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levacccp [35]
3 years ago
14

In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw

ay. If the 3.5 kg book is pushed from rest through a distance of 0.91 m by the horizontal 25 N force from the broom and then has a speed of 1.53 m/s, what is the coefficient of kinetic friction between the book and floor?
Physics
1 answer:
RSB [31]3 years ago
5 0

Answer:

The coefficient of kinetic is

u_{k}=0.59

Explanation:

The forces in the axis 'x' and 'y' using law of Newton to find coefficient of kinetic friction

ΣF=m*a

ΣFy=W-N=0

ΣFy=Fn-Fu=m*a

F_{u} =u_{k} *N\\F_{N}=25N\\N=W\\N=3.5kg*9.8\frac{m}{s^{2} }=34.3N

F_{N}-F_{u}=m*a\\F_{N}-u_{k}*N=m*a\\u_{k}*N=F_{N}-m*a\\u_{k}=\frac{F_{N}-m*a}{N}

Now to find the coefficient can find the acceleration using equation of uniform motion accelerated

v_{f} ^{2}=v_{o}^{2}+2*a(x_{f}-x_{o})\\x_{o}=0\\v_{o}=0\\v_{f} ^{2}=2*a*x_{f}\\a=\frac{v_{f} ^{2}}{2*a*x_{f}}\\ a=\frac{(1.53\frac{m}{s} )^{2}}{2*0.91m}\\a= 1.28 \frac{m}{s^{2} }

So replacing the acceleration can fin the coefficient:

u_{k}=\frac{F_{N}-m*a }{N}\\u_{k}=\frac{25N-(3.5kg*1.28\frac{m}{s^{2}} }{34.3N} \\u_{k}=0.59

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Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel
castortr0y [4]

Answer:

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

\vec{V_{AB}} =\vec{V_A} - \vec{V_B}

\vec{V_A} = 22 km/hr

\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}

                 = 31.52\hat{i} + 24.62 \hat{j}

putting respective value to get velocity of  A with respect to B

\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})

\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

6 0
3 years ago
4. (a) A tank containing methanol has walls 2.50 cm thick made of glass of refractive index 1.550. Light from the outside air st
zhenek [66]

Answer:

29.775°

1.911

Explanation:

Law of refraction(Snell's law):  

n_1 sin Ф_1 = n_2 sin Ф_2  

n_1 = The index of refraction for material with incident light,

n_2 = The index of refraction for material with refracted light

Ф_1= angle of incidence,

Ф_2 = angle of refraction.  

Note: the angle is measured from the normal: a line drawn perpendicular to the surface at the point where the incident ray strikes the surface.

The Part of the light which continues into the second medium is transmitted rather than reflected, but the transmitted ray changes direction as it crosses the boundary. The transmission of light from one medium to another, but with a change in direction, is called refraction. The angles of the incidence and reflected light is described by (Snells law) which is shown above.  

a)

n_3 = 1.329 —> for metalmol —> from table

1 —> in air

2 —> in the glass

3 —> in methanol  

from air to glass

n_1 sin Ф_1 = n_2 sin Ф_2  

1 x sin 41.3° =  1.550  sin Ф_2  

Ф_2 = arcsin(1*sin41.3°/1.550) = 25.201°

from glass to methanol:

n_2 sin Ф_2 = n_3 sin Ф_3  

1.550 sin 25.201° = 1.329 * sin Ф_3

Ф_3 = arcsin(1.550 sin 25.201°/1.329) = 29.775

from glass to unknown:

n_2 sin Ф_2 = n_3 sin Ф_3  

1.550 sin 25.201° = n_unknown * sin 20.2

n_unknown = 1.550 sin 25.201°/sin 20.2

                     = 1.911

5 0
3 years ago
Amy has a mass of 50 kg, and she is riding a skateboard traveling 10 m/s. What is her momentum?
german
P=mv
=50*10
=500kg m/s
7 0
3 years ago
Please help!!!!!!!! 25 points
Archy [21]

Answer:

B: beaks, finches.

Explanation:

Beaks, finches. This is because he noticed that fruit-eating finches had beaks similar to parrots, while finches that ate insects had narrow beaks. The difference in their beaks is due to adapting to different environments which causes them to evolve into different species of finches.

7 0
3 years ago
Read 2 more answers
A 13 cm long tendon was found to stretch 3.7 mm by a force of 12.1 N . The tendon was approximately round with an average diamet
liq [111]

Answer:

young's modulus=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2

Explanation:

We have given length of tendon L= 13 cm =0.13 m

Change in length \Delta L=3.7mm=3.7\times 10^{-3}m

Force = 12.1 N

Average diameter d =8.8 mm

So r=\frac{d}{2}=\frac{8.8}{2}=4.4mm=4.4\times 10^{-3}m

Area A=\pi\times  (4.4\times 10^{-3})^2=60.79\times 10^{-6}m^2

Now stress =\frac{force}{area}=\frac{12.1}{60.79\times 10^{-6}}=0.1990\times 10^6N/m^2

Strain =\frac{change\ in\ lenght}{length}=\frac{3.7\times 10^{-3}}{0.13}=28.46\times 10^{-3}

Now young's modulus=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2

8 0
3 years ago
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