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Alekssandra [29.7K]

3 years ago

10

A smooth wooden 40.0 N block is placed on a smooth wooden table a force of 14.0N is required to keep the block moving at a const

ant velocity what is the coefficient of sliding friction between the block and the table top

1 answer:

dsp733 years ago

3 0

Answer:

0.35 is the friction coefficient.

Explanation:

The equation for the friction coefficient is friction coefficient = force/normal force. The force required here is 14N, and the normal force is the weight, 40N. 14/40=0.35.

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an airplane flies for 4 hours with a constant speed of 696 km/h and then for another 35 minutes with a constant sppeed of 936 km

Nastasia [14]

Answer:

Total distance covered by airplane is 3330 km

Explanation:

Distance covered in speed of v= 696 km /hr for t = 4 hours is

$x_{1}=v \times t\\x_{1}=696 \times 4\\x_{1}=2784 km$

Convert given speed in km/min, then distance covered in 35 minutes at speed of 936 km/hr is

$x_{2}=\frac{936}{60} \times 35\\x_{2}==546 km$

Total distance covered by airplane is

$x = x_{1}+x_{2}\\x =2784 +546\\x=3330 km$

Total distance covered by airplane is 3330 km

8 0

3 years ago

An electrostatic paint sprayer has a 0.17 m-diameter metal sphere at a potential of 25.0 kV that repels charged paint droplets o

Troyanec [42]

Answer:

$q=0.236uC$

Explanation:

From the question we are told that:

Diameter $d=0.17m$

Radius $r=0.17/2=>0.085$

Potential $E=25.0kV$

Generally the equation for Potential on spere is mathematically given by

$E=\frac{1}{4 \pi e_0}*\frac{q}{r}$

Therefore

$q=\frac{25*10^3*0.085}{\frac{1}{4 \pi e_0}}$

Where

$\frac{1}{4 \pi e_0}=9*10^9$

Therefore

$q=\frac{25*10^3*0.085}{(9*10^9}}$

$q=0.236uC$

4 0

3 years ago

A young child hold a string attached to a balloon. What is the reaction force to the balloon pulling up on the earth?

Alina [70]

Answer:

As the Ballon pulls up, the distance between the Ballon and the center of the earth increases, the gravitational pull reduces and the gravity potential energy increases.

3 0

3 years ago

Captain Kirk (80.0 kg) beams down to a planet that is the same size as Uranus and finds that he weighs

Archy [21]

The mass of the planet is $1.51\cdot 10^{26}kg$

Explanation:

The weight of Captain Kirk on the surface of the planet is equal to the gravitational force between him and the planet, which is:

$F=G\frac{Mm}{R^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

M is the mass of the planet

m is the mass of Captain Kirk

R is the radius of the planet

In this problem, we have:

m = 80.0 kg is the mass of Kirk

$R=25,362 km = 2.54\cdot 10^7 m$ is the radius of the planet (same as Uranus)

F = 1250 N is the magnitude of the gravitational force between Kirk and the planet

Solving for M, we find the mass of the planet:

$M=\frac{FR^2}{Gm}=\frac{(1250)(2.54\cdot 10^7)^2}{(6.67\cdot 10^{-11})(80.0)}=1.51\cdot 10^{26}kg$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

8 0

3 years ago

Thinking Mathematically: Explore the quantitative dependencies of the acceleration upon the speed and the radius of curvature. T

solong [7]

The expression for the centripetal acceleration allows to find the results for the questions are:

A) The acceleration varies INVERSELY with the radius of curvature.

B) The acceleration varies DIRECTLY with the speed.

C) The acceleration becomes four times greater.

D) The acceleration becomes nine times greater.

E) The acceleration is reduced to half.

F) The acceleration is reduced to a third.

In circular motion there must be an acceleration towards the center of the circle, it is called cenripetal acceleration, in this case all the energy supplied to the system is used to change the direction of the speed even when its magnitude remains constant.

$a_c = \frac{v^2}{r}$

Where $a_c$ the centripetal acceleration, v is the speed and r the radius of curvature of the circle.

Now we can answer the questions about centripetal acceleration.

A) For the same speed, the acceleration varies INVERSELY with the radius of curvature.

B) For the same radius of curvature the acceleration varies DIRECTLY with the speed.

C) The speed is doubled

v = 2 v₀

$a_c = \frac{(2v_o)^2 }{r}$

$a_c = 4 \ \frac{v_o^2 }{r}$

The acceleration becomes four times greater than the original value

D) The speed is tripled

v = 3 v₀

$a_c = 9 \frac{ v_o^2}{r}$

Acceleration becomes nine times greater than the original

E) the radius of curvature is doubled

r = 2 r₀

$a_c = \frac{v_o^2}{2 r_o }$

$a_c = \frac{1}{2} a_o$

Acceleration is reduced to half the original value

F) The radius of curvature is tripled

r = 3 r₀

$a_c = \frac{v_o^2 }{3 r_o} \\ \\a_c = \frac{1}{3} a_o$

The acceleration is reduced to a third of the initial one.

In conclusion using the expression for the centripetal acceleration we can find the answers for the questions are:

A) The acceleration varies INVERSELY with the radius of curvature.

B) The acceleration varies DIRECTLY with the speed.

C) The acceleration becomes four times greater.

D) The acceleration becomes nine times greater.

E) The acceleration is reduced to half.

F) The acceleration is reduced to a third.

Learn more here: brainly.com/question/6082363

6 0

3 years ago