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Tamiku [17]
3 years ago
11

A uniform crate with a massof 30 kg must be moved up along the 15° incline without tipping. Knowing that force P is horizontal,

determine (a) the largest allowable coefficient of static friction between the crate and the incline, (b) the corresponding magnitude of force P.
Physics
1 answer:
lakkis [162]3 years ago
8 0

Answer:

coefficient of friction =0.268

magnitude of force P=289.78N

Explanation:

The coefficient of friction is obtained by mgsinФ/mgcosФ=tanФ=tan15=0.268

force P is horizontal as stated in the question, horizontal component of P=mgcosФ=30*10*cos15=289.78

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Answer:

The average force exerted on the window due to two snowballs is 6 N

Explanation:

Given:

Mass of snowballs m = 300 \times 10^{-3} Kg

Velocity of snowball v = 10 \frac{m}{s}

For finding the average force,

Force is equal to the change in momentum,

   F = \frac{dP}{dt}

Here, final velocity is zero so we write,

 F = \frac{mv}{1}

Where dt = 1 sec

 F = 300 \times 10^{-30} \times 10

F = 3 N

Above value of force is due to one ball, but here given in question there are two ball,

F = 3 \times 2

F = 6 N

Therefore, the average force exerted on the window due to two snowballs is 6 N

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Find the change in kinetic energy of a 1.0 kg fish leaping to the right at 12.0 m/s.
sp2606 [1]

Answer:

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Explanation:

Given parameters:

Mass of fish  = 1kg

Velocity  = 12m/s

Unknown:

Change in kinetic energy = ?

Solution:

Kinetic energy is the energy due to the motion of a body. It is mathematically given as:

       K.E  = \frac{1}{2}  m v²  

Now, insert the parameters and solve;

  K.E  =  \frac{1}{2}  x 1 x 12  = 6J

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4 0
3 years ago
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Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec
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The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
E= \frac{\sigma}{2\epsilon_0}
where
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\epsilon_0 is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
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7 0
3 years ago
Consider electrons of kinetic energy 6.0 eV and 600 keV. For each electron, find the de Broglie wavelength, particle speed, phas
irinina [24]

Answer:

For 6.0 eV

0.5 nm, 1.45*10^6 m/s, 6.17*10^10 m/s, 1.45*10^6 m/s

For 600 eV

1.26*10^-3 nm, 2.66*10^8 m/s, 3.37*10^8 m/s, 2.66*10^8 m/s

Explanation:

See attachment for calculation

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3 years ago
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