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kiruha [24]
3 years ago
12

Two men decide to use their cars to pull a truck stuck in mud. They attach ropes and one pulls with a force of 821 N at an angle

of 32◦ with respect to the direction in which the truck is headed, while the other car pulls with a force of 1140 N at an angle of 26◦ with respect to the same direction.
What is the net forward force exerted on the truck in the direction it is headed?
Physics
1 answer:
ratelena [41]3 years ago
5 0

Answer:

1737.8 N

Explanation:

Given information

F_1= 821 N

F_2= 1140 N

\theta_1=32^{\circ}

\theta_2=26^{\circ}

Net force, F_{net}=F_1 cos \theta_1+F_2 cos \theta_2=821cos 32^{\circ}+ 1140 cos 26^{\circ}=1737.833662 N \approx 1737.8 N

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A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force.  We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as  588 newtons  or as 
132.3 pounds.  That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
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That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
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if heat energy is transferred from direct contact between a warm and a cold object it has been transferred by what ?
Katarina [22]

When heat energy is transferred from direct contact between a warm and a cold object , it is known as heat transfer by conduction.

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5 0
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A mass of 8 kg moves at a rate of 14.3 m/sec. what is the KE developed by the mass?
Lynna [10]
Kinetic energy is the energy associated with the motion of an object. It's a scalar quantity, there is no direction associated with KE and it has no components. KE =  \frac{mv ^{2} }{2} =  \frac{8 *14.3 ^{2} }{2} = 4 *204.49 = 817.96J. Therefore Kinetic energy is 817.96J.
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What trend does the reactivity of nonmetals show in a periodic table? random changes without any trends on the periodic table ch
Naddik [55]

Answer:

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7 0
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Read 2 more answers
What force must the worker exert to get the box moving & what force must the worker exert to accelerate the box at 0.1 meter
photoshop1234 [79]

Since static friction is the minimum force required to just start the motion of a stationary object.

Here if we need to start an object from rest then we required F = 700 N

So for the first part of the above problem Force will be F = 700 N

Now if the box is already moving then we will have to use kinetic friction force between box and floor

now we can write the equation of net force as

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here

F_k = kinetic friction = 220 N

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a = acceleration = 0.1 m/s^2

now we will have

F - 220 = 500* 0.1

F = 220 + 50 = 270 N

3 0
3 years ago
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