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vovangra [49]
2 years ago
13

Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction

between the boxes and the horizontal surface.
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
5 0
<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>
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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

J=\frac{E}{\rho}      (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

J=\frac{I}{A}        (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}

Next, you replace for all variables:

I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A

hence, the current in the wire is 4.75A

4 0
1 year ago
A container explodes and breaks into three fragments that fly off 120° apart from each other, with mass ratios 1: 4: 2. If the f
RSB [31]

Answer:

V₂ = 1.5 m/s

Explanation:

given,

speed of the first piece = 6 m/s

speed of the third piece = 3 m/s

speed of the second fragment = ?

mass ratios = 1 : 4 : 2

fragment break  fly off = 120°

α = β = γ  = 120°

sin α = sin β = sin γ = 0.866

using lammi's theorem

\dfrac{A}{sin\alpha}=\dfrac{B}{sin\beta}=\dfrac{C}{sin\gamma}

A,B and C is momentum of the fragments

\dfrac{m\times 6}{0.866}=\dfrac{4m\times v_2}{0.866}=\dfrac{2m\times 3}{0.866}

4 x V₂ = 2 x 3

V₂ = 1.5 m/s

3 0
2 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
valentina_108 [34]

Answer:

v = 69.82 ms^-1

Explanation:

As we know,

R = vi2 sin2Ꝋ / g  

vi2 =R g / sin2 Ꝋ where R is range R = 52m, Ꝋ = 3 Degrees

vi2 = 52 x 9.8 / sin 2(3) = 4875.227

v = 69.82 ms^-1

3 0
2 years ago
If the net force acting on an object increases by 50 percent, then the acceleration of the object will
ki77a [65]
<span>If the net force acting on an object increases by 50 percent, then
the acceleration of the object will also increase by 50 percent.

This answer is not offered among the list of choices.

So the correct response is "D. none of the above"</span>
8 0
2 years ago
A sprinter accelerates from rest to a velocity of 12m/s in the first 6 seconds of the 100 meter dash .
GREYUIT [131]

Answer:

a) 36 m

b) 64 m

Explanation:

Given:

v₀ = 0 m/2

v = 12 m/s

t = 6 s

Find: Δx

Δx = ½ (v + v₀) t

Δx = ½ (12 m/s + 0 m/s) (6 s)

Δx = 36 m

The track is 100 m, so the sprinter still has to run another 64 m.

5 0
2 years ago
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