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Licemer1 [7]
2 years ago
10

Can I pass 6th grade with these grades:

Physics
2 answers:
Reika [66]2 years ago
7 0

Answer:

No you can't so sorry

Explanation:

I hope this helps.

coldgirl [10]2 years ago
5 0

Answer:

You can't

Explanation:

I'm sorry but with a 25% and a 50% it's not enough. Hope this helps you out and your teachers are nice enough to work with you.

You might be interested in
Two balls of mass m1 and m2, with velocities v1 and v2 collide head on. Is there any way for both balls to have zero velocity af
nordsb [41]

Answer:

Explanation:

As the final Kinetic energy is zero or less than initial kinetic energy, the collision must be inelastic.  

In Inelastic collision both the bodies must stick together as final velocity is zero for both the bodies.

To conserve the momentum, momentum associated before the collision of first must be equal and opposite to the momentum associated with the second ball.

i.e.

m_1v_1=m_2v_2

8 0
4 years ago
To see if your results are reasonable, you can compare the final velocity of the stone as it falls down unwinding the wire from
pentagon [3]

Answer:

The velocity of the stone is 2.57 m/s.

Explanation:

Given that

Height = 0.337 m

We need to calculate the velocity of the stone

Using equation of motion

v^2-u^2=2gh

Where, v = velocity of stone

u = initial velocity

g = acceleration due to gravity

h = height

Put the value into the formula

v^2-0=2\times9.8\times 0.337

v=\sqrt{2\times9.8\times0.337}

v=2.57\ m/s

Hence, The velocity of the stone is 2.57 m/s.

7 0
4 years ago
What is the relative energy expense of galloping rather than trotting at 3.5 m/s?
labwork [276]

Answer: Trotting uses only 75 percent of the energy as galloping

Explanation: Trotting is only 300 J/m, whereas galloping is roughly 400 J/m

7 0
3 years ago
A 3.0 kg ball moving at 8 m/s to the right collides with a 1.0 kilogram ball at rest. After the collision, the 3
amid [387]

Let m₁ = 3.0 kg and v₁ = + 8 m/s (so right is positive), and m₂ = 1.0 kg and v₂ = 0. The total momentum of the two balls before and after collision is conserved, so

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

where v₁' = + 5 m/s and v₂' are the velocities of the two balls after colliding, so

(3.0 kg) (8 m/s) = (3.0 kg) (5 m/s) + (1.0 kg) v₂'

Solve for v₂' :

24 kg•m/s = 15 kg•m/s + (1.0 kg) v₂'

(1.0 kg) v₂' = 9 kg•m/s

v₂' = (9 kg•m/s) / (1.0 kg)

v₂' = + 9 m/s

which is to say, the second ball is given a speed of 9 m/s to the right after colliding with the first ball.

7 0
3 years ago
Can someone help me ?​
Paraphin [41]

Explanation:

300m÷60s=5m/s

hope this can help maybe

8 0
2 years ago
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