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2 years ago

I WILL MARK YOU TOP IF YOU HELP ME I NEED THIS ASAP!!!!

1 answer:

6 0

The basketballs and racquetball eventually stopped bouncing due to the first law. the first law states that an object at rest will stay at rest unless acted upon by an external force. this means that once the 2 balls lose friction, they will remain at rest until acted upon by an external force. once the 2 balls lose friction(energy), they come to a complete stop.

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A bullet of mass m=26g is fired into a wooden block of mass M=4.7kg. The block is attached to a string of length 1.5m. The bulle

vodka [1.7K]

Answer:

u = 449 m/s

Explanation:

Given,

Mass of the bullet, m = 26 g

Mass of the wooden block,M = 4.7 Kg

height of the block,h = 0.31 m

initial speed of the block, u = ?

Using conservation of energy

$(M+ m)gh = \dfrac{1}{2}(M+m)v^2$

$gh = \dfrac{1}{2}v^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.81\times 0.31}$

v = 2.47 m/s

Now, using conservation of momentum to calculate the speed of the bullet.

m u + M u' = (M+m)v

m u = (M+m)v

0.026 x u = (4.7+0.026) x 2.47

u = 449 m/s

Hence, the speed of the bullet is equal to 449 m/s.

5 0

2 years ago

In an insulated vessel, a quantity of hot water at temperature T1 is mixed with a different quantity of cold water at temperatur

erastovalidia [21]

Answer:Water Only

Explanation:

Given

vessel is insulated therefore no heat can be added or removed i.e. heat exchange is zero

If hot water at $T_1$ is mixed with cold water at $T_2$ then at equilibrium vessel contains only water and final temperature of water will be between $T_1$ and $T_2$

Heat released by hot water is equal to heat gain by cold water .

4 0

2 years ago

a constant force acts on an object of mass 10 kg for a duration of 2 seconds it increases the object velocity from 4metre per se

inn [45]

Answer:

F=ma

a=v/t

a=vf-vi/t

a=8-4/2

a=4÷2

a=2msec*2

F=10×2

F=20N

3 0

2 years ago

Question 10(Multiple Choice Worth 2 points)

Svetach [21]

$P = mv$

P momentum
m mass
v velocity

3 0

2 years ago

Read 2 more answers

Suppose that 4.4 moles of a monatomic ideal gas (atomic mass = 7.9 × 10-27 kg) are heated from 300 K to 500 K at a constant volu

timurjin [86]

Answer:

1) ΔQ₁ = 10.97 x 10³ J = 10.97 KJ

2) W₁ = 0 J

3) P = 41.66 x 10³ Pa = 41.66 KPa

4) v = 1618.72 m/s

5) ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ

6) W₂ = - 7.33 KJ

Explanation:

The heat transfer for a constant volume process is given by the formula:

ΔQ₁ = ΔU = n Cv ΔT

where,

ΔQ₁ = Heat transfer during constant volume process

ΔU = Change in internal energy of gas

n = No. of moles = 4.4 mol

Cv = Molar Specific Heat at Constant Volume = 12.47 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 500 k - 300 k = 200 k

Therefore,

ΔQ₁ = (4.4 mol)(12.47 J/mol.k)(200 k)

ΔQ₁ = 10.97 x 10³ J = 10.97 KJ

Since, work done by gas is given as:

W₁ = PΔV

where,

ΔV = 0, due to constant volume

Therefore,

W₁ = 0 J

The average kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

but, K.E is also given by:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

mv² = 3KT

v = √(3KT/m)

where,

v = average speed of gas molecue = ?

K = Boltzman Constant = 1.38 x 10⁻²³ J/k

T = Absolute Temperature = 500 K

m = mass of a molecule = 7.9 x 10⁻²⁷ kg

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/k)(500 k)/(7.9 x 10⁻²⁷ kg)]

v = 1618.72 m/s

From kinetic molecular theory, we know that or an ideal gas:

P = (1/3)ρv²

where,

P = pressure of gas = ?

m = Mass of Gas = (Atomic Mass)(No. of Atoms)

m = (Atomic Mass)(Avogadro's Number)(No. of Moles)

m = (7.9 x 10⁻²⁷ kg/atom)(6.022 x 10²³ atoms/mol)(4.4 mol)

m = 0.021 kg

ρ = density = mass/volume = 0.021 kg/0.44 m³ = 0.0477 kg/m³

Therefore,

P = (1/3)(0.0477 kg/m³)(1618.72 m/s)²

P = 41.66 x 10³ Pa = 41.66 KPa

The heat transfer for a constant pressure process is given by the formula:

ΔQ₂ = n Cp ΔT

where,

ΔQ₂ = Heat transfer during constant pressure process

n = No. of moles = 4.4 mol

Cp = Molar Specific Heat at Constant Pressure = 20.79 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 300 k - 500 k = -200 k

Therefore,

ΔQ₂ = (4.4 mol)(20.79 J/mol.k)(-200 k)

ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ

Negative sign shows heat flows from system to surrounding.

From Charles' Law, we know that:

V₁/T₁ = V₂/T₂

V₂ = (V₁)(T₂)/(T₁)

where,

V₁ = 0.44 m³

V₂ = ?

T₁ = 500 K

T₂ = 300 k

Therefore,

V₂ = (0.44 m³)(300 k)/(500 k)

V₂ = 0.264 m³

Therefore,

ΔV = V₂ - V₁ = 0.264 m³ - 0.44 m³ = - 0.176 m³

Hence, the work done , will be:

W₂ = PΔV = (41.66 KPa)(- 0.176 m³)

W₂ = - 7.33 KJ

Negative sign shows that the work is done by the gas

4 0

3 years ago