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xxTIMURxx [149]
3 years ago
9

In a tug-of-war game on one campus, 15 students pull on a rope at both ends in an effort to displace the central knot to one sid

e or the other. Two students pull with force 196 N each to the right, four students pull with force 98 N each to the left, five students pull with force 62 N each to the left, three students pull with force 150 N each to the right, and one student pulls with force 250 N to the left. Assuming the positive direction to the right, express the net pull on the knot in terms of the unit vector. How big is the net pull on the knot? In what direction?
Physics
1 answer:
mojhsa [17]3 years ago
6 0

Answer:

Net pull = 110 N to the left

Explanation:

Group the different pulls according to the direction (right or left)

2 pull 196 N each to the right

4 pull 98 N each to the left

5 pull 62 N each to the left

3 pull 150 N each to the right

1 pull 250 N to the left

Since positive direction is to the right, the pulls to the left will have a minus (-)

Net Force= 2(196)+4(-98)+5(-62)+3(150)+1(-250) \\Net Force = -110

The resulting force is negative, meaning the direction is to the left

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In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0
3 years ago
A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The
marusya05 [52]

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

3(670nm) /5 =402nm

∴λ = 402nm

Explanation:

4 0
3 years ago
Juan makes an adjustment to an electromagnet that causes the electromagnet to lose some of its strength. What did Juan most like
Assoli18 [71]
Reducing the amount of loops will cause a loss of strength, as the loops make the magnet stronger.
5 0
3 years ago
Read 2 more answers
Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of
kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

W=2300\ kg\times 32\ ft/s^2=73600\ lb

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

W=0.022\ kg\times 32\ ft/s^2=0.704\ lb

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5 0
3 years ago
. Each valve A, B, and C, when open, releases water into a tank at its own constant rate. With all three valves open, the tank f
olasank [31]

Answer:

Time taken by A and B is 1.2 hr.

Explanation:

Given that

Time taken by tank when all(A+B+C) are open = 1 hr

Time taken by tank when A+C are open = 1.5 hr

Time taken by tank when B+C are open = 2 hr

If we treat as filling of tank is a work then

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Lets take work is 1 unit

1 = 1(1/a+1/b+1/c)          ---------1

1 = 1.5(1/a+1/c)           ----------2

1 = 2(1/b+1/c)             --------3

From equation 1 and 3

1=1(1/a+1/2)

a=2

Form equation 2

1 = 1.5(1/2+1/c)

c=6

From equation 3

 1 = 2(1/b+1/6)  

b=3

So time taken by

A is alone to fill tank is 2 hr

B is alone to fill tank is 3 hr

C is alone to fill tank is 6 hr

So time\ taken\ by\ A\ and\ B =\dfrac{2\times 3}{2+3} hr

Time taken by A and B is 1.2 hr.

7 0
3 years ago
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