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slamgirl [31]
2 years ago
6

A 2,493-kg car is moving down a road with a slope (grade of 14% at a constant speed of 13 m/s. what is the direction and magnitu

de of the frictional force? (define positive x in the forward direction, i.e., down the slope?
Physics
1 answer:
sweet [91]2 years ago
3 0
To calculate for the frictional force, we calculate first for the normal force (forcer perpendicular to the surface)
                               Fn = (2,493 kg) (9.8 m/s²) = 24243.51 N
Then, we multiply this value with the frictional constant which is not given in this problem. If the frictional coefficient is equal to x then, the frictional force is equal to 24243.51x. 
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A spring of spring constant 25 N/m is hung vertically and a 0.300 kg mass is attached to one end, causing a displacement of the
Mamont248 [21]

Answer:k=1175

Explanation:thank  you for asking

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Will give correct answer brainliest<br><br>5 kg m/s<br>8kg m/s<br>80 kg m/s<br>200 kg m/s​
o-na [289]

Answer: Here this will help you..

Explanation:

1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second

5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second

10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second

20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second

30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second

40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second

50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second

75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second

100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second

8 0
2 years ago
Question<br> In order for work to be done, what three things are necessary
Misha Larkins [42]
Perseverance, good mind set, and work ethic
5 0
3 years ago
Read 2 more answers
A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?
Yanka [14]

Answer:

\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 380.2 \times  {11}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 190.1 \times 121

\sf \implies KE = 190.1 \times 121

\sf \implies KE = 23002.1 \: J

\therefore

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0
3 years ago
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