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zysi [14]
3 years ago
14

The force needed to overcome static friction is usually less than that needed to overcome kinetic friction.

Physics
1 answer:
xenn [34]3 years ago
3 0

Answer:

True

Explanation:

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You will be colliding two carts in this lab. Select all the systems for which you predict the total momentum of the system will
baherus [9]

From the momentum conservation we know that the initial momentum is equal to the final momentum. The momentum in a singular way can be defined as the product between the mass and the velocity of an object. In the presented system, however, there are two objects, therefore the mass of both and the speed of both, before and after the collision must be taken into account. Mathematically we could describe this as

m_1u_1+m_2u_2 = m_1v_1+m_2v_2

Here,

m_{1,2} = Mass of each object

u_{1,2} = Initial velocity of each object

v_{1,2} = Final velocity of each object

From here we can realize that it is necessary to use the system on both cars to be able to predict what will happen either with their masses, or their speeds.

The correct answer is C.

6 0
3 years ago
What is 12 and 1 half % expressed as a fraction
kiruha [24]
12 1/2 i would assume
6 0
3 years ago
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What is the correct name for this formula BaCI 2?
KIM [24]
BaCI2 stands for Barium Chloride. 
3 0
3 years ago
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A charged sphere with 1 × 10 8 units of negative charge is brought near a neutral metal rod. The half of the rod closer to the s
jeyben [28]

Answer:

-4*10⁴ units.

Explanation:

As the metal rod was initially neutral (which means that it has the same quantity of positive and negative charges), after being close to the charged sphere, as charge must be conserved, the total charge of the metal rod must  still remain to be zero.

So, if due to the influence of the negative charge in the sphere, the half of the road closer to the sphere has a surplus charge of +4*10⁴ units, the charge on the half of the rod farther from the sphere must be the same in magnitude but of the opposite sign, i.e., -4*10⁴ units.

5 0
3 years ago
-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a
bulgar [2K]

Answer:

<em>The velocity of the carts after the event is 1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

\displaystyle v'=\frac{2*5+8*0}{2+8}=\frac{10}{10}=1

The velocity of the carts after the event is 1 m/s

3 0
3 years ago
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