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Soloha48 [4]
3 years ago
5

a 2000 kg truck is traveling at a velocity of 30 m/s. What velocity must a 1000 kg car have in order to have the same momentum a

s the truck?
Physics
1 answer:
Molodets [167]3 years ago
4 0

The car should have a velocity of 60 m/s to attain the same momentum as that of the truck of 2000 kg.

Answer:

Explanation:

Momentum is measured as the product of mass of object with the velocity attained by that object.

Momentum of 2000 kg truck = Mass × Velocity

Momentum of 2000 kg truck = 2000×30 = 60000 N

Similarly, the momentum of 1000 kg car will be 1000× velocity of the 1000 kg car.

Since, it is stated that momentum of 2000 kg truck is equal to the momentum of 1000 kg of car, then the velocity of 1000 kg of car can be determined by equating the momentum of car and truck.

Momentum of 2000 kg truck = Momentum of 1000 kg car

60000=1000×velocity of 1000 kg car

Velocity of 1000 kg car = 60000/1000=60 m/s

So, the car should have a velocity of 60 m/s to attain the same momentum as that of the truck of 2000 kg.

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Explain about ohm's law.​
belka [17]

Answer:

Statement:

The electric current passing through a conductor is directly proportional to the potential difference across its ends provided temperature and other physical conditions remain constant.

Explanation:

Current is directly proportional to voltage loss through a resistor. That is, if the current doubles, then so does the voltage. To make a current flow through a resistance there must be a voltage across that resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and resistance (R).

V∝I or I∝V⇒V=IR.

4 0
3 years ago
Read 2 more answers
a hiker walks 200 m west and then walks 100 m north in what direction is her resulting displacement draw to show your answer​
Sliva [168]

Answer:

The direction of the displacement is in North-West.

Explanation:

Resultant displacement D is

=\sqrt{(200)^{2} + (100)^{2} } \\=223.60m                        

Here the direction is

\Theta =tan^{^{-1}}\left ( \frac{100}{200} \right )\\\Theta =26.6^{o}

Then the direction is 26.6^{o} North-west.            

5 0
2 years ago
A disc is thrown through the air for 1.5 min with a power output of 12.5 W. How much work is done when throwing the disc?
klasskru [66]

Answer:

work = 1125 [J]

Explanation:

To solve this problem we must remember the definition of power, which is defined as the relationship between work and time. The power can be calculated using the following equation:

Power = work/time

Power = 12.5 [w]

work = joules [J]

time = 1.5 [min] = 90 [s]

work = 12.5*90

work = 1125 [J]

7 0
3 years ago
Veronica claims that she can throw a dart at a dartboard from a distance of 2 m and hit the 5 cm wide bullseye if she throws the
wariber [46]
First, find the amount of time for the dart to hit the board using this equation: t = d/v

t = 2 m/ 15 m/s = 0.133 s

Then, find the height the dart has fallen from its initial point using this equation: h = 0.5gt²

h = 0.5(9.81 m/s²)(0.133 s)² = 0.0872 m or 8.72 cm

Since the diameter of the bull's eye is only 5 cm, and you started at the same level of the top of the bull's eye, that means the maximum allowance would only be 5 cm. Since it exceeded to 8.72 cm, it means that <em>Veronica will not hit the bull's eye.</em>
3 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
                                       M = (0.08333 / 0.1755) (9.8) =  4.65 meters .

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
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