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1 year ago

When Trinity pulls on the rope with her weight, Newton's Third Law of Motion tells us that the rope will _____

2 answers:

7 0

<span>A force is a push or a pull that acts upon an object as a results of its interaction with another object. ... These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction.</span>

Bingel [31]1 year ago

6 0

When Trinity pulls on the rope with her weight, Newton's Third Law of Motion tells us that the rope will <u>"pull back".</u>

Newton's third law of motion expresses that, at whatever point a first question applies a power on a second object, the first object encounters a power meet in extent however inverse in heading to the power that it applies.

Newton's third law of movement reveals to us that powers dependably happen in sets, and one question can't apply a power on another without encountering a similar quality power consequently. We once in a while allude to these power matches as "action-reaction" sets, where the power applied is the activity, and the power experienced in kind is the response (despite the fact that which will be which relies upon your perspective).

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A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity

Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches = 0.0508 m

n = 1750 rpm

$H_{nom}=2hp=1491.4W$

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

$w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m$

$V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s$

$F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N$

$(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N$

$T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm$

$F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N$

$F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N$

$H_a=1491*1.25=1863.75W$

$n_f_s=\frac{H_a}{H_{nom}K_S }=1$

dip = $\frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm$

7 0

1 year ago

Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for

g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0

2 years ago

Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th

katovenus [111]

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

3 0

1 year ago

Read 2 more answers

U-238 has protons and146 neutrons. A particular isotope of plutonium has 94 protons, neutrons, and a mass number of 241. Thorium

enyata [817]

$^{238}U$

so mass number = 238

mass number = protons + neutrons

given that

neutrons = 146

238 = protons + 146

protons = 92

$^{241}Pu$

so mass number = 241

mass number = protons + neutrons

given that

Protons = 94

241 = 94 + neutrons

neutrons = 147

$^ATh$

A = mass number

Protons = 90

Neutrons = 137

A = protons + Neutrons

A = 90 + 137 = 227

5 0

2 years ago

Read 2 more answers

To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device for heating a cup of water in a car connects to

jeyben [28]

Answer:

The time to boil the water is 877 s

Explanation:

The first thing we must do is calculate the external resistance (R) of the circuit, from the description we notice that it is a series circuit, by which the resistors are added

V = i (r + R)

We replace we calculate

r + R = V / i

R = v / i - r

R = 10/12 -0.04

R = 0.793 Ω

We calculate the power supplied

P = V i = I² R

P = 12² 0.793

P = 114 W

This is the power dissipated in the external resistance

We use the relationship, that power is work per unit of time and that work is the variation of energy

P = E / t

t = E / P

t = 100 10³/114

t = 877 s

The time to boil the water is 877 s

4 0

2 years ago