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liq [111]
2 years ago
8

A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma

ss m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.
1- Find an expression for the maximum tension T max for which the small box rides on top of the large box without slipping.Express your answer in terms of the variablesM ,m , ms, and appropriate constants.2-A horizontal rope pulls a 10 kg wood sled across frictionless snow. A 6.0 kg wood box rides on the sled. What is the largest tension force for which the box doesn't slip? Assume that Mk= 0.50.
Physics
1 answer:
Nesterboy [21]2 years ago
5 0

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

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3 0
2 years ago
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current
sladkih [1.3K]

Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero  U = 0.1355 \ m

Explanation:

From the question we are told that

    The distance of wire one from two along the y-axis is    y = 0.340 m

   The current on the first wire is  I_1 =  (27.5i) A

    The force per unit length on each wire is  Z =  295 \mu N/m = 295*10^{-6}  N/m

Generally the force per unit length is mathematically represented as

         Z = \frac{F}{l}  =  \frac{\mu_o I_1I_2}{2\pi y}

=>      \frac{\mu_o I_1I_2}{2\pi y}  =  295

Where  \mu_o is the permeability of free space with a constant value of  \mu_o  =  4\pi *10^{-7} \ N/A2

substituting values

       \frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340}  =  295 *10^{-6}

=>    I_2 =  18.23 \ A

Let U  denote the  line in the xy-plane where the total magnetic field is zero

So  

      So the force per unit length of  wire 2  from  line  U is equal to the force per unit length of wire 1  from  line  (y - U)      

   So  

         \frac{\mu_o  I_2  }{2 \pi U} =  \frac{\mu_o  I_1  }{2 \pi(y -  U) }

substituting values

          \frac{  18.23  }{ U} =  \frac{ 27.5 }{(0.34 -  U) }

         6.198 -18.23U = 27.5U

          6.198=45.73U

          U = 0.1355 \ m              

5 0
2 years ago
The transmutation of a radioactive uranium isotope, 234/92 U, into a radon isotope, 222/86 Rn, involves a series of three nuclea
BaLLatris [955]

Answer:

The answer is 26/98 how i did this is i divided them mulitiplyed well i cant really explain it but im pretty dure its right

Explanation:

3 0
2 years ago
Read 2 more answers
A skateboarder rides swiftly up the edge of a bowl-shaped surface and leaps into the air. While in the air, the skateboarder fli
Katen [24]

Answer:

Option(a) is the correct answer to the given question .

Explanation:

The main objective of the angular momentum is evaluating however much the rotational movement as well as the angular velocity in the entity does have.The angular momentum is measured in terms of kgm^{2 }\  / s.

  • In the given question the skateboarder rides quickly up the bottom of a bowl-shaped surface and climb into the air.it means it is rotational movement also it is not touching anything so it is angular momentum.
  • All the other option is incorrect because it is not follows the given scenario
3 0
2 years ago
1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational
sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = \frac{m v^{2} }{r}

==> 1.1 × 10^{19} = (9.4 × 10^{21} × v^{2} ) ÷ 1.5 × 10^{8}

==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

==> v= 0.419188323 × 10^{3} m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×10^{8} ÷ 0.419188323 × 10^{3}

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0
2 years ago
Read 2 more answers
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