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3 years ago

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A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it st

ruck the wall. Has the momentum of the ball changed, and if so, by how much?

1 answer:

jeka57 [31]3 years ago

3 0

Answer:

The magnitude of change in momentum is (2mv).

Explanation:

The momentum of an object is given by the product of mass and velocity with which it is moving.

Let the mass of ball is m. A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it struck the wall.

Initial speed of the ball is v and final speed, when it rebounds, is (-v). The change in momentum is given by :

p = final momentum - initial momentum

$p=-mv-mv\\\\p=-2mv$

So, the magnitude of change in momentum is (2mv).

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A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

$m_{m}$ = mass of metal = 400 g

$c_{m}$ = specific heat of metal = ?

$T_{mi}$ = initial temperature of metal = 100 °C

$m_{a}$ = mass of aluminum cup = 100 g

$c_{a}$ = specific heat of aluminum cup = 900.0 J/kg ∙ K

$T_{ai}$ = initial temperature of aluminum cup = 15 °C

$m_{w}$ = mass of water = 500 g

$c_{w}$ = specific heat of water = 4186 J/kg ∙ K

$T_{wi}$ = initial temperature of water = 15 °C

$T$ = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

$m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}$

7 0

3 years ago

Give a combination of four quantum numbers that could be assigned to an electron occupying a 5p orbital.

Elden [556K]

Answer:

n=5, l=1, m(l) = -1, m(s)= + 1/2

Explanation:

Quantum number are used to describe the position and spin of an electron inside an atom. There are four types of quantum number for describing an electron inside an atom. They are: the principal quantum number, spin quantum number, magnetic quantum number and angular momentum quantum number.

(1).PRINCIPAL QUANTUM NUMBER: denoted by n, and has possible values of n= 1,2,3,4,.... IN HERE, n= 5

(2).ANGULAR MOMENTUM QUANTUM NUMBER: it is denoted by l, and has possible values of l= 0,1,2,3,...,(n-1).

Our l here is one( that is, s-orbital=0, p-orbital=1, d-orbital= 3 and so on)

(3).MAGNETIC QUANTUM NUMBER: The magnetic quantum number, which is denoted by m subscribt l, specifies the exact orbital in which you can find the electron. It has values ranging from -l,...,-1,0,1,...,l.

Here, our value is -1 that is m(l)= -1

(4).SPIN QUANTUM NUMBER: describes the orientation of electrons. Electrons can only have two values here, either a positive one and the half(+1/2) that is the spin up electron or the negative one and half(-1/2) that is the spin down electron.

8 0

3 years ago

our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three

Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is $\tau = 35.63\ N \cdot m$

b

The distance of the single for equilibrium to occur is $r_s =1.62 \ m$

Explanation:

From the question we are told that

The mass of the left rock is $m_s = 2.25 \ kg$

The mass of the rock on the right $m_p = 10.1 kg$

The distance from fulcrum to the center of the pile of rocks is $r_p = 0.360 \ m$

Generally the torque produced by the pile of rock is mathematically represented as

$\tau = m_p * g * r_p$

Substituting values

$\tau = 10.1 * 9.8 * 0.360$

$\tau = 35.63\ N \cdot m$

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

The torque due to the single rock is

$\tau = m_s * g * r_s$

At equilibrium the both torque are equal

$35.63 = m_s * r_s * g$

Making $r_s$ the subject of the formula

$r_s = \frac{35.63 }{m_s * g}$

Substituting values

$r_s = \frac{35.63 }{2.25 * 9.8}$

$r_s =1.62 \ m$

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lawyer [7]

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