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uysha [10]
3 years ago
6

The surface of an incline is coated with an experimental substance that is intended to reduce the frictional force between a blo

ck and the
surface of the incline. A 2 kg block is placed at the top of the incline at a height of 1.8 m, as shown in the figure. After the block is released
from rest, the block slides down the incline and a motion detector at the bottom of the incline measures the block's speed as 5.8 after the
block is no longer on the incline. Which of the following claims is correct about the experimental substance?
The experimental substance reduced all the frictional force because all the gravitational potential energy of the Earth-block
system at the top of the incline was converted into the kinetic energy of the block at the bottom of the incline.
The experimental substance did not reduce all the frictional force because some of the gravitational potential energy of the
Earth-block system at the top of the incline was converted into nonmechanical energy
с
The effectiveness of the experimental substance cannot be determined because the speed of the block at the bottom of the
inline, as measured by the motion detector, indicates that the block has more energy at this location than the Earth-block
system had at the top of the incline.
D
The effectiveness of the experimental substance cannot be determined without knowing the magnitude of the frictional
force between the block and the incline before and after the experimental substance was applied to the incline.
Lo

Physics
2 answers:
ankoles [38]3 years ago
6 0

Answer:

b

Explanation:

nikklg [1K]3 years ago
4 0

Answer:

粗糙錯愕額外此時自由KK預約為我

Explanation:

因為嗚嗚嗚粗糙此次四月蘇澳碩果僅存有一位我也在鶯鶯燕燕

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Find the net force and acceleration. 15 points. Will give brainliest!
gladu [14]

Answer:

\boxed{F_{net} = 28.7 \ N}

\boxed{a = 2.1 \ m/s^2}

Explanation:

<u><em>Finding the net force:</em></u>

<u><em>Firstly , we'll find force of Friction:</em></u>

F_{k} = (micro)_{k}mg

Where (micro)_{k} is the coefficient of friction and m = 13.6 kg

F_{k} = (0.16)(13.6)(9.8)\\

F_{k} = 21.32 \ N

<u><em>Now, Finding the net force:</em></u>

F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\

F_{net} = 28.7 \ N

<u><em>Finding Acceleration:</em></u>

a = \frac{F_{net}}{m}

a = \frac{28.7}{13.6}

a = 2.1 \ m/s^2

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4 years ago
Of the more than 2,000 known minerals approximately 95% of them belong to what mineral group?
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5 0
3 years ago
804 n of force are applied to a 51.7 kg. What is the acceleration that the object experiences?
Andreyy89

We can use Newton II here  (where F=m*a), that F is the net (or resultant) force on the object, m is the mass of the object and a is the acceleration the object experiences.

This means, in this case there would be no friction and absolutely no other force which gives a component in the plane of motion, only then can you assume that F=804N.

Now using F= m*a

804 = 51.7*a

Therefore a = 804/51.7 = 15.55 m/s²


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