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viva [34]
2 years ago
5

The following pairs of surfaces are pushed together with the same amount of force. Between which pair of surfaces will friction

be the greatest?
A. two pieces of sandpaper
B. two pieces of waxed paper
C. a painted surface and a piece of sandpaper
D. a painted surface and a piece of waxed paper


Juan and his friends want to build their own go-karts and race down their street. According to Newton’s Second Law of Motion, which of these will help Juan to win the race?
A. increasing the mass of his go-kart
B. increasing the amount of inertia the go-kart has
C. decreasing the net force applied to the go-kart
D. decreasing the mass of his go-kart
Physics
2 answers:
Aleks [24]2 years ago
5 0
A.  The friction between two pieces of sandpaper is greater than
the friction between any of the pairs of surfaces.

D.  Juan should decrease the mass of his go-kart.  Then any force
that pushes it forward will give it greater forward acceleration.
ozzi2 years ago
5 0

c. a painted surface and a piece of sandpaper

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A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th
icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

\theta= \theta_0 +\omega t

From our given data we know that,

\omega = 76\frac{rev}{min}

\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})

\omega = 7.958rad/s

Moreover we know that

\theta_0 = 0.47 rad

Therefore for time t=8.1s we have,

\theta= \theta_0+ \omega t

\theta= 0.47+(7.958)(8.1)

\theta = 64.9298rad

That number in revolution is:

\theta = 64.9298rad(\frac{1rev}{2\pi})

\theta = 15.108 Revolutions

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

\theta_{net} = 0.108*2\pi=0.216\pi

Therefore the angle of the speck at a time 8.1s is 0.216\pi

4 0
2 years ago
Susan and Hannah are each riding a swing. Susan has a mass of 25 kilograms, and Hannah has a mass of 30 kilograms. Susan’s swing
Charra [1.4K]

Answer:

Kinetic energy is given by:

K.E. = 0.5 m v²

Susan has mass, m = 25 kg

Velocity with which Susan moves is, v = 10 m/s

Hannah has mass, m' = 30 kg

Velocity with which Hannah moves is, v' = 8.5 m/s

<u>Kinetic energy of Susan:</u>

0.5 m v² = 0.5 × 25 kg × (10 m/s)² = 1250 J

<u>Kinetic energy of Hannah:</u>

0.5 m v'² = 0.5 × 30 kg × (8.5 m/s)² = 1083.75 J

Susan's kinetic energy is <u>1250 J </u>and Hannah's kinetic energy is <u>1083.75 J</u>.

Since kinetic energy is dependent on mass and square of speed. Thus, speed has a greater effect than mass. As it is evident from the above example. Susan has greater kinetic energy due to higher speed than Hannah.

4 0
2 years ago
A small glass bead charged to 8.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm long glass rod and is 4.0 cm
Assoli18 [71]

Answer:

71nC is the total charge of the rod

Explanation:

See attached file

8 0
1 year ago
Which points are most efficient for the utilization of resources on a production possibilities diagram?
KonstantinChe [14]

Answer: most effective way is to practice reduce reuse and recycle for utilisation of resources

3 0
1 year ago
What is its characteristic wavelength? [Hint: Recall that the kinetic energy of a moving object is E=12mv2, where m is the mass
Alchen [17]

Answer:

λ =  1.06 * 10^{-11} m

Explanation:

Using the De Broglie equation, the characteristic wavelength is given by:

λ = \frac{h}{p}

where

h = Planck's constant = 6.626 * 10^{-34} Js.

p = momentum

Momentum, p, can be calculated using:

p =  \sqrt{2Em}

where

m = mass of the electron = 9.11 * 10^{-31} kg

E = Energy of the electron = 13.4 keV = 13.4 * 10^3 * 1.6 * 10^{-19} J = 2.144 * 10^{-15} J

=> p = \sqrt{2 * 2.144 * 10^{-15} * 9.11 * 10^{-31}}

p = \sqrt{3.906 * 10^{-45}}

p = 6.250 * 10^{-23} kgm/s

Therefore, characteristic wavelength, λ, is:

λ =  \frac{6.626 * 10^{-34}}{6.250 * 10^{-23}}

λ =  1.06 * 10^{-11} m

4 0
2 years ago
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