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Furkat [3]
3 years ago
14

Emboldened by the success of their late night keg pull in Exercise 61 above, our intrepid young scholars have decided to pay hom

age to the chariot race scene from the movie ‘Ben-Hur’ by tying three ropes to a couch, loading the couch with all but one of their friends and pulling it due west down the street. The first rope points N80◦W, the second points due west and the third points S80◦W. The force applied to the first rope is 100 pounds, the force applied to the second rope is 40 pounds and the force applied (by the non-riding friend) to the third rope is 160 pounds. They need the resultant force to be at least 300 pounds otherwise the couch won’t move. Does it move? If so, is it heading due west?
Physics
1 answer:
alexdok [17]3 years ago
5 0

Answer is answer

XD                                ssssss

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A defibrillator passes 14.0 A of current through the torso of a person for 0.0300 s. How much charge moves in coulombs?
krek1111 [17]
<h2>Answer:</h2>

4.2 C

<h2>Explanation:</h2>

The charge (Q) moving is the product of the current(I) flowing through the torso of the person and the time taken (t) for the flow.

i.e

Q = I x t

Where;

I = current = 14.0A

t = time taken  = 0.0300s

Substituting the values of I and t into the equation above gives

Q = 14.0 x 0.0300

Q = 4.2 C

Therefore quantity of charge moving is 4.2 C

4 0
3 years ago
A box is pulled 6 meters across the ground at a constant velocity by a horizontally applied force of 50 newtons. At the same tim
Vikentia [17]

Answer:

(a) Friction force = 50 N

(b) Work done by friction = 300 j

(c) Net work done = 0 j

Explanation:

We have given that the box is pulled by 6 meter so d = 6 m

Force applied on the box F = 60 N

We have have given that velocity is constant so acceleration will be zero

So to applied force will be utilized in balancing the friction force

So friction force F_{friction}=50N

Work done by friction force W_{friction}=F_{friction}\times d=50\times 6=300j

Work done by applied force W=F\times d=50\times 6=300j

So net work done = 300-300 = 0 j

7 0
3 years ago
A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total)
levacccp [35]

In component form, the displacement vectors become

• 350 m [S]   ==>   (0, -350) m

• 400 m [E 20° N]   ==>   (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W]   ==>   (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

5 0
3 years ago
Help on 9 and 10! Please help me!
Ivan

9 is D I believe, I don't know about 10

6 0
3 years ago
SP: Calculate the moment
ipn [44]

Answer:

Moment of the force is 20 N-m.

Explanation:

Given:

Force exerted by the person is, F=80\ N

Distance of application of force from the point about which moment is needed is, d=25\ cm=\frac{25}{100}\ m=0.25\ m

Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.

Therefore, the moment of the force about the end of the claw hammer is given as:

M=F\times d\\\\M=(80\ N)(0.25\ m)\\\\M=20\textrm{ N-m}

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.

6 0
3 years ago
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