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Mekhanik [1.2K]
3 years ago
7

a force 10N drags a mass 10 kg on a horizontal table with acceleration 0.2m\s. If the acceleration due to gravity is 10m\s2, a c

oefficient of friction between the moving mass and a table is?​
Physics
1 answer:
kakasveta [241]3 years ago
7 0

Answer:

Explanation: 0.02

Recall: μ mg = ma

μ = ?

m = 10N

g = 10m\s2

a = 0.2m\s

Substituting the values , we have

μ x 10 x 10 = 10 x 0.2

100  μ = 2

μ = 2/100

μ = 1/50

μ = 0.02

μ = 0.02

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The answer is 0.245N.

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(b) 0.100

For the block on the left, f_{k} =u_{k} n= 0.100(2.45N)=0.245N.

∑F_{x}=ma_{x}

–0.308N+0.245N=(0.250kg)a

a=−0.252m/s^{2} if the force of static friction is not too large.

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To learn more about kinetic energy, refer to:

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7 0
2 years ago
What force causes a falling object to gain kinetic energy?
hoa [83]

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5 0
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What is the power of a hair dryer if it uses 72,000 joules energy in 60.0 seconds
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1. I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground
ladessa [460]

The acceleration of gravity on Earth is  9.8 m/s² .
The speed of a falling object keeps increasing smoothly,
in such a way that the speed is always 9.8 m/s faster than
it was one second earlier.

If you 'drop' the penny, then it starts out with zero speed. 
If you also start the clock at the same instant, then

         After  1.10 sec,  Speed = (1.10 x 9.8) = 10.78 meters/sec


         After  1.85 sec,  Speed = (1.85 x 9.8) = 18.13 meters/sec

But you want this second one given in a different unit of speed.
OK then:

     =  (18.13 meter/sec) x (3,600 sec/hr) x (1 mile/1609.344 meter)

     =    (18.13 x 3,600 / 1609.344)  (mile/hr)  =  40.56 mph  (rounded)

We did notice that in an apparent effort to make the question
sound more erudite and sophisticated, you decided to phrase
it in terms of 'velocity'.  We can answer it in those terms, if we
ASSUME that there is no wind, and the penny therefore doesn't
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With that assumption in force, we are able to state unequivocally
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3 0
3 years ago
A 5.0-nC point charge is embedded at the center of a nonconducting sphere (radius = 2.0 cm) which has a charge of -8.0 nC distri
Igoryamba

Answer:

3.6 × 10⁵ N/C = 360 kN/C

Explanation:

Let R = 2.0 cm be the radius of the sphere and q = -8.0 nC be the charge in it. Let q₁ be the charge at radius r = 1.0 cm. Since the charge is uniformly distributed, the volume charge density is constant. So, q/4πR³ = q₁/4πr³

q₁ = q(r/R)³. The electric field due to q₁ at r is E₁ = kq₁/r² = kq(r/R)³/r² = kqr/R³

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So, the magnitude of the total electric field at r = 1.0 cm is

E = E₁ + E₂ = kqr/R³ + kq₂/r² = k(qr/R³ + q₂/r²)

E = 9 × 10⁹(-8 × 10⁻⁹ C × 1 × 10⁻² m/(2 × 10⁻² m)³ + 5 × 10⁻⁹ C/(1 × 10⁻² m)²)

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E = 9 × 10⁹(4 × 10⁻⁵)

E = 36 × 10⁴ N/C = 3.6 × 10⁵ N/C = 360 kN/C

6 0
3 years ago
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