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Gnoma [55]
3 years ago
5

What type of reaction occurs when a candle is burned?

Physics
1 answer:
mojhsa [17]3 years ago
8 0
Answer is B - Exothermic because energy is released
You might be interested in
The initial and final velocities of two blocks experiencing constant acceleration are respectively −7.45 m/s and 14.9 m/s. (a) T
ikadub [295]

Answer:

a) a_{1}=3.7 m/s^{2}

b) a_{2}=3.68 m/s^{2}

Explanation:

a) The displacement of the first object is 22.5 m, so we can use the next equation:

v_{f}^{2}=v_{i}^{2}+2a\Delta x

a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}

a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.5}

a_{1}=3.7 m/s^{2}

positive acceleration.

b) Using the same equation we can find the second value of the acceleration:

a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}

a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.6}

a_{2}=3.68 m/s^{2}

positive acceleration.

I hope it helps you!

8 0
3 years ago
The force of gravity considers which two parts?
attashe74 [19]
Mass of object/source
Gravity
Fg=mg
3 0
3 years ago
An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe
nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

f'=\frac{(v+v_o)}{(v+v_s)}f

Where f is the actual frequency, f' is the observed frequency, v is the velocity of the sound waves, v_o the velocity of the observer (which is negative if the observer is moving away from the source)  and v_s the velocity of the source  (which is negative if is moving towards the observer). For this problem:

f=3.72kHz\\v=342m/s\\v_o=0m/s\\v_s=-114m/s

f'=\frac{(342+0)}{(342-114)}3.72\times10^3\\f'=\frac{342}{228}3.72\times10^3\\f'=(1.5)3.72\times10^3\\f'=5580Hz=5.58kHz

5 0
2 years ago
You observe a plane approaching overhead and assume that its speed is 600 miles per hour. The angle of elevation of the plane is
scZoUnD [109]

Answer:4.34 miles

Explanation:

first  Elevation =19^{\circ}

After 1 minute Elevation changes to 59^{\circ}

Ditsance travelled in 1 minute =600\times \frac{1}{60}=10 mile

Now

tan59=\frac{H}{x}

H=xtan59

tan19=\frac{H}{10+x}

H=\left ( 10+x\right )tan19

Equating H

we get

1.319x=10tan19

x=2.61 miles

H=2.61\times tan59=4.34 miles

4 0
3 years ago
A box mass of 24kg is being pulled horizontally on a rough surface by an applied force of 585N. The coefficient of kinetic frict
Elan Coil [88]

the normal force is the force applied opposite to the weight of the was box. So the normal force is equal to the weight of the box = 24 kg *(9.81 m/s2) = 235.44 N

the acceleration of the box be solve using newtons 2nd law of motion:

F = ma

a = F/ m = 585 N/ 24 kg = 24.38 m/s2

8 0
3 years ago
Read 2 more answers
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