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2 years ago

Baking cake is what a physical or chemical change

Physics

1 answer:

xenn [34]2 years ago

8 0

Answer:

It's a chemical change.

Explanation:

The ingredients will react together once mixed and heated. Also the ingredients can not be pulled apart.

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If the frequencies of two component waves are 24 Hz and 20 Hz, they should produce _______ beats per second.

horrorfan [7]

This can be answered using the beat frequency formula, which is simply the difference between 2 frequencies.

Let: <span>fᵇ = beat frequency
</span>f₁ = first frequency
f₂ = second frequency

fᵇ = |f₁ - f₂|

substituting the values:
fᵇ = |24Hz - 20Hz|
fᵇ = 4Hz

The unit Hz also means beats per second, therefore:
<span>fᵇ = 4 beats per second
</span>
Therefore, the answer is C. 4

8 0

3 years ago

Read 2 more answers

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu

dalvyx [7]

Answer:

The frictional torque is $\tau = 0.2505 \ N \cdot m$

Explanation:

From the question we are told that

The mass attached to one end the string is $m_1 = 0.341 \ kg$

The mass attached to the other end of the string is $m_2 = 0.625 \ kg$

The radius of the disk is $r = 9.00 \ cm = 0.09 \ m$

At equilibrium the tension on the string due to the first mass is mathematically represented as

$T_1 = m_1 * g$

substituting values

$T_1 = 0.341 * 9.8$

$T_1 = 3.342 \ N$

At equilibrium the tension on the string due to the mass is mathematically represented as

$T_2 = m_2 * g$

$T_2 = 0.625 * 9.8$

$T_2 = 6.125 \ N$

The frictional torque that must be exerted is mathematically represented as

$\tau = (T_2 * r ) - (T_1 * r )$

substituting values

$\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )$

$\tau = 0.2505 \ N \cdot m$

5 0

2 years ago

A secret agent is locked in a room. He pushes against the door but cannot open it. Finally, he falls to the floor exhausted. Has

Vitek1552 [10]

Answer: There is not work done at the door because the door did not move.

Explanation: Work is defined as the movement done by a force.

So if you move to apply a force F in an object and you move it a distance D, the work applied on the object is

W = F*D

In this case, the secret agent pushes against the door, so there is a force, but the agent does not move the door, so D = 0, so there is no motion of the door, which implies that there is no work done at the door.

8 0

2 years ago

Does this look like a good circuit diagram?

AnnyKZ [126]

It's hard to tell what's going on down there in the corner with the resistor and the ammeter. There seems to be as many as 3 or 4 wires in and out of the ammeter, which would be wrong. A real ammeter only has two ... one in and one out. (Same for a resistor.)

It's hard to say whether this circuit works, until we can clearly understand how everything is hooked up in that corner of the drawing.

6 0

2 years ago

The 1.5-kg collar C starts from rest at A and slides with negligible friction on the fixed rod in the vertical plane. Determine

arlik [135]

Answer:

Explanation:

check attached image for figure, there is supposed to be a figure for this question containing a distance(height of collar at position A) but i will assume 0.2m or 200mm

Consider the energy equilibrium of the system

$U_{A-B}=\bigtriangleup T\\\\F\cos 30°\times h_A - F\sin30°\times h_A + Wh_A=\frac{1}{2}m(v^2_B-v^2_A)\\\\v_B=\sqrt{\frac{2Fh_A(\cos 30° - \sin30°)+mgh_A}{m+v^2_A}}$

Here, F is the force acting on the collar, $h_A$ is the height of the collar at position A, m is the mass of the collar C, g is the acceleration due to gravity, $v_B$ is the velocity of the collar at position B, and $v_A$ is the velocity of the collar at A

Substitute 14.4N for F, 0.2m for $h_A$ , 1.5kg for m, $9.81m/s^2$ for g and 0 for $v_A$

$v_B=\sqrt{\frac{2(14.4\times 0.2(\cos 30° - \sin30°)+1.5\times 9.81\times 0.2}{1.5+0}}\\\\=\sqrt{\frac{6.618}{1.5}}\\\\=4.412m/s$

Therefore, the velocity at which the collar strikes the end B is 4.412m/s

8 0

2 years ago