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allsm [11]
4 years ago
15

A one-dimensional conservative force is given by the function F(x) = (2.00 N/m) x +(1.00 N/m3) x3. What is the change in potenti

al energy of the object as it moves from x = 1.00 m to x = 2.00 m? Enter your answer in Joules.
Physics
1 answer:
topjm [15]4 years ago
8 0

Answer:

Explanation:

Given that

F(x) = (2.00 N/m) x +(1.00 N/m3) x3.

Which can be written as

F(x) = 2x +x³ N/m

Potential energy from x=1 to x=2

P.E is given as

P.E=∫F(x)dx

P.E=∫2x+x³ dx. From x=1 to x=2

P.E= 2x²/2 + x⁴/4 from x=1 to x=2

P.E= x² + x⁴/4 from x=1 to x=2

P.E=(2²+2⁴/4)-(1² +1⁴/4)

P.E=(4+4)-(1+1/4)

P.E, =8-5/4

P.E=27/4

P.E=6.75J

Since it is moving from lower position to upper position, then P.E should be negative

P.E = - 6.75J

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2. A car accelerates uniformly from +10.0 m/s to +50.0 m/s over a distance of 225 m. How long did it take to go that distance? S
artcher [175]
Let's call the constant acceleration a.
At a time t, its speed will thus be v(t)=a*t+v0 where v0 is its initial speed, here 10 m/s. Hence v(t)=a*t+10.

From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.

Hence P(t)=a*t^2/2+10t

Let's call T the time at which it's at 50 m/s, we know that P(T)=225m and that v(T)=50 m/s hence a*T+10=50 thus a=40/T and P(T)=(40/2+10)T=30T

Hence T=225/30=7.5

It took 7.5 seconds


7 0
3 years ago
A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
3 years ago
Which segment represent slowing down
lakkis [162]

Answer:

A downward sloping line

Explanation:

6 0
3 years ago
Read 2 more answers
on aircraft carriers, catapults are used to accelerate jet air craft to flight speeds in a short distance. One such catapult tak
sineoko [7]

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning)

The jet's change in speed = (70 m/s) - (zero) = 70 m/s

So acceleration = (70 m/s) / (2.5 s)

Acceleration = (70 / 2.5) m/s²

<em>Acceleration = 28 m/s²</em>  

That's about 2.9 G's .  Jet pilots can endure a lot more than that, but maybe the catapult or the hook on the airplane can't.  Let's look a little closer:

F = m A (Newton #2)

The force on the airplane = (18,000 kg) x (28 m/s²)

Force on the airplane = 504,000 Newtons

That's about 113,000 pounds !  Maybe the part of the airplane that the catapult pushes on can't handle any more force than that.  Or maybe that's the most force the catapult can deliver.

Also, the REACTION force on the catapult is the same 113,000 pounds.  Maybe the hooks or the chains or the struts on the catapult can't handle any more force than that.

That's almost 57 tons for gosh sakes !  Maybe the DECK of the carrier can't handle more force than that, and that's why they can't launch the airplane with acceleration of more than 2.9 G's .

8 0
3 years ago
For the following types of electromagnetic radiation, how do the wavelength, frequency, and photon energy change as one goes fro
KIM [24]

Answer:

Wavelength, frequency and the photon energy changes as the one goes across the ranges of the electro-magnetic radiations.

Explanation:

Electro-magnetic radiations may be defined as the form of energy that is radiated or given by the electro-magnetic radiations. The visible light that we can see is the one of the electro-magnetic radiations. Other forms are the radio waves, gamma waves, UV rays, infrared radiations, etc.

The wavelength of the radiations decreases as we go from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.

The frequency of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.

The photon energy of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.

5 0
3 years ago
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