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Cerrena [4.2K]
3 years ago
13

A 0.07-kg lead bullet traveling 258 m/s strikes an armor plate and comes to a stop. If all of the bullet's energy is converted t

o heat that it alone absorbs, what is its temperature change?
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
7 0

Answer:

temperature change is 262.06°K

Explanation:

given data

mass = 0.07 kg

velocity = 258 m/s

to find out

what is its temperature change

solution

we know here

heat change Q is is equal to kinetic energy that is

KE = 0.5 × m× v²   ...........1

here m is mass and v is velocity

KE = 0.5 × 0.07 × 258²

KE = 2329.74 J

and we know

Q = mC∆t     .................2

here m is mass and ∆t is change in temperature and C is 127J/kg-K

so put here all value

2329.74 = 0.07 × 127 × ∆t

∆t = 262.06

so temperature change is 262.06°K

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a 100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. When the mixture reaches equillibriu
Artyom0805 [142]

Answer:

The latent heat of fusion of water is 334.88 Joules per gram of water.

Explanation:

Let the latent heat of ice be 'x' J/g

1) Thus heat absorbed by 100 gram of ice to get converted into water equals

Q_1=100\times x

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

Q_2=100\times 4.186\times 11=4604.6Joules

Thus total energy needed equals Q_1+Q_2=100x+4604.6

3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

Q_3=650\times 4.186\times (25-11)\\\\Q_{3}=38092.6Joules

Now by conservation of energy we have

Q_1+Q_2=Q_3\\\\100x+4604.6=38092.6\\\\\therefore x=\frac{38092.6-4604.6}{100}=334.88J/g

6 0
3 years ago
2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo
seropon [69]

Answer:

The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=2\pi f

Put the value into the formula

\omega=2\times\pi\times60

\omega=376.9\ rad/s

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}

B_{max}=2.901\times10^{-13}\ T

Hence, The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

7 0
3 years ago
A ball rolls down an incline with an acceleration of 10 cm/s^2. If it starts with an initial velocity of 0 cm/s and has a velocy
11Alexandr11 [23.1K]
Given a = 10 cm/s²
          u = 0 cm/s
          v = 50 cm/s
we know that 
         v²=u²+2aS
        2500=2×10×S
        2500÷20 = S
        S= 125 cm
The ramp is 125 cm
8 0
3 years ago
A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25
Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress  = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter  d₂ =12in  d₁= 12 -4in = 8in

4 -in.-thick  

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

           = 66/ (П* (d₂²-d₁²)/4

           =66/ (3.142* (12²-8²)/4

          = 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

           = 66* cos 25/ (П* (d₂²-d₁²)/4

           =59.82/ (3.142* (12²-8²)/4

          = 59.82/62.84 = 0.952kips/in²

shearing stress  = tangential stress /2

                           = T/2 = 0.952/2 = 0.476 kips/in²

8 0
3 years ago
Explain why a 10 kg stone and 1 kg stone dropped from the same height
11111nata11111 [884]
The acceleration of a body in a gravitational field is independent of its mass. Both the stones will fall with the same acceleration through the same height and hence they will strike the ground simultaneously.
GIVE ME POINTS
3 0
3 years ago
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