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Sergeu [11.5K]
2 years ago
8

50 N

Physics
1 answer:
Mariana [72]2 years ago
5 0

The work is done by the force pushing the mower is determined as 500 J.

<h3>Work done by the force</h3>

The work done by the force is calculated as follows;

W = Fsinθ x d

where;

  • Fsinθ = Fx  is the applied force parallel to the ground
  • d is the distance
  • θ is the angle of inclination

W = Fx(d)

W = 25 x 20

W = 500 J

Thus, the work is done by the force pushing the mower is determined as 500 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

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The type of conversion that is taking place when natural gas is


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Please help me**** i need some answers now
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Output can not be greater than input because the conversion of energy can not be greater than 100%.
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An electron of mass 9.11 1031 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to
elena55 [62]

Explanation:

It is given that,

Mass of an electron, m=9.11\times 10^{-31}\ kg

Initial speed of the electron, u=3\times 10^5\ m/s

Final speed of the electron, v=7\times 10^5\ m/s

Distance, d = 5 cm = 0.05 m

(a) The acceleration of the electron is calculated using the third equation of motion as :

a=\dfrac{v^2-u^2}{2d}

a=\dfrac{(7\times 10^5)^2-(3\times 10^5)^2}{2\times 0.05}

a=4\times 10^{12}\ m/s^2

Force exerted on the electron is given by :

F=m\times a

F=9.11\times 10^{-31}\times 4\times 10^{12}

F=3.64\times 10^{-18}\ N

(b) Let W is the weight of the electron. It can be calculated as :

W=mg

W=9.11\times 10^{-31}\times 9.8

W=8.92\times 10^{-30}\ N

Comparison,

\dfrac{F}{W}=\dfrac{3.64\times 10^{-18}}{8.92\times 10^{-30}}

\dfrac{F}{W}=4.08\times 10^{11}

Hence, this is the required solution.

8 0
3 years ago
A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s
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Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0
2 years ago
High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of th
murzikaleks [220]

Answer:

Lifetime = 4.928 x 10^-32 s  

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s  

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2 years ago
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