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laiz [17]
2 years ago
13

What happens to the velocity of a sound wave in air if the temperature of the air increases?

Physics
2 answers:
Yuki888 [10]2 years ago
6 0
The answer is D :) hope this helps
Burka [1]2 years ago
5 0
Decreases ( no B is answer)
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What are ways to improve the design of this experiment? Check all that apply
Serhud [2]

Answer:

B,D,E

Explanation:

I got you

B. Experiment with a wider range of materials.

D.Use a laboratory galvanometer to make precise measurements.

E. Test the strength of the electromagnet by varying the number of wire coils.

3 0
3 years ago
Two blocks of masses 3.0 kg and 5.0 kg are connected by a spring and rest on a frictionless surface. They are given velocities t
miskamm [114]

Answer:

-0.7 m/sec

Explanation:

Mass of first block = m1 =3.0 kg

Mass of second block = m2= 5.0 kg

Velocity of first block = V1= 1.2 m/s

Velocity of second block = V2 = ?

Momentum of Center of mass MVcom  is sum of both blocks momentum and is given by

MVcom= m1v1+m2v2

Where

M= mass of center of mass

Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)

Putting values, we get;

0= 3×1.2+5v2

==> v2=  3.6/5= - 0.7 m/s

-ve sign indicates that block 2 is moving in opposite direction of block 1

3 0
3 years ago
Which factors affect the resistance of a material? Check all that apply.
shutvik [7]

Answer: Length, thickness, and temperature

Explanation:

I did it

6 0
2 years ago
The pirate ship tie at the amusement park is a giant pendulum that riders sit in. It swings back and forth, with a maximum veloc
kkurt [141]

Here as we know that there is no loss of energy

so we can say that maximum kinetic energy will become gravitational potential energy at its maximum height

So here we have

\frac{1}{2}mv^2 = mgh

here we have

v = 20 m/s

m = 8000 kg

now from above equation we have

\frac{1}{2}(8000)(20^2) = (8000)(9.8)h

h = \frac{200}{9.8)

h = 20.4 m

so maximum height is 20.4 m

4 0
3 years ago
Read 2 more answers
How much heat is needed to warm 0.072kg of gold from 20 celsius and 90 celsius if the specific heat of gold 136 joules
dybincka [34]

Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.

Mass of the gold (m) = 0.072 kg

Temperature change (ΔT) = 90 - 20 = 70 degree Celsius

Specific heat capacity of the gold (c) = 136 J/kg C

Heat supplied = m × c × ΔT

Heat supplied = 0.072 × 136 × 70

Heat supplied = 685.44 Joules

Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules

5 0
3 years ago
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