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sergij07 [2.7K]
3 years ago
15

The lower the value of the coefficient of friction,the_the resistance to sliding​

Physics
1 answer:
malfutka [58]3 years ago
7 0

Answer:

lower

Explanation:

The lower the value of the coefficient of friction, the lower the resistance to sliding.

The coefficient of friction is the ratio of the frictional force and the normal force pressing two surfaces in contact together.

               U  = \frac{F}{N}  

U is the coefficient of friction

F is the frictional force

N is the normal force

 We see that coefficient of friction is directly proportional to frictional force.

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Cobaltâ’60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted by an atom of this
Crank

Energy of gamma rays is given by equation

E = h\nu

here we know that

h = Planck's constant

\nu = frequency

now energy is given as

E = 4.70 MeV = 4.70 \times 10^6 \times 1.6 \times 10^{-19}

E = 7.52 \times 10^{-13} J

now by above equation

E = h\nu

7.52 \times 10^{-13} = 6.6 \times 10^{-34} \nu

\nu = 1.14 \times 10^{21} Hz

now for wavelength we can say

\lambda = \frac{c}{\nu}

\lambda = \frac{3\times 10^8}{1.14 \times 10^{21}}

\lambda = 2.63 \times 10^{-13} m

3 0
2 years ago
If a 1-megaton nuclear weapon is exploded
Elenna [48]

Answer:

1,780,000 N

Explanation:

0.2 atm × (1.013×10⁵ Pa/atm) = 20,260 Pa

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F = 20,260 Pa × (3.89 m × 22.6 m)

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4 0
2 years ago
Two satellites are in circular orbits around Earth. Satellite A has speed vA. Satellite B has an orbital radius nine times that
Pani-rosa [81]

Answer:

option B

Explanation:

given,

Satellite B has an orbital radius nine times that of satellite A.

R' = 9 R

now, orbital velocity of the satellite A

........(1)

now, orbital velocity of satellite B

from equation 1

hence, the correct answer is option B

8 0
2 years ago
Which of the following best describe the particles in a liquid.
VashaNatasha [74]
A. fixed volume, changeable shape.
4 0
3 years ago
A compact disc (CD) is played by a cd player, which uses a laser to read the tracks on the disc. The disc spins initially at app
uranmaximum [27]

Answer:

a. The laser tracking mechanism experiences a changing tangential velocity

c. The laser tracking mechanism experiences a non-zero angular acceleration

d. The laser tracking mechanism experiences a non-zero tangential acceleration

Explanation:

a. The laser tracking mechanism experiences a changing tangential velocity

This is because the tangential velocity v = rω where r = radius of disc and ω  = angular speed of discs. Since r is constant, v ∝ ω.

Since the angular speed changes from 200 rpm to 500 rpm, thus, the tangential velocity would also change.

So, the laser tracking mechanism experiences a changing tangential velocity

c. The laser tracking mechanism experiences a non-zero angular acceleration

Since angular acceleration, α = Δω/Δt where Δω = change in angular speed and Δt = change in time.

Since there is a change in angular speed from 200 rpm to 500 rpm in time Δt, there is thus a non-zero angular acceleration.

So, The laser tracking mechanism experiences a non-zero angular acceleration

d. The laser tracking mechanism experiences a non-zero tangential acceleration

Since tangential acceleration, a = rα where r = radius of disc and α = angular acceleration.

Since there is an angular acceleration of the disc, there is thus going to be a tangential acceleration given by a = rα.

So, the laser tracking mechanism experiences a non-zero tangential acceleration

Statement b is false because, the disc experiences a changing angular speed from 200 rpm to 500 rpm.

4 0
2 years ago
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