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Mars2501 [29]
3 years ago
8

An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass

1.2 kg is attached to the free end of the spring. The glider is pulled toward the right along a frictionless air track, and then released. Now the glider is moving in simple harmonic motion with amplitude 0.045 m. The motion is horizontal (one-dimensional).
Suddenly, Slimer holding an apple flies in and approaches the glider. Slimer drops the apple vertically onto the glider from a very small height. The apple sticks to the glider. The mass of the apple is 0.18 kg.
Recall that the total mechanical energy is E = 1/2 mv^2 + 1/2 kx^2 = 1/2 kA^2 = constant
(a) Calculate the new amplitude of the motion of the glider with apple if the apple is dropped at the moment when the glider passes through its equilibrium position, x = 0 m.
Hints: The total energy of the glider just before the collision is E = 1/2mglider v^2 = 1/2KA^2
The apple sticks to the glider in a completely inelastic collision. The glider is now moving with the apple but at a lower speed. The linear momentum is conserved. Write a corresponding equation.
Also, assume that the collision is very short, so just before the collision the glider is at x = 0 m, and just after the collision the glider and apple are still at x = 0 m. Therefore, the total energy of the glider just after the collision is Enew = 1/2mglider + applev^2new = 1/2kA^2new
(b) Calculate the period of the motion of the glider and the period of the motion of the glider with apple.
Hint: it's a very simple question.

Physics
1 answer:
solniwko [45]3 years ago
4 0

Answer:

A) The new amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please find the attached files for the solution

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3. A large passenger ship required a force of 1,600,000 N to move 2000 m. How much work is done on the ship?​
Dovator [93]

Answer:

3,200,000,000 J

Explanation:

Work is defined as the amount of energy transferred as an object is moved a certain distance with a certain force. Mathematically, we express this with the equation

W=Fs

where W is work (measured in joules), F is the force applied (in Newtons), and s is the distance, also called the <em>displacement </em>(in meters).

Here, we have F = 1,600,000 N and s = 2000 m, so our work will be

W=1.600.000(2.000)=3.200.000.000 J

7 0
3 years ago
To water the yard you use a hose with a diameter of 3.2 cm. Water flows from the hose with a speed of 1.1 m/s. If you partially
cupoosta [38]

Answer:

The speed of water flow inside the pipe at point - 2 = 34.67 m / sec

Explanation:

Given data

Diameter at point - 1 = 3.2 cm

Velocity at point - 1 = 1.1 m / sec = 110 cm / sec

Diameter at point - 2 = 0.57 cm

Velocity at point - 2 = ??

We know that from the continuity equation the rate of flow is constant inside  a pipe between two points.

Thus

⇒ A_{1} × V_{1} = A_{2} × V_{2}

⇒  \frac{\pi }{4} × d_{1} ^{2} × V_{1} =

⇒  d_{1} ^{2} × V_{1} =  d_{2} ^{2}  × V_{2}

⇒  (3.2)^{2} × 110 = (0.57)^{2} × V_{2}

⇒ V_{2} = 3467 cm / sec

⇒ V_{2} = 34.67 m / sec  

Thus the speed of water flow inside the pipe at point - 2 = 34.67 m / sec

3 0
3 years ago
The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ
LenaWriter [7]

The orbital period of the satellite[T] is given as 2*10^{6} S.

The radius of the satellite is given [R] 2.5*10^{12} m.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V=\frac{2\pi R}{T}

where T is the time period of the satellite and R is the radius of the satellite.

                        V=\frac{2*3.14*10^{12} }{2*10^{6} }

                               = 7.85*10^{6} m/s

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V=\sqrt{\frac{GM}{R^{2} } }

where G is the gravitational constant and M is the mas of central object.

But we know that g=\frac{GM}{R^{2} }

                               ⇒GM=gR^{2}  where g is the acceleration due to gravity of that central object.


Hence    V=\sqrt{\frac{gR^{2} }{R} }

               ⇒   V=\sqrt{gR}

By knowing the value of g due to that central object we can also calculate its tangential speed.

                           

 




7 0
3 years ago
Read 2 more answers
What is the maximum mass of ethyl alcohol you could boil with 2000 j of heat, starting from 19 ∘c?
RSB [31]
The boiling point of ethanol is at 78.37°C. So, the energy must include sensible heat to raise 19°C to the boiling point and latent heat to change liquid to gas. The equation would be

Energy = Sensible heat + Latent heat
Energy = mCpΔT + mΔH

For ethanol, 
Cp = 46.068 + 102,460T - 139.63T² - 0.030341T³ + 0.0020386T⁴ J/kmol·K
ΔH = 38,560 J/mol

Integrate the Cp expression to determine CpΔT:

CpΔT = ∫₂₉₂³⁵²(46.068 + 102,460T - 139.63T² - 0.030341T³ + 0.0020386T⁴ )dT
The upper limit is (78.37+273) = 352 K, while the lower limit is (19 + 273) = 292.
CpΔT = 2384857192 J/kmol·K

2,000 J = m(2384857192 J/kmol)(1 kmol/1000 mol) + m(38,560 J/mol)
m = 8.253×10⁻⁴ moles of ethanol
Since the molar mass of ethanol is 46.07 g/mol,
Mass = (8.253×10⁻⁴ mol)(46.07 g/mol)
Mass = 0.038 g ethanol
8 0
4 years ago
An eagle is flying horizontally at a speed of 4.10 m/s when the fish in her talons wiggles loose and falls into the lake 6.20 m
Sonbull [250]

Answer:

V=11.74m/s, 69.59°

Explanation:

From newtons equation of motion ,we know that

V^2 = u^2+2gh

for the rt

For the vertical component of the speed

Vy^2=V0^2+ 2gh

Vy= the final speed in the vertical axis

V0= initial speed, 0m/s

g= acceleration due to gravity 9.81m/s

h=height/distance between the eagle and the lake

Vy^2= 2*9.81*6.2

Vy =√121.644

Vy=11.02m/s

The resultant speed will be

V=(Vy^2+Vx^2)^0.5

V=(11.02^2+4.1^2)^0.5

V=137.7241^0.5

V=11.74m/s

Direction

Tan^-1(11.02/4.1)

β=69.59°

V=11.74m/s, 69.59°

Answer =11.74m/s, b°

4 0
3 years ago
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