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Harlamova29_29 [7]
3 years ago
7

If the force of kinetic friction on a crate is 50 N and the weight of the crate is

Physics
1 answer:
stealth61 [152]3 years ago
5 0

Answer:

The coefficient of kinetic friction μ = 0.278

Explanation:

The frictional force on the crate is given by F = μN where μ = the coefficient of kinetic friction and N = normal force. The normal force equals the weight of the crate W = 180 N.

Since the kinetic friction F = 50 N, and the normal force which is the weight of the crate equals N = W = 180 N, the coefficient of kinetic friction μ is given by μ = F/N

= 50 N/180 N

= 0.278

So, the coefficient of kinetic friction μ = 0.278

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A typical garden hose has an inner diameter of 5/8". Let's say you connect it to a faucet and the water comes out of the hose wi
castortr0y [4]
Since speed (v) is in ft/sec, let's convert our diameters from inches to feet:
1) 5/8in = 0.625in
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2) 0.25in × 1ft/12in = 0.021ft
Equation:
v = 4q \div ( {d}^{2} \pi) \: where \: q = flow \\ v = velocity \: (speed) \: and \:  \\ d = diameter \: of \: pipe \: or \: hose \\ and \: \pi = 3.142
we \: can \: only \: assume \:that \\  flow \: (q) \:stays \: same \: since \: it \\  isnt \: impeded \: by \:  anything \\ thus \:it  \: (q)\:  stays \: the \: same \:  \\ so \: 4q \: can \: be \: removed \: from \:  \\ the \: equation
then \: we \: can \: assume \: that \: only \\ v \: and \: d \: change \: leading \:us \: to >  >  \\ (v1 \times {d1}^{2} \pi) = (v2  \times   {d2}^{2}\pi)
both \: \pi \: will \: cancel \: each \: other \: out \:  \\ as \: constants \:since \: one \: is \: on \\ each \: side \: of \: the \:  =

(v1  \times   {d1}^{2}) = (v2 \ \times {d2}^{2}) \\ (7.0 \times   {0.052}^{2}) = (v2  \times   {0.021}^{2}) \\ divide \: both \: sides \: by \:  {0.021}^{2} \\ to \: solve \: for \: v2 >  >
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44 ft/sec
4 0
3 years ago
N which device is chemical energy transformed into electrical energy?
densk [106]
The answer is batteries

7 0
3 years ago
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A proton travels with a speed of 1.8×106 m/s at an angle 53◦ with a magnetic field of 0.49 T pointed in the +y direction. The ma
Likurg_2 [28]

Answer:

Magnetic force, F=1.12\times 10^{-13}\ N

Explanation:

It is given that,

Velocity of proton, v=1.8\times 10^6\ m/s

Angle between velocity and the magnetic field, θ = 53°

Magnetic field, B = 0.49 T

The mass of proton, m=1.672\times 10^{-27}\ kg

The charge on proton, q=1.6\times 10^{-19}\ C

The magnitude of magnetic force is given by :

F=qvB\ sin\theta

F=1.6\times 10^{-19}\ C\times 1.8\times 10^6\ m/s\times 0.49\ Tsin(53)

F=1.12\times 10^{-13}\ N

So, the magnitude of the magnetic force on the proton is 1.12\times 10^{-13}\ N. Hence, this is the required solution.

3 0
3 years ago
What is the escape velocity from a planet with a mass that is 10.8% Earths mass and a size that is 53% Earths radial size?
DENIUS [597]

The escape velocity from the planet is 5052 m/s

Explanation:

The formula to calculate the escape velocity from a planet is:

v=\sqrt{\frac{2GM}{R}}

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M is the mass of the planet

R is the radius of the planet

For Earth,

M_E=5.98\cdot 10^{24} kg is the mass

R_E=6.37\cdot 10^6 m is the radius

Here we have a planet that:

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M=0.108M_E

- its radius is 53% of that of Earth, so

R=0.53 R_E

Substituting everything into the first equation, we find the escape velocity from this planet:

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#LearnwithBrainly

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3 years ago
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