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m_a_m_a [10]
3 years ago
7

APEX!!!!

Physics
1 answer:
Bogdan [553]3 years ago
7 0
the answer to your question is D:8,000KG M/S
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A lawn roller is rolled across a lawn by a force of 107 N along the direction of the handle, which is 13.5 ◦ above the horizonta
-BARSIC- [3]

Answer:

22.02 m

Explanation:

given,

Force, F = 107 N

angle made with horizontal = 13.5◦

Power develop by the lawn roller = 69.4 W

time = 33 s

distance = ?

Force along horizontal= F cos θ

          = 107 cos 13.5°= 104 N

Power = \dfrac{work\ done}{time}

69.4 = \dfrac{W}{33}

W = 2290.2 J

Work done= Force x displacement

displacement= \dfrac{2290.2}{104}

                      = 22.02 m

6 0
3 years ago
Which of the following is an example of kinetic energy?
Elza [17]
The answer is C) a rolling bowling ball because kinetic energy is the energy of movement and potential energy is the energies of the others, since there are not in movement. i hope this helps.
7 0
3 years ago
How could a homework machine help students all around the world?
slamgirl [31]

Answer:

Hello There!!

Explanation:

The machine can help you alot and do all your homework while you rest and let the homework machine do it and get good grades because of it with no worries but if the machine does it then you won't learn how to do them.

hope this helps,have a great day!!

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8 0
3 years ago
A force of 400-N pushes on a 25-kg box horizontally. The box accelerates at 9 m/s? Find the coefficient of kinetic friction betw
umka2103 [35]

Answer:

<h3>0.69</h3>

Explanation:

Using the Newtons law of motion;

\sum Fx = ma_x\\Fm - Ff = ma_x

Fm is the moving force = 400N

Ff is the frictional force = μR

μ is the coefficient of kinetic friction

R is the reaction = mg

m is the mass

a is the acceleration

The equation becomes;

Fm - \mu R = ma_x\\Fm - \mu mg = ma_x\\400- \mu (25)(9.8) = 25(9)\\400 - 254.8 \mu = 225\\- 254.8 \mu = 225 - 400\\- 254.8 \mu = -175\\ \mu = \frac{-175}{- 254.8} \\\mu = 0.69

Hence the coefficient of kinetic friction between the box and floor is 0.69

7 0
3 years ago
Calculate the solubility (in m units) of ammonia gas in water at 298 k and a partial pressure of 2.50 bar . the henry's law cons
Eduardwww [97]
Henry's Law (formulated in 1803 by William Henry) states that aa constant temperature, the amount of gas dissolved in a liquid is directly proportional to the partial pressure exerted by that gas on the liquid.
 Mathematically it can be formulated as
 C = H⨯P
 being:
 C: the molar concentration of dissolved gas A,
 P: the partial pressure of it
 H: Henry's constant
 Substituting:
 C = P * H
 C = (2.50 * 0.9869) * 58.0
 C = 143.1
 Answer:
 the solubility (in m units) is
 C = 143.1
8 0
3 years ago
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