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3 years ago

How much energy, in joules, is released when 17 kilograms of granite is cooled from 45°C to 21°C?

2 answers:

4 0

The final answer is -322, 320 Joules. The solution for the problem is:

The equation that must be used in this problem is:

U = mCp(Tf-Ti)

where:
U = energy released or absorbed in Joules
m = mass in kg
Cp = specific heat of material in J/kg-C
Tf = final temperature, C
Ti = initial temperature, C

Looking up the Cp of granite gives a value of 790 J/kg-C.

U = 17kg (790 J/kg-C) (21-45)
<span>U = - 322, 320 Joules (negative means heat is released)</span>

pashok25 [27]3 years ago

4 0

The awnser would be D

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In a physics lab experiment, a spring clamped to the table shoots a 22g ball horizontally. When the spring is compressed 21cm ,

worty [1.4K]

a square and a triangle I think

6 0

4 years ago

Calculate the wave length of a water wave with a speed of 20 m/s and a frequency of 2.5 Hz

12345 [234]

Wavelength of the water wave is 8 m

Explanation:

Wavelength measures the distance between two successive crests or troughs of the wave. It is given by the following equation

λ = v/f, where f is the frequency, v is the velocity of the wave

Here, v = 20 m/s and f = 2.5 Hz

⇒ λ = 20/2.5

= 8 m

5 0

3 years ago

A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi

Alex_Xolod [135]

Magnitude: 12.1 N.
Direction: 17.0° to the 8 N force.

<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

$\displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .
$\displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .

The sum of forces on each direction will be the resultant force on that direction:

Resultant force parallel to the 8 N force: $(8 + \dfrac{5\sqrt{2}}{2})\;\text{N}$ .
Resultant force normal to the 8 N force: $\dfrac{5\sqrt{2}}{2}\;\text{N}$ .

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

$\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N}$ (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

$\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491$ .

Find the size of the angle using inverse tangent:

$\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree$ .

In other words, the resultant force is 17.0° relative to the 8 N force.

4 0

3 years ago

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago

Does a thicker core make an electromagnet stronger?

goldenfox [79]

Answer:

Yes

Explanation:

The core of an electromagnet serves to stabilize the magnetic field created by the wire. The thicker the core, the more metal there is to amplify the current. Therefore, a thicker core does make an electromagnet stronger. Hope this helps!

5 0

4 years ago