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hoa [83]
3 years ago
6

A 0.50-kg block slides across a tabletop with an initial velocity of 20 cm/s and comes to rest in a distance of 70 cm. Find the

friction force that act on the block.​
Physics
2 answers:
mars1129 [50]3 years ago
6 0

Answer:

FAE= 0.014 N

Explanation:

The KE of block is decreased because of the slowing action of the friction force .

Change in KE of block = work done on block by friction ƒ

⠀ ➪ ½mu²ƒ - ½mu²i = Fƒs cos θ

Because the friction force on the block is opposite in direction to the displacement , cos θ = -1

➢ Using Uƒ = 0 , Vƒ = 0.20 m/s , and s = 0.70 m

✒ We find ,

➪½mu²ƒ - ½mu²i = Fƒs cos θ

➪0-½ (0.50 kg) (0.20 m/s)² = (Fƒ) (0.70 m) (-1)

➪ Fƒ = 0.014 N

Hope this helped, can i pls have brainliest

Alexeev081 [22]3 years ago
5 0
0.014 is the answer. Hope that helps!
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Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

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Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
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