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3 years ago

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

1 answer:

4 0

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

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What types of alternative fuels did bp pursue after browne became ceo in 1995?

Sliva [168]

Answer:solar energy

Explanation:

This is the energy gotten from sunlight to knock off electrons from fee atom.

This panel contains photovoltaic cells

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2 years ago

(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh

makvit [3.9K]

Answers:

(a) $2509.98 m/s$

(b) 397042.215 m

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is $700 km$ and whose gravitational acceleration at the surface is $a_{g}=4.5 m/s^{2}$

Knowing this, let's begin:

a) In this part we need to find the escape speed $V_{e}$ on the asteroid:

$V_{e}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the universal gravitational constant

$M$ is the mass of the asteroid

$R=700 km=700(10)^{3} m$ is the radius of the asteroid

On the other hand we know the gravitational acceleration is $a_{g}=4.5 m/s^{2}$ , which is given by:

$a_{g}=\frac{GM}{R^{2}}$ (2)

Isolating $GM$ :

$GM=a_{g}R^{2}$ (3)

Substituting (3) in (1):

$V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R}$ (4)

$V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)}$ (5)

$V_{e}=2509.98 m/s$ (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

$E_{o}=E_{f}$ (7)

Being:

$E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R}$ (8)

$E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h}$ (9)

Where:

$E_{o}$ is the initial mechanical energy

$E_{f}$ is the final mechanical energy

$K_{o}$ is the initial kinetic energy

$K_{f}=0$ is the final kinetic energy

$U_{o}$ is the initial gravitational potential energy

$U_{f}$ is the final gravitational potential energy

$m$ is the mass of the object

$V=1510 m/s$ is the radial speed of the object

$h$ is the distance above the surface of the object

Then:

$\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h}$ (10)

Isolating $h$ :

$h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R$ (11)

$h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m$ (11)

$h=397042.215 m$ (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance $h=981.8 km=981.8(10)^{3} m$ . This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

$\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2}$ (13)

Isolating $V$ :

$V=\sqrt{2a_{g} R(1-\frac{R}{R+h})}$ (14)

$V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})}$ (15)

Finally:

$V=1917.76 m/s$

7 0

2 years ago

Why does an ice cube melt in your hand? Question 3 options: The ice remains in a solid state even though heat is applied The tem

iVinArrow [24]

Answer:

3rd choice

Explanation:

heat is transfered causing it to melt

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2 years ago

Read 2 more answers

Find the following answer based on the image.

saul85 [17]

We are given:

Mass of Neptune = 1.03 * 10²⁶ kg

Distance from the center of Neptune (r) = 2.27 * 10⁷

now, computing the value of the acceleration due to gravity (g)

Finding g:

We know the formula:

g = G(mass of planet) / (r)²

g = [6.67 * 10⁻¹¹ * 1.03*10²⁶] / (2.27*10⁷) [since G is 6.67*10⁻¹¹]

g = (6.87 * 10¹⁵) / (5.15 * 10¹⁴)

which can be rewritten as:

g = (6.87 * 10¹⁵ * 10⁻¹⁴) / 5.15

g = (6.87 * 10¹⁵⁻¹⁴) / 5.15

g = (6.87/5.15) * 10

g = 1.34 * 10

g = 13.4 m/s² (approx)

5 0

2 years ago

Describe the advantages and disadvantages of keeping the road clear of ice

harina [27]

A advantage is less car accidents and a disadvantage is, in order to keep the roads clear of ice is a chemical. Like salt. Which is bad for the animals.

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3 years ago