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kow [346]
2 years ago
13

A 2.0 kg block has a rope attached to the block on a table and is pulled with a force of 8.0 N. The block accelerated at 2.5m/s^

2. The coefficient of kinetic friction between the block and the table is ______. (Round to the nearest hundred.)
Physics
1 answer:
Masja [62]2 years ago
3 0

Answer:

0.15

Explanation:

Assuming the rope is horizontal, sum the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum the forces in the x direction:

∑F = ma

F − Nμ = ma

Substitute:

F − mgμ = ma

mgμ = F − ma

μ = (F − ma) / (mg)

Plug in values:

μ = (8.0 N − 2.0 kg × 2.5 m/s²) / (2.0 kg × 9.8 m/s²)

μ = 0.15

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The acceleration of a particle traveling along a straight line isa=14s1/2m/s2, wheresis in meters. Ifv= 0,s= 1 m whent= 0, deter
Yuliya22 [10]

Answer:

0.78m/s

Explanation:

We are given that

Acceleration=a=\frac{1}{4}s^{\frac{1}{2}}m/s^2

v=0, s=1 when t=0

We have to find the particle's velocity at s=2m

We know that

a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v

vdv=ads

\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds

\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}

By using formula:\int x^ndx=\frac{x^{n+1}}{n+1}+C

\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1

Substitute s=2

\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)

\frac{v^2}{2}=\frac{0.50}{3}\times 1.83

v^2=2\times 0.305=0.61

v=\sqrt{0.61}=0.78m/s

Hence, the velocity of particle at s=2m=0.78m/s

3 0
3 years ago
Which method is not a technique currently used by ocean scientists to map the topography of the ocean floor?
raketka [301]

Answer:

Magnetometer

Explanation:

Magnetometer technique is not using by scientists for studying the ocean floor.The scientists currently is using SONAR ( sound navigation and ragging) technique for studying the ocean floor.SONAR is used sound waves sound waves for studying the ocean floor or we can say that SONAR is based on sound propagation.

Therefore answer is Magnetometer

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Answer:

The liquid formed from a melted solid has the same mass as the solid has.

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As long as no water can escape, the mass of the ice before melting must equal the mass of the liquid water after.

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olga_2 [115]
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