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FinnZ [79.3K]
3 years ago
15

A 5.0-kg clay putty ball and a 10.0-kg medicine ball are headed towards each other. Both have the same speed of 20 m/s. If they

collide perfectly inelastically, what approximately is the speed of the blob of clay and ball immediately after the collision?
Physics
1 answer:
babymother [125]3 years ago
8 0

Answer:

- 3.33 m /s

Explanation:

m1 = 5 kg

m2 = 10 kg

u1 = 10 m/s

u2 = - 10 m/s

Let the velocity of the combination of the blob is v.

Use the conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=\left ( m_{1}+m_{2} \right )v

5 x 10 - 10 x 10 = (5 + 10)v

50 - 100 = 15 v

v = - 3.33 m /s

Thus, the velocity of the blob after sticking together is - 3.33 m/s.

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A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe
Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet =  327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf  = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

     = 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

   = 0.00353 × 128 × (327-40)

   = 0.00353 × 128 × 287

   = 129.68 J

Quantity of heat required to melt the bullet:

Q = mH_f

Q = 0.00353 × 24500

   = 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

                                                                      = 216.17 J

3 0
3 years ago
A long cylindrical capacitor is made of a central wire of radius a = 2.50 mm surrounded by a conducting shell of radius b = 7.50
asambeis [7]

Answer:

The capacitance per unit length is 5.06\times10^{-11}\ F/m

(b) is correct option.

Explanation:

Given that,

Radius a= 2.50 mm

Radius b=7.50 mm

Dielectric constant = 3.68

Potential difference = 120 V

We need to calculate charge per length for the capacitance

Using formula of charge per length

\lambda=\dfrac{4\pi\epsilon_{0}\Delta V}{2 ln(\dfrac{r_{2}}{r_{1}})}

Put the value into the formula

\lambda=\dfrac{120}{9\times10^{9}\times2 ln(\dfrac{7.50\times10^{-3}}{2.50\times10^{-3}})}

\lambda=6.068\times10^{-9}\ C/m

We know that,

\lambda=\dfrac{Q}{L}

We need to calculate the capacitance per unit length

Using formula of capacitance per unit length

C=\dfrac{\dfrac{Q}{L}}{\Delta V}

C=\dfrac{6.068\times10^{-9}}{120}

C=5.06\times10^{-11}\ F/m

Hence, The capacitance per unit length is 5.06\times10^{-11}\ F/m

7 0
3 years ago
(a) A long, straight solenoid has N turns, uniform cross-sectional area A, and length l. Show that the inductance of this soleno
Paul [167]

Answer:

a. L = μ₀AN²/l b. 1.11 × 10⁻⁷ H

Explanation:

a. The magnetic flux through the solenoid, Ф = NAB where N = number of turns of solenoid, A = cross-sectional area of solenoid and B = magnetic field at center of solenoid = μ₀ni where μ₀ = permeability of free space, n = number of turns per unit length = N/l where l = length of solenoid and i = current in solenoid.

Also, Li = Ф where L = inductance of solenoid.

So, Li = NAB

= NA(μ₀ni)

= NA(μ₀Ni/l)

Li = μ₀AN²i/l

dividing both sides by i, we have

So, L = μ₀AN²/l

b. The self- inductance, L = μ₀AN²/l where

A = πd²/4 where d = diameter of solenoid = 0.150 cm = 1.5 × 10⁻³ m, N = 50 turns, μ₀ = 4π × 10⁻⁷ H/m and l = 5.00 cm = 5 × 10⁻² m

So, L = μ₀AN²/l

L = μ₀πd²N²/4l

L = 4π × 10⁻⁷ H/m × π(1.5 × 10⁻³ m)²(50)²/(4 × 5 × 10⁻² m)

L = 11,103.3 × 10⁻¹¹ H

L = 1.11033 × 10⁻⁷ H

L ≅ 1.11 × 10⁻⁷ H

6 0
3 years ago
What must happen to an atom of magnesium in order to become a magnesium ion Mg+2?
igomit [66]

Answer:

Answer is: c. It must lose two electrons and become an ion.

Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

Explanation:

4 0
3 years ago
Read 2 more answers
When monochromatic light illuminates a grating with 7000 lines per centimeter, its second order maximum is at 62.4°. what is the
zhannawk [14.2K]

When red light illuminates a grating with 7000 lines per centimeter, its second maximum is at 62.4°. What is the wavelength of this light?

ans: 633nm

6 0
3 years ago
Read 2 more answers
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