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3 years ago

Does sound travel faster in a warm room or a cold room? Explain your answer.

Physics

2 answers:

notsponge [240]3 years ago

4 0

Warm air because cold air makes the molecules more close

Mashcka [7]3 years ago

3 0

A warm room. sound travels faster in places with more excited molecules.

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in a lever, a load of 600N is lifted by using 400 effort. If the load is at the distance of 20cm and the effort at the distance

koban [17]

Explanation:

Load (l) = 680N

Effort (E) = 500N

Length slope (l) = 12m

Height slope (h) = 8 m

Output = load * height

680 *8 = 5.44 *103 J

The Input = effort * length = 500 *12 = 6000J

the Mechanical advantage (M.A) = load effort= 600500=1.36

the Velocity ratio (V.R) =lh=128 = 1.5

the Efficiency =M.A100%V.R= 90.6%

8 0

3 years ago

Use Hooke's Law, which states that the distance a spring stretches (or compresses) from its natural, or equilibrium, length vari

Phantasy [73]

Answer:

706.68 N

Explanation:

By Hooke's law,

$F = ke$

$k=\dfrac{F}{e}$

Using the values in the question,

$k=\dfrac{265\text{ N}}{0.15 \text{ m}}=1766.7\text{ N/m}$

When e = 0.4 m,

$F = 1766.7\text{ N/m}\times0.4\text{ m}=706.68\text{ N}$

6 0

3 years ago

Plzzz Help

m_a_m_a [10]

Answer:

The electromagnetic waves appear more blue in color.

Explanation:

Doppler's Effect: When a source moves with respect to the observer the frequency of the wave emitted from the source changes. If the source moves away from the observer, the frequency decreases and wavelength increases and vice versa.

Here the light source is moving towards the observer so the frequency will increase and wavelength will decrease. Thus the spectrum will shift towards the blue part. This is known as blue shift. The light wave will appear blue in color.

8 0

3 years ago

a shopper pushes a cart 40.0m south down one aisle and then turns 90.0 degrees and moves 15.0m. He then makes another 90.0 degre

valentinak56 [21]

Answer:

Explanation:

The displacement is the distnce of the shopper from the starting point.

Sum of movement along the vertical = 40-20 = 20m

Movement along the horizontal (x direction) = 15.0m

Displacement will be gotten using the pythagoras theorem.

d = √20²+ 15²

d = √400+225

d = √625

d = 25.0m

Hence the shoppers total displacement is 25.0m

8 0

3 years ago

Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera

torisob [31]

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

No intersection: There's nothing that blocks the camera's view of the top of the building.
Two intersections: The planet blocks the camera's view of the top of the building.
One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle $\angle \mathrm{B\hat{C}D}$ which corresponds to this minor arc.

This angle comes can be split into two parts:

$\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}$ .

Also,

$\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}$ .

The radius of this circle is:

$\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}$ .

The lengths of segment DC, AC, BC can all be found:

$\rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m$ ;
$\rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m$ ;
$\rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m$ .

In the two right triangles $\triangle\mathrm{DAC}$ and $\triangle \rm BAC$ , the value of $\angle \mathrm{B\hat{C}A}$ and $\angle \mathrm{A\hat{C}D}$ can be found using the inverse cosine function:

$\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}$

$\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}$

$\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}$ .

The length of the minor arc will be:

$\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km$ .

5 0

3 years ago