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Aleksandr-060686 [28]
3 years ago
13

How does medium affect the level/amplitude of sound?

Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
7 0
Stop drunk drive fast go
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Why do blades come in different lengths in a jig saw sander?​
trasher [3.6K]

Answer:

To determine the minimum blade length, add 1" to the workpiece thickness. One type of material, and some materials can be cut by more than one type of blade. No matter the material, there's likely a jigsaw blade designed specifically for. Armed with the right blade, follow these pointers to make your work go (and cut) .

Explanation:

6 0
2 years ago
In longitude waves, the medium moves parallel to the direction of energy transport
adelina 88 [10]
Yep that's correct
And transverse waves move perpendicular to the direction of energy transport
7 0
3 years ago
A 4kg book sits on a table and your pet hamster puts his front paws on the book and pushes down with a force of 3N. What is the
Natali [406]
There are three forces acting on the book. 
1. Force due to gravity
2. Force exerted downward by the hamster
3. Normal Force in reaction to the downward forces

Since the book is not moving, the net force is zero. The summation of all forces must be zero. Then we could find the normal force which is unknown (denoted as x).

∑F = -(4 kg)(9.81 m/s2) - 3 N + x =0
∑F = -39.24 N - 3N + x =0
x = 42 N

Therefore, the normal force is 42 N.
4 0
3 years ago
nan moved 18m to the right and then another 22m to the right if the motion takes 20 seconds what is nans velocity
IrinaVladis [17]
Step 1: list known info

distance(change in position (Δx))= 18m+22m= 40m
time= 20 seconds

Step 2 :solve for velocity

velocity= Δx÷time
v= 40/20= 2m/s

Answer: the velocity is 2 meters per a second (m/s)


7 0
3 years ago
A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the
podryga [215]

Answer:

6.9\times 10^{28}m^{-3}

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=rr=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m

1mm=10^{-3} m

Current=I=8 A

Drift velocity=v_d=5.4\times 10^{-5} m/s

We have to find the density of free electrons in the metal

We know that

Density of electron=n=\frac{I}{v_deA}

Using the formula

Density of free electrons=\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}

By using Area of wire=\pi r^2

\pi=3.14\\e=1.6\times 10^{-19} C

Density of free electrons=6.9\times 10^{28}m^{-3}

3 0
3 years ago
Read 2 more answers
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