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3 years ago

9.5 g bullet has a speed of 1.5 km/s what is the kinetic energy of the bullet

Physics

2 answers:

kicyunya [14]3 years ago

8 0

Answer:

The kinetic energy of the bullet is approximately 10688 Joules, or about 10.7kJ.

Explanation:

Use the kinetic energy formula:

$E_k=\frac{1}{2}mv^2= \frac{1}{2}9.5\cdot10^{-3}kg\cdot 1.5^2\cdot 10^6 \frac{m^2}{s^2}\approx 10688J$

The kinetic energy of the bullet is approximately 10688 Joules, or about 10.7kJ.

Amiraneli [1.4K]3 years ago

3 0

The answer is 0.000824653J

You need to use the formula Mass * Velocity^2 over 2

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During crystallisation the crystals separate out from the hot ________solution of a substance on cooling

alekssr [168]

Answer:

The process of separation or deposition of crystals from a hot saturated solution on gentle cooling of the solution is called 'crystallisation'.

Explanation:

6 0

3 years ago

Steam at 700 bar and 600 oC is withdrawn from a steam line and adiabatically expanded to 10 bar at a rate of 2 kg/min. What is t

Setler79 [48]

Answer:

Final temperature of the steam = 304.29 K = 31.14°C

Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T = 0 J/K.

But 0.01 J/k.s was obtained though, which is approximately 0.

Explanation:

For an adiabatic system, the Pressure and temperature are related thus

P¹⁻ʸ Tʸ = constant

where γ = ratio of specific heats. For steam, γ = 1.33

P₁¹⁻ʸ T₁ʸ = P₂¹⁻ʸ T₂ʸ

P₁ = 700 bar

P₂ = 10 bar

T₁ = 600°C = 873.15 K

T₂ = ?

(700⁻⁰•³³)(873.15¹•³³) = (10⁻⁰•³³)(T₂¹•³³)

T₂ = 304.29 K = 31.14°C

b) Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T

To prove this

Entropy of the process

dQ - dW = dU

dQ = dU + dW

dU = mCv dT

dW = PdV

dQ = TdS

TdS = mCv dT + PdV

dS = (mCv dT/T) + (PdV/T) =

PV = mRT; P/T = mR/V

dS = (mCv dT/T) + (mRdV/V)

On integrating, we obtain

ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

To obtain V₁ and V₂, we use PV = mRT

V/m = specific volume

Pv = RT

R for steam = 461.52 J/kg.K

For V₁

P = 700 bar = 700 × 10⁵ Pa, T = 873.15 K

v₁ = (461.52 × 873.15)/(700 × 10⁵) = 0.00570 m³/kg

For V₂

P = 10 bar = 10 × 10⁵ Pa, T = 304.29 K

v₂ = (461.52 × 304.29)/(10 × 10⁵) = 0.143 m³/kg

ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

m = 2 kg/min = 0.0333 kg/s, Cv = 1410.8 J/kg.K

ΔS = [0.03333 × 1410.8 × In (304.29/873.15)] + (0.03333 × 461.52 × In (0.143/0.00570)

ΔS = - 49.56 + 49.57 = 0.01 J/K.s ≈ 0 J/K.s

3 0

4 years ago

A helicopter’s speed increases from 30 m/s to 40 m/s in 5 seconds.

Mandarinka [93]

7 m over s² or 7ms⁻¹

5 0

3 years ago

A ball held 2.0 meters above the floor has 59 joules of potential energy the bald mass is about

PilotLPTM [1.2K]

Answer:

Mass of ball = 3.0102 Kg.

Explanation:

Potential energy at height h is given by

PE = mgh

where m is the mass of the body

g is the the acceleration due to gravity whose value is 9.8 m/s^2

h is the height from surface level

PE is potential energy

In the problem given h = 2 meters

PE = 59 joules

59 = m*9.8*2

=> m = 59/9.8*2 = 3.0102 Kg

Thus, mass of ball is 3.0102 Kg.

4 0

4 years ago

Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 5.30 m/

Tju [1.3M]

Answer:

a) v_{p} = 2.83 m / s , b) 50.5º north east

Explanation:

This is a vector problem.

$v_{bg} = v_{bm} + v_{mg}$

The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground

To make the sum we decompose the speed of the ball in its components

The angle of 30 east of the south, measured from the positive side of the x axis is

θ = 30 + 270 = 300

$v_{bx}$ = $v_{b}$ cos 300

$v_{by}$ = v_{b} sin. 300

v_{bx} = 3.60 cos 300 = 1.8 m / s

v_{by} = 3.60 sin 300 = -3,118 m / s

Let's add speeds on each axis

X axis

vₓ = v_{bx}

vₓ = 1.8 m / s

Y Axis

$v_{y}$ = v1 - vpy

v_{y} = 5.30 - 3.118

v_{y} = 2.182 m / s

The magnitude of the velocity can be found using the Pythagorean theorem

$v_{p}$ = √ (vₓ² + v_{y}²)

v_{p} = √ (1.8² + 2.182²)

v_{p} = 2,829 m / s

v_{p} = 2.83 m / s

b) for direction use trigonometry

tan θ = $v_{y}$ / vₓ

θ = tan ⁺¹ v_{y} / vₓ

θ = tan⁻¹ 2.182 / 1.8

Tea = 50.48º

This address is 50.5º north east

8 0

3 years ago