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Neporo4naja [7]
3 years ago
15

True or false. Air masses mix at fronts?

Physics
1 answer:
vodka [1.7K]3 years ago
6 0
Hello there!

The answer would be- False. Air masses do not mix at fronts.

Hope this helps... tell me if i am incorrect! Thanks!

~DL ☆♡☆♡
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A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north
sergiy2304 [10]

Answer:

Acceleration of the car will be a=0.1375m/sec^2

Explanation:

We have given mass of the ball m = 1600 kg

Force in north direction F= 7560 N

Resistance force which opposes the movement of car F_R=7340N

So net force on the car F_{net}=F-F_R=7560-7340=220N

According to second law of motion we know that F=ma

So 220=1600\times a

a=0.1375m/sec^2

7 0
3 years ago
What is the relationship between electric field lines and equipotential lines that you observed in doing the lab
lidiya [134]

Answer:

Explained below

Explanation:

Generally speaking, we know in physics that Electric field lines are lines which usually start at positive charges and deflect away from them to terminate at the negative charges. Meanwhile Equipotential lines are lines that are used to connect points located on the same electric potential.

Finally, in conclusion, electric field lines are usually lines that go through in a perpendicular manner across every equipotential lines.

8 0
3 years ago
a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car
Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 282 kg is the mass of the bumper car

u_1 = 3.50 m/s is the initial velocity of the bumper car (we take the right as positive direction)

v_1 = 0.800 m/s is the final velocity of the bumper car

m_2 = 155 kg is the mass of the second bumper car

u_2 = -1.88 m/s is the initial velocity of the second car (moving to the left)

v_2 is the final velocity of the second car

Solving for v_2,

v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s (to the right)

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

7 0
3 years ago
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