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denis-greek [22]
2 years ago
9

A 4 kg block has a coefficient of friction of 0.3 between itself and the surface

Physics
1 answer:
Deffense [45]2 years ago
4 0

Answer:

C - 11.76 N (Newtons)

Explanation:

2 steps are needed,

First find the Force being applied by the box on the surface.

F = M(a) [Force equals: mass X Acceleration (of gravity)]

You get 39.2 N of Force.

Plug in the remaining values into the equation for Frictional Force

F(f) = μ(F) [Coefficient of Friction X Force applied by the Box]

So

F(f) = 0.3(39.2) = 11.76 N (of Frictional Force)

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6 0
3 years ago
A certain lightbulb has a tungsten filament with a resistance of 26 Ω when cold and 170 Ω when hot. If the equation R = R0 [1 +
iris [78.8K]

Answer:

Explanation: The equation that relates resistance of tungsten at different temperatures is as follows

R = R₀ [1 + α ∆T]  , R₀ is resistance at lower temperature , R is resistance at higher temperature . α is temperature coefficient of resistivity and ∆T is rise in temperature .

Putting the values

170 = 26 [1 + .0045 ∆T]

∆T = 1230.75

lower temperature = 40◦C

higher temperature = 1230 + 40

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5 0
3 years ago
A conducting spherical shell with inner radius a and outer radius b has a positive point charge +Q at the center in the empty re
Andreas93 [3]

Answer:

a) q_inner = -Q

, q_outer = -2Q

b)    E₁ = k Q / r²        r<a

       E₂ = 0               a<r<b

       E₃ = - k 2Q/r²     r>b

d)   the charge continues inside the spherical shell, the results do not change

Explanation:

a) The point load in the center induces a load on the inner surface of the shell with constant opposite sign  

q_inner = -Q  

the outer shell of the shell the load is  

q_outer = -3 Q + Q  

q_outer = -2Q

b) To find the electric field again, use Gauss's law,  

We define as a Gaussian surface a sphere  

Ф = E. dA = q_{int}/ε₀

in this case the electric field lines and the radii of the sphere are parallel, so the sclar product is reduced to the algegraic product  

E A = q_{int}/ε₀

the area of ​​a sphere is  

A = 4 π r²  

E = 1 / 4πε₀  Q/ r²  

k = 1 / 4πε₀  

let's apply this expression to the different radii

i) r <a  

in this case the load inside is the point load  

q_{int}= + Q  

E₁ = k Q / r²

ii) the field inside the shell  

a <r <b  

As the sphere is conductive, so that it is in electrostatic equilibrium, there can be no field within it.  

E₂ = 0

iii)  r>b

   q_{int}  = Q- 3Q = -2Q

    E₃ = k (-2Q/r²)

     E₃ = - k 2Q/r²

c) see  attached

d) as the charge continues inside the spherical shell, the results do not change, since the lcharge inside remains the same and it does not matter its precise location, but remains within the Gaussian surface

6 0
2 years ago
Block m1 of mass 2m and velocity v0 is traveling to the right (+x) and makes an elastic head-on collision with block m2 of mass
Oksana_A [137]
1) In any collision the momentum is conserved

(2*m)*(vo) + (m)*(-2*vo) = (2*m)(v1') + (m)(v2')

candel all the m factors (because they appear in all the terms on both sides of the equation)

2(vo) - 2(vo) = 2(v1') + (v2') => 2(v1') + v(2') = 0 => (v2') = - 2(v1')

2) Elastic collision => conservation of energy

=> [1/2] (2*m) (vo)^2 + [1/2](m)*(2*vo)^2 = [1/2](2*m)(v1')^2 + [1/2](m)(v2')^2

cancel all the 1/2 and m factors =>

2(vo)^2 + 4(vo)^2 = 2(v1')^2 + (v2')^2 =>

4(vo)^2 = 2(v1')^2 + (v2')^2

now replace (v2') = -2(v1')

=> 4(vo)^2 = 2(v1')^2 + [-2(v1')]^2 = 2(v1')^2 + 4(v1')^2 = 6(v1')^2 =>

(v1')^2 = [4/6] (vo)^2 =>

(v1')^2 = [2/3] (vo)^2 =>

(v1') = [√(2/3)]*(vo)

Answer: (v1') = [√(2/3)]*(vo)




 
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3 years ago
Two identical vertical springs S1 and S2 have masses m1 = 400 g and m2 = 800 g attached to them. If m1 causes spring S1 to stret
OlgaM077 [116]

Answer:

potential energy = mgh

= 400÷1000 × 10× 4÷100

= 0.4 × 10 × 0.04

=4/10 ×10×4/100

= 4/10 × 4/10

=16/100

= 0.16 joules

m1 (400) stretches 4cm

m1 (100g) stretches 1cm

so, m2(800g) stretches 8 cm

potential energy of m2 = mgh

= 800/1000 ×10×8/100

= 0.8 × 0.8

=8/10 ×8/10

= 64/100

=0.64 joules

Ratio of s1 to s2

16/100 ÷ 64/100

= 1:4 ( answer)

6 0
2 years ago
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