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TEA [102]
2 years ago
11

Help please!

Physics
1 answer:
JulsSmile [24]2 years ago
7 0

3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:

(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v

where v is the velocity of the combined players. Solve for v :

450 kg•m/s - 320 kg•m/s = (155 kg) v

v = (130 kg•m/s) / (155 kg)

v ≈ 0.84 m/s

4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that

(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v

where v is the new velocity of the 4-kg ball. Solve for v :

30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v

v = (16.4 kg•m/s) / (4 kg)

v = 4.1 m/s

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I need to name 20 gases in physics
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4 0
4 years ago
Pleaseeee Please help, I will love you forever and ever
matrenka [14]

Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J  = - 63000 J

Cp = 4184 J / kg°C

Formula

                        Q = mCp(T2 - T1)

                         T2 = T1 + Q/mCp    

Substitution

                       T2 = 35 - 63000/(0.5 x 4184)

                        T2 = 35 - 63000/2092

                        T2 = 35 - 30.1

                         T2 = 4.9 °C

6 0
3 years ago
Which equation correctly relates mechanical energy, thermal energy and total of energy when there is friction present in the sys
agasfer [191]

Answer:

E total = ME + E thermal

Explanation:

APEX

4 0
3 years ago
A child throws a baseball upward with an initial velocity of 20 m/s. The child wants to throw the baseball at least as high as t
Umnica [9.8K]
When the ball starts its motion from the ground, its potential energy is zero, so all its mechanical energy is kinetic energy of the motion:
E= \frac{1}{2}mv^2
where m is the ball's mass and v its initial velocity, 20 m/s.

When the ball reaches its maximum height, h, its velocity is zero, so its mechanical energy is just gravitational potential energy:
E=mgh

for the law of conservation of energy, the initial mechanical energy must be equal to the final mechanical energy, so we have
\frac{1}{2}mv^2 = mgh
From which we find the maximum height of the ball:
h= \frac{v^2}{2g}= \frac{(20 m/s)^2}{2 \cdot 9.81 m/s^2}=20.4 m

Therefore, the answer is yes, the ball will reach the top of the tree.

5 0
3 years ago
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