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dangina [55]
2 years ago
12

A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and betwe

en the lower box and the table is 0.600, while the coefficient of static friction between these same surfaces is 0.800. A horizontal pull to the right is exerted on the lower box, as shown in the figure, and the boxes move together. What is the friction force on the UPPER box?

Physics
1 answer:
natta225 [31]2 years ago
4 0
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.


Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
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Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
1 year ago
The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Exp
olganol [36]

Answer:

Ft

Explanation:

We are given that

Initial velocity=u=0

We have to find the magnitude of p of the momentum of the particle at time t.

Let mass of particle=m

Applied force=F

Acceleration, a=\frac{F}{m}

Final velocity , v=a+ut

Substitute the values

v=0+\frac{F}{m}t=\frac{F}{m}t

We know that

Momentum, p=mv

Using the formula

p=m\times \frac{F}{m}t=Ft

6 0
2 years ago
What is the resistance ofa wire made of a material with resistivity of 3.2 x 10^-8 Ω.m if its length is 2.5 m and its diameter i
Katarina [22]

R = 0.407Ω.

The resistance  R of a particular conductor is related to the resistivity ρ of the material by  the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of ​​the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4.  Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

5 0
2 years ago
g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb
Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber  = v x 1 x g

Buoyant force on lead =  .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g  = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115  / 33.5

= .36  

= 36 %

4 0
2 years ago
When two objects are in contact, moving together, which of the following statements must be true? Choose all that apply. When tw
Setler [38]

Answer:

The objects must have the same acceleration and the objects must exert the same magnitude force on each other.

Explanation:

The objects must have the same weight: FALSE. This is not needed, any two object can move together in contact no matter their mass.

The objects must have the same acceleration: TRUE. If they have different accelerations, they will separate since the distance each of them travel at a given time will be different.

The objects must have the same net force acting on them: FALSE. This is not needed, since what matters is acceleration, and a=F/m, so if both objects have different net force acting on them, they could have different masses also to compensate and result in the same acceleration.

The objects must exert the same magnitude force on each other: TRUE, this is the 3rd Newton Law, an action must follow the same reaction.

7 0
2 years ago
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