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Darya [45]
3 years ago
12

A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m

by a trained dog who exerts ahorizontal force with magnitude 39.0N.
(a) Use the work-energy theorem to find the final speedof the 12-pack if there is no friction between the 12-pack and thefloor.
m/s
(b) Use the work-energy theorem to find the final speed of the12-pack if the coefficient of kinetic friction between the 12-packand the floor is 0.30.
m/s
Physics
1 answer:
erastovalidia [21]3 years ago
8 0

Answer:

a)  W = 46.8 J  and b)   v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

    W = ΔK = k_{f} -K₀

a) work is the scalar product of force by distance

    W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

    W = F d cos θ

    W = 39.0 1.20 cos 0

    W = 46.8 J

b) zero initial kinetic language because the package is stopped

    W -W_{fr} = k_{f} -K₀

    W - fr d= ½ m v² - 0

    W - μ N d = ½ m v

   on the horizontal surface using Newton's second law

     N-W = 0

     N = W = mg

 

     W - μ mg d = ½ m v

    v² = (W -μ mg d) 2/m  

    v = √(W -μ mg d) 2/m

    v = √[(46.8 -  0.30 4.30 9.8 1.20) 2/4.3 ]

    v = √(31.63 2/4.3)

    v = 3.84 m/s

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Which of the following is a consequence of the special theory of relativity?
aleksandr82 [10.1K]

Answer:

D.

Explanation:

Specifically, Special Relativity showed us that space and time are not independent of one another but can be mixed into each other and therefore must be considered as the same object, which we shall denote as space-time. The consequences of space/time mixing are: time dilation. and length contraction.

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2 years ago
What is the atomic mass of Jupiter?
lana [24]
The mass of Jupiter is 1.9 x 1027 kg.
4 0
3 years ago
A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

5 0
3 years ago
A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang
olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: s_c = \omega_ct = 0.233t

For the horse: s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

s_c = s_h

0.233t = 1.57 + 0.0068t^2

0.0068t^2 - 0.233t + 1.57 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}

t= \frac{0.233\pm0.11}{0.0136}

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0
3 years ago
A silver wire with resistivity 1.59 × 10-8 Ω-m carries a current density of 4.0 A/mm2, What is the magnitude of the electric fie
Nat2105 [25]

Answer:

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire, \rho=1.59\times 10^{-8}\ \Omega-m

Current density of the wire, J=4\ A/mm^2=4\times 10^6\ A/m^2

We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

E=J\times \rho

E=4\times 10^6\times 1.59\times 10^{-8}

E = 0.0636 V/m

or

E = 0.064 V/m

So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

6 0
3 years ago
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