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Vesna [10]
3 years ago
9

Lighting a match results in light and heat. Since energy was not created by the match, what transformation took place?

Physics
1 answer:
Yuki888 [10]3 years ago
3 0
Chemical energy stored in the compounds caked on the head of the match, and later the
chemical energy stored in the wood, was released in the form of heat and light when the
chemical compounds got hot enough.
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The velocity of sound in air at 300C is approximately :
kumpel [21]

The velocity of sound in at 300C is 511.3 m/s.

Explanation:

The equation that gives the speed of sound in ar as a function of the air temperature is the following:

v=(331.3+0.6T) m/s

where

T is the temperature of the air, measured in Celsius degrees

In this problem, we want to find the speed of sound in ar for a temperature of

T=300^{\circ}C

Substituting into the equation, we find:

v=331.3 + 0.6(300)=511.3 m/s

So, the velocity of sound in at 300C is 511.3 m/s.

Learn more about sound waves:

brainly.com/question/4899681

#LearnwithBrainly

6 0
3 years ago
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Answer:

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Explanation:

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6 0
3 years ago
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The best way to cool soft and thick foods (such as beans, sauce or chili) when using the refrigerator is?
AnnZ [28]
<span>The best way to cool soft and thick foods when using the refrigerator is by having them to be placed and poured on a pan or another way is by having them to be placed in one container in which they are in a water bath, to be heated of.</span>
3 0
3 years ago
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A rock with a mass of 8 kg falls straight down from a height of 7 m. What work is done?
fenix001 [56]
F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps
7 0
3 years ago
Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da
valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0
3 years ago
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