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3 years ago

Calculate the heat required, in calories, to raise

Physics

1 answer:

Goshia [24]3 years ago

5 0

Answer: $Q=4450.74\ J$

Explanation:

Given

mass of tin $m=100 gm$

Also we know that

specific heat of tin $c_{tin}=0.21 J/gm-^{\circ}C$

Melting Point of Tin $T=231.94^{\circ}C$

Heat required to raise the temperature of 100 gm tin is given by

$Q=mc\Delta T$

where $Q=heat\ added$

$c=specific\ heat\ of\ tin$

$\Delta T=change\ in\ temperature$

$Q=100\times 0.21\times (231.94-20)$

$Q=4450.74\ J$

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3 years ago

The crew of an enemy spacecraft attempts to escape from your spacecraft by moving away from you at 0.259 of the speed of light.

Vladimir79 [104]

If you do not have to use relative physics but classic physics, this is how you solve it:

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Speed of your foe respect to you: 0.259c

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Speed of the torpedo respect your foe: 0.349c - 0.259c = 0.09c

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3 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$

If we manipulate the equation we have that:

$k=\frac{2mgh}{d^{2} }$

$k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}$

$k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}$

$k=243\ \frac{N}{m}$

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3 years ago

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Natalka [10]

She is traveling at a constant speed.

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3 years ago

Read 2 more answers

The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The r

ira [324]

Given Information:

Number of turns = N = 84

Area of Rectangular coil = 2.61x3.64 cm = 0.0261x0.0364 m

Magnetic field = B = 0.80 T

Current = I = 10.5 mA = 0.0105 A

Angular speed = ω = 3.54x10³ rev/min

Required Information:

(a) Maximum torque = τmax = ?

(b) Peak output power = Ppeak = ?

(d) Average power = Pavg?

Answer:

(a) Maximum torque = 0.00067 N.m

(b) Peak output power = 0.248 W

(d) Average power = 0.1115 W

Explanation:

(a) The toque τ acting on the rotor is given by,

τ = NIABsin(θ)

Where N is the number of turns, I is the current, A is the area of rectangular coil and B is the magnetic field

A = 0.0261x0.0364

A = 0.00095 m²

The maximum toque τ is achieved when θ = 90°

τmax = NIABsin(90°)

τmax = 84*0.0105*0.00095*0.80*1

τmax = 0.00067 N.m

(b) The peak output power of the motor is given by,

Pmax = τmax*ω

ω = 3.54x10³ x 2π/60

ω = 370.7 rad/sec

Pmax = 0.00067*370.7

Pmax = 0.248 W

W = 2∫NIABωsin(ωt) dt

W = -2NIABcos(ωt)

Evaluating limits,

W = -2NIABcos(π) - (-2NIABcos(0))

W = 2NIAB + 2NIAB

W = 4NIAB

W = 4*84*0.0105*0.00067*0.80

W = 0.00189 J

(d) Average power of the motor is given by

Pavg = W/t

t = 2π/ω

t = 2π/370.7

t = 0.01694 sec

Pavg = W/t

Pavg = 0.00189/0.01694

Pavg = 0.1115 W

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3 years ago