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Katena32 [7]
3 years ago
7

A 6.2 kg object moving in the +x direction at 5.3 m/s collides head-on with an 7.8 kg object moving in the −x direction at 2.5 m

/s.
Find the final velocity of each mass for each of the following situations. (Take the positive direction to be +x.)
(a) The bodies stick together.
1 m/s (6.2 kg mass)
2 m/s (7.8 kg mass)
(b) The collision is elastic.
3 m/s (6.2 kg mass)
4 m/s (7.8 kg mass)
(c) The 6.2 kg body is at rest after the collision.
5 m/s (6.2 kg mass)
6 m/s (7.8 kg mass)
(d) The 7.8 kg body is at rest after the collision.
7 m/s (6.2 kg mass)
8 m/s (7.8 kg mass)
(e) The 6.2 kg body has a velocity of 4.0 m/s in the -x direction after the collision.
9 m/s (6.2 kg mass)
10 m/s(7.8 kg mass)
Physics
1 answer:
Crank3 years ago
8 0

Answer:

10ms kg 88.2

Explanation:

Toook test

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Three 7 0hm resistors are connected in series across a 10 V battery. What is the equivalent resistance of the circuit?
Tomtit [17]

Given

Three 7 ohm resistor are in series.

The battery is V=10V

To find

The equivalent resistance

Explanation

When the resistance are in series then the resistance are added to find its equivalent.

Thus the equivalent resistance is:

R=7+7+7=21\Omega

Conclusion

The equivalent resistance is 21 ohm

4 0
1 year ago
If a box is pulled with a force of 100 N at an angle of 25
love history [14]

The X and Y components of the force are 90.63 Newton and 42.26 Newton respectively.

<u>Given the following data:</u>

  • Force = 100 Newton.
  • Angle of inclination = 25°

To determine the X and Y components of the force:

<h3>The horizontal component (X) of a force:</h3>

Mathematically, the horizontal component of a force is given by this formula:

F_x = Fcos \theta\\\\F_x =100 \times cos25\\\\F_x =100 \times 0.9063

Fx = 90.63 Newton.

<h3>The vertical component (Y) of tensional force:</h3>

Mathematically, the vertical component of a force is given by this formula:

F_y = Fsin \theta\\\\F_y =100 \times sin25\\\\F_y =100 \times 0.4226

Fy = 42.26 Newton.

Read more on horizontal component here: brainly.com/question/4080400

6 0
2 years ago
Help<br> How much current will flow when a 120 V power supply is connected to a 30<br> resistor ?
AVprozaik [17]
Current= voltage divided by resistance
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7 0
3 years ago
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
3 years ago
A capacitor is connected to an AC generator. As the generator's frequency is increased, what happens to the current in the capac
Licemer1 [7]

Answer:

The current increases.

Explanation:

A capacitor can be defined as an electronic component used in electrical circuits to store charge temporarily.

A capacitor is connected to an AC generator. As the generator's frequency is increased, the current in the capacitor increases as well.

This ultimately implies that, when a capacitor is connected to an AC generator, the frequency of the circuit is directly proportional to the amount of current flowing through it.

8 0
3 years ago
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