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2 years ago

9

When you push a 2.00 kg book resting on a tabletop it takes 4.60 N to start the book sliding. What is the coefficient of static

friction between the book and the tabletop? g

1 answer:

natali 33 [55]2 years ago

3 0

The coefficient of static friction is 0.234.

Answer:

Explanation:

Frictional force is equal to the product of coefficient of friction and normal force acting on any object.

So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.

Normal force = mass * acceleration due to gravity

Normal force = 2 * 9.8 = 19.6 N.

And the frictional force is given as 4.6 N, then

$Coefficient of static friction = Frictional force/Normal force$

Coefficient of static friction = 4.6 N / 19.6 N = 0.234

So the coefficient of static friction is 0.234.

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r-ruslan [8.4K]

Answer:

1- b: 2- a : 3- c : 4- d

Explanation:

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2 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

2) The <em>potential energy</em> at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The <em>kinetic energy</em> of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

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2 years ago

UGH PLSSSS HELP WILL GIVE BRAINLIEST, AND 20 POINTS

Svetradugi [14.3K]

Answer:

false

Explanation:

the answer is false bc for a non-polar bond to form the electrons are shared equally...

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3 years ago

An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1

Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

4 0

2 years ago

The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i

Sergio [31]

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius= $r=\frac{d}{2}=\frac{2.3}{2}=1.15 m$

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency= $\frac{1200}{60}=20 Hz$

1 minute=60 seconds

Angular velocity= $\omega=2\pi f$

Angular velocity= $2\times \frac{22}{7}\times 20=125.7 rad/s$

b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

$v=r\omega=1.15\times 125.7=144.56 m/s$

c.Centripetal acceleration= $\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2$

Centripetal acceleration== $\frac{18170.56\times g}{9.81}=1852.25 g m/s^2$

7 0

3 years ago