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Natasha2012 [34]

2 years ago

7

7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained

by the hammer head, how much will its temperature increase

1 answer:

marta [7]2 years ago

8 0

The increase in temperature of the metal hammer is 0.028 ⁰C.

The given parameters:

<em>mass of the metal hammer, m = 1.0 kg</em>
<em>speed of the hammer, v = 5.0 m/s</em>
<em>specific heat capacity of iron, 450 J/kg⁰C</em>

The increase in temperature of the metal hammer is calculated as follows;

$Q = K.E\\\\mc \Delta T = \frac{1}{2} mv^2\\\\\Delta T = \frac{v^2}{2 c}$

where;

<em>c is the </em><em>specific heat capacity</em><em> of the metal hammer</em>

<em />

Assuming the metal hammer is iron, c = 450 J/kg⁰C

$\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C$

Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.

Learn more about heat capacity here: brainly.com/question/16559442

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Answer:

$r_{cm} = 2520.5 km$

Explanation:

As we know that mass is the product of volume and density

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so we will have

$\frac{M_p}{M_c} = (\frac{r_p}{r_c})^3$

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3 years ago

Which of the following is a good example of a physical change?

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3 years ago

A horizontal bend in a pipeline conveying 1 m3/s of water gradually reduces from 600 mm to 300 mm in diameter and deflects the f

Brums [2.3K]

Answer:

The force excerted on the bend = 45.3 kN

This happens under an angle of tan^-1(1.7/4.2) = 22° to the x-direction

Explanation:

Step 1: Data given

A horizontal bend in a pipeline conveying 1m³/s of water

Diameter reduces from 600 mm to 300 mm

angle = 60°

At the larger end, the pressure = 170 KN/m²

Step 2:

1 m³/s = A1 *V1 = A2*V2

⇒ with A1 = the area at side 1

⇒ with A2 = the area at side 2

V1 = 1/((π/4)(0.6²)) = 3.537 m/s

V2 = 1/((π/4)(0.3²)) = 14.147 m/s

p1/p*g + (V1)²/2g = p2/p*g + (V2)²/2g

⇒ with p1 =170 *10³ N/m²

⇒ with p = 10³

⇒ with g = 9.81 m/s²

⇒ with p2 = TO BE DETERMINED

⇒ with V1 = 3.537 m/s

⇒ with V2 = 14.147 m/s

p2/fg = (170*10³)/(10³*9.81) + (3.537²)/(2*9.81) - (14.147²)/(2*9.81)

p2/fg = 7.767

p2 = 7.767 * 9810

p2 = 7.62*10^4 N/m²

Gravity forces are 0 along the horizontal plane. The only forces acting on the fluid mass = pressure and momentum forces.

Let's consider Fx and Fy as 2 components of total force F excerted by the bent boundary surface on the fluid mass.

In x-direction we have: p1*A1 + Fx - p2A2cos∅ = pQ(V2cos∅ - V1)

In y-direction we have: 0 + Fy - p2A2sin∅ = pQ(V2sin∅-0)

Fx = 10³*(14.147cos60° - 3.537) + 7.62 * 10^4 * π/4 *(0.3²) * cos60° - 170*10^3 * π/4 *(0.6²)

Fx = -4.2 *10^4 N (The negative sign shows the direction to the left)

Fy = 10³*(14.147sin60°) + 7.62 *10^4 * π/4 *(0.3²) *sin 60°

Fy = 1.7*10^4 N ( The positive sign shows the direction is upwards)

The law of motion says: the forces Rx and Ry excerted by the fluid on the bend will be equal and opposite to Fx and Fy:

Rx = -Fx = 4.2 *10^4 N ( positive sign means direction to the right)

Ry = -Fy = -1.7 *10^4 N (Negative sign means direction downwards)

The resultant force on the bend:

R = √((Rx)² + (Ry)²

R = √((4.2 * 10^4)² + (-1.7*10^4)²)

R = 45310 N = 45.3 * 10³ N = 45.3 kN

The force excerted on the bend = 45.3 kN

This happens under an angle of tan^-1(1.7/4.2) = 22° to the x-direction

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3 years ago

To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi

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Answer:

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Explanation:

a) The acceleration of the positron can be found as follows:

$F = q*E$ (1)

Also,

$F = ma$ (2)

By entering equation (1) into (2), we have:

$a = \frac{F}{m} = \frac{qE}{m}$

<u>Where:</u>

F: is the electric force

m: is the particle's mass = 9.1x10⁻³¹ kg

q: is the charge of the positron = 1.6x10⁻¹⁹ C

E: is the electric field = 286 N/C

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b) The positron's speed can be calculated using the following equation:

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<u>Where</u>:

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I hope it helps you!

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2 years ago