф= \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) minimun Ф = - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
Explanation:
For this exercise let's use kinematics to find the time it takes for the mass to reach the floor
y = y₀ + v₀ t - ½ g t²
as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)
0 = y₀ - ½ g t²
t =
The bucket-spring system has a simple harmonic motion, which is described by
x = A cos wt
in this expression we assumed that the phase constant (Ф) is zero
let's replace the time
x = A cos (w \sqrt{2y_{o}/g})
this is the distance where the system must be for the mass to fall into it.
a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes
w =
In the initial state
w = \sqrt{k/2}
When the mass changes
w ’= \sqrt{k/3}
the displacement in each case is
x = A cos (wt)
for the new case
x ’= A cos (w’t + Ф)
the phase constant is included to take into account possible changes due to the collision of the mass.
we see that this maximum expressions when the cosine is maximum
cos (w´t + Ф) = 1
w’t + Ф = 0
Ф = -w ’t
Ф = -
\sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) the function is minimun if
cos (w’t + fi) = 0
w’t + Ф = π / 2
Ф = π / 2 - w ’t
Ф = - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }