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2 years ago

7

The specific heat capacity of copper is 385 J/kg/°C. Calculate the energy needed to increase the temperature of a 4 kg block of

copper from 15°C to 33°C.

1 answer:

Vesnalui [34]2 years ago

6 0

Answer:

heat energy =mc∆t

Q = 4×385×[33-15]

Q = 4×385×18

Q = 27720joule

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3 years ago

A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline

balandron [24]

Answer:

<em>2.78m/s²</em>

Explanation:

Complete question:

<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>

According to Newton's second law of motion:

$\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\$

Where:

$\mu$ is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

$\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2$

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

$gsin\theta - \mu g cos\theta = a_x\\$

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>

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3 years ago

A truck starts off 151 miles directly north from the city of Hartville. It travels due east at a speed of 41 miles per hour. Aft

Bess [88]

Answer:

9.51

Explanation:

The distance s is given by:

$s(t) = \sqrt{151^2 + (vt)^2}$

The change in distance is given by the time derivative of s:

$\frac{ds}{dt} = \frac{v^2t}{\sqrt{151^2 + (vt)^2}}$

For the time t you solve the equation of distance x for time:

$x = vt => t = \frac{x}{v}$

Plugging in for t:

$\frac{ds}{dt}(t=\frac{x}{v})=\frac{vx }{\sqrt{151^2 + x^2}}=9.51$

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2 years ago