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gregori [183]
3 years ago
13

Carbon monosulfide ionic or covalent

Physics
1 answer:
taurus [48]3 years ago
3 0

Answer:

Covalent.

Explanation:

Carbon and Sulfur are both nonmetals.

Nonmetal+nonmetal= covalent compound.

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Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they hook toget
Nikitich [7]

Answer:

<em>Second option</em>

Explanation:

<u>Linear Momentum</u>

The linear momentum of an object of mass m and speed v is

P=mv

If two or more objects are interacting in the same axis, the total momentum is

P_t=m_1v_1+m_2v_2+...

Where the speeds must be signed according to a fixed reference

The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right

P_1=-2mv

The second cart of mass m goes to the right at a speed v

P_2=mv

The total momentum before the impact is

P_t=-2mv+mv=-mv

The total momentum after the collision is negative, both carts will join and go to the left side

The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven

The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either

The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".

The second option is the correct one because the mass m_2 has a negative momentum and then the sum of both masses keeps being negative

3 0
4 years ago
Read 2 more answers
Explain how Law 1 applies to the image to the left.
alisha [4.7K]

Answer:

12

Explanation:

3 0
3 years ago
Read 2 more answers
Consider an object sliding at constant velocity along a frictionless surface. Which of the following best describes the forces o
Aliun [14]
-- The net vertical force on the object is zero.
Otherwise it would be accelerating up or down.

-- The net horizontal force on the object is zero.
Otherwise it would be accelerating horizontally,
that is, its 'velocity' would not be constant.  That
would contradict information given in the question.

The total net force on the object is the resultant of the
net vertical component and net horizontal component.

Total net force =  √(0² + 0²)

                         =  √(0 + 0)

                         =  √0

                         =  Zero.

The correct answer is the last choice on the list.

Also, you know what ! ?  It doesn't even matter whether the surface it's
sliding on is frictionless or not. 

If the object's velocity is constant, then the NET force on it must be zero. 
If it's sliding on sandpaper, then something must be pushing it with constant
force, to balance the friction force, and make the net force zero.  If the total
net force isn't zero, then the object would have to be accelerating ... either
its speed, or its direction, or both, would have to be changing.
3 0
3 years ago
Janelle stands on a balcony, two stories above Michael. She throws one ball straight up and one ball straight down, but both wit
antoniya [11.8K]

Answer:

Both balls have the same speed.

Explanation:

Janelle throws the two balls from the same height, with the same speed. Both balls will have the same potential and kinetic energy. Energy must be conserved. When the balls pass Michael, again they must have the same potential and kinetic energy.

4 0
3 years ago
Heat Q flows spontaneously from a reservoir at 404 K into a reservoir at 298 K. Because of the spontaneous flow, 2740 J of energ
aleksklad [387]

Answer:

The heat is 10458 J and 15480 J.

Explanation:

Given that,

Temperature T₁ = 404 K

Temperature T₂ = 298 K

Work done = 2740 J

If the temperature T₁ =298 k

Temperature T₂ = 245 K

We need to calculate the heat

Using efficiency formula

\eta=\dfrac{W}{Q}...(I)

We need to calculate the efficiency

Using formula of efficiency

\eta=1-\dfrac{T_{2}}{T_{1}}

\eta=1-\dfrac{298}{404}

\eta=0.262

Put the value of efficiency in equation (I)

0.262=\dfrac{2740}{Q}

Q=\dfrac{2740}{0.262}

Q=10458\ J

Again, we need to calculate the efficiency

\eta=1-\dfrac{T_{2}}{T_{1}}

\eta=1-\dfrac{245}{298}

\eta=0.177

We need to calculate the heat

Using equation (I)

Q=\dfrac{2740}{0.177}

Q=15480\ J

Hence, The heat is 10458 J and 15480 J.

8 0
3 years ago
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