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ohaa [14]
3 years ago
10

. The formation of carbon dioxide from carbon and oxygen is an example of a reaction

Physics
2 answers:
Reika [66]3 years ago
8 0
Photosynthesis maybe
andre [41]3 years ago
3 0
An oxidation reaction
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A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surfa
Burka [1]

Answer:

(a) = 0 N

(b) = 2.4 N

Explanation:

given

box of banana weight = 40.0 N

coefficient of static friction  μ = 0.40

coefficient of kinetic friction = 0.20

a).  when no horizontal force is applied .

according to Newton 's third law of motion If there is no force applied to the box,so the frictional  force exerted is 0 N

b) magnitude of friction force

box and the box is initially at rest

    friction force =.Static friction coefficient × weight of the box

      friction force = 0.40 × 6

       friction force =  2.4 N

8 0
3 years ago
A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its
Savatey [412]

Answer:

a). E_{kA}=1.2635 J

b). V_{B}=4.535\frac{m}{s}

c). ΔE_{t}=8.4635 J

Explanation:

ΔE=kinetic energy

a).

E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J

b).

E_{kB}=\frac{1}{2}*m*v_{B} ^{2}

V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}

c).

net work= EkA+EkB

E_{t}=E_{kA}+ E_{kB}\\E_{t}=1.2635J+7.2J\\E_{t}=8.4635J

3 0
3 years ago
a small asteroid of mass 125 kg is orbiting a planet that has a mass of 3.52x 10^13 what is the radial distance between the aste
asambeis [7]

Answer:

r = 2.031 x 10⁶ m = 2031 km

Explanation:

In order for the asteroid to orbit the planet, the centripetal force must be equal to the gravitational force between asteroid and planet:

Centripetal Force = Gravitational Force

mv²/r = GmM/r²

v² = GM/r

r = GM/v²

where,

r = radial distance = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Planet = 3.52 x 10¹³ kg

v = tangential speed = 0.034 m/s

Therefore,

r = (6.67 x 10⁻¹¹ N.m²/kg²)(3.52 x 10¹³ kg)/(0.034 m/s)²

<u>r = 2.031 x 10⁶ m = 2031 km</u>

7 0
3 years ago
Marys airplane trip took 5.8 hours for one-half of that time, the airplane flew at a constant speed of 640 miles per hour and fo
Olenka [21]
Distance is speed x time.  Half of the trip is 5.8/2 = 2.9hrs.
640 x 2.9 = 1856mi
580 x 2.9 = 1682mi
1856mi+1682mi=3538mi.

You could also calculate her average speed.  This is easy since it was divided in two equal time slices.  Average Speed = (640+580)/2 = 610mi/hr
Now 610mi/hr x 5.8hrs = 3538mi
7 0
3 years ago
What is the speed of a wave that has a frequency of 25Hz and a wavelength of 3m
Arada [10]
The answer to the question is:

75m/s

Just do 25*3
4 0
3 years ago
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