How many layers does the world have?

To calculate the density of an object which of the following values do you need to know a. Buoyancy b. Volume c. Weight d. Mass

dem82 [27]

B. volume and d. mass

3 0

3 years ago

How would decreasing the volume of the reaction vessel affect each of the following equilibria?2NOBr(g)⇌2NO(g)+Br2(g)

mafiozo [28]

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

5 0

3 years ago

If you had started with a larger mass, how would the half-life change?

iogann1982 [59]

Answer:

There is no change, unless your mass is somehow at the quantum level, at which the concept of half-life breaks down.

Half life is a property of the specific radioactive isotope...NOT of the initial sample's mass.

3 0

2 years ago

magine two carts, one with twice the mass of the other, that are going to have a head-on collision. In order for the two carts t

scoray [572]

Answer:

Twice as fast

Explanation:

Solution:-

- The mass of less massive cart = m

- The mass of Massive cart = 2m

- The velocity of less massive cart = u

- The velocity of massive cart = v

- We will consider the system of two carts to be isolated and there is no external applied force on the system. This conditions validates the conservation of linear momentum to be applied on the isolated system.

- Each cart with its respective velocity are directed at each other. And meet up with head on collision and comes to rest immediately after the collision.

- The conservation of linear momentum states that the momentum of the system before ( P_i ) and after the collision ( P_f ) remains the same.

$P_i = P_f$

- Since the carts comes to a stop after collision then the linear momentum after the collision ( P_f = 0 ). Therefore, we have:

$P_i = P_f = 0$

- The linear momentum of a particle ( cart ) is the product of its mass and velocity as follows:

m*u - 2*m*v = 0

Where,

( u ) and ( v ) are opposing velocity vectors in 1-dimension.

- Evaluate the velcoity ( u ) of the less massive cart in terms of the speed ( v ) of more massive cart as follows:

m*u = 2*m*v

u = 2*v

Answer: The velocity of less massive cart must be twice the speed of more massive cart for the system conditions to hold true i.e ( they both come to a stop after collision ).

8 0

3 years ago

The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top

vaieri [72.5K]

Answer:

v = 44,16 m/s

Explanation:

We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:

$v_y^2 =v_0^2 +2ay$

Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:

$v_y^2 =2ay \rightarrow v_y = \sqrt{2ay}$

Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.

v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s $v_y =- \sqrt{2*(-9,81 m/s^2)*(-99,4 m)} = 44,1613 m$

So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.

I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D

7 0

3 years ago