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german
2 years ago
8

Two blocks of masses 1.0 kg and 2.0 kg, respectively, are pushed by a constant applied force F across a horizontal frictionless

table with constant acceleration such that the blocks remain in contact with each other, as shown above. The 1.0 kg block pushes the 2.0 kg block with a force of 2.0 N. The acceleration of the two blocks is
0

1.0 m/s2

1.5 m/s2

2.0 m/s2

3.0 m/s2
Physics
1 answer:
BaLLatris [955]2 years ago
6 0

Answer:

1.0 m/s^2

Explanation: happy to help :)

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Which phrase is the best definition of energy
raketka [301]

Answer:

The capacity for doing work.

Explanation:

It has the forms kinetic, potential, thermal, electric, nuclear or other forms of energy.

5 0
3 years ago
A 150-W lamp is placed into a 120-V ac outlet. What is the peak current?
devlian [24]

Answer:

Explanation:

Formula

W = I * E

Givens

W = 150

E = 120

I = ?

Solution

150 = I * 120   Divide by 120

150/120 = I

5/4  = I

I = 1.25

Note: This is an edited note. You have to assume that 120 is the RMS voltage in order to go any further. That means that the peak voltage is √2 times the size of 120. The current has the same note applied to it. If the voltage is its rms value, then the current must (assuming the properties of the bulb do not change)

On the other hand, if the voltage is the peak value at 120 then 1.25 will be correct.

However I would go with the other answerer's post and multiply both values by  √2

3 0
3 years ago
Read 2 more answers
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
Consider two soap bubbles having radii r1 and r2 (r1 &lt; r2) connected via a valve. What happens if we open the valve?
Alex17521 [72]

Answer:

Explanation:

The radius of the smaller bubble, r1 will decrease and that of the bigger bubble, r2 will increase.

The pressure that is present in the smaller bubble usually is greater than the pressure that exists inside that of the bigger bubble. This then makes air to flow from r1 to r2 thereby making the radius of the smaller bubble r1, to decrease while keeping that of the bigger bubble r2 higher.

4 0
3 years ago
Describe an Activity to show magnetic poles doesn't seperated<br>​
mojhsa [17]

Answer:

We know that pole is the point where the strength of the magnet is maximum. So more and more iron particles will be attracted at poles of a magnet when we bring a magnet near the iron particles. We will observe the crowdness of particles at the ends of magnet. This indicates the presence of two poles in a magnet. Hence poles are present in a magnet in pair. If a magnet is divided into two parts, each part also possesses a pair of poles.

Two properties of a magnet are: A magnet always has two poles: north pole and south pole. Like magnetic poles repel each other and unlike magnetic poles attract each other.

7 0
3 years ago
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