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Ratling [72]
3 years ago
8

The black lines that wrap themselves around a basketball represent an example of what type of geometry?

Physics
1 answer:
Dmitry [639]3 years ago
5 0
In Euclidean geometry parallel lines never intersect. But in non-Euclidean geometry parallel lines can either curve away from each other, or curve towards each other. Example : the black lines that wrap themselves around the basketball.
Answer: B ) non-Euclidean 
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As sound waves move from air to glass, the wave property of ________ will cause the wave to bend in that direction
Zarrin [17]
<span>The amplitude because that controls the height of the wave. Correct answer: Amplitude.</span>
5 0
4 years ago
A toy cannon uses a spring to project a 5.38-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force
yawa3891 [41]

(a) 1.43 m/s

We can solve this problem by using the law of conservation of energy.

The initial total energy stored in the spring-mass system is

E=U=\frac{1}{2}kx^2

where

k = 7.91 N/m is the spring constant

x=5.08 cm = 0.0508 m

Substituting,

E=\frac{1}{2}(7.91)(0.0508)^2=0.0102 J

The final kinetic energy of the ball is equal to the energy released by the spring + the work done by friction:

E+W_f=K

where

K_f=\frac{1}{2}mv^2 is the kinetic energy of the ball, with

m=5.38 g = 5.38\cdot 10^{-3} kg being the mass of the ball

v being the final speed

W_f = -F_f d is the work done by friction (which is negative since the force of friction is opposite to the motion), where

F_f = 0.0323 N is force of friction

d = 14.5 cm = 0.145 m is the displacement

Substituting,

W_f = -(0.0323)(0.145)=-4.68\cdot 10^{-3} J

So, the kinetic energy of the ball as it leaves the cannon is

K_f = E+W_f = 0.0102 - 4.68\cdot 10^{-3}=0.00552 J

And so the final speed is

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00552)}{0.00538}}=1.43 m/s

(b) +5.08 cm

The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.

The elastic potential energy of the spring is

U=\frac{1}{2}kx^2

And this has all been released when it becomes zero, so when x = 0 (equilibrium position of the spring). However, the spring was initially compressed by 5.08 cm, so the ball has maximum speed when

x = +5.08 cm

with respect to the initial point.

(c) 1.78 m/s

The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.

As we calculated in part (a), the total energy released by the spring is

E = 0.0102 J

The work done by friction here is just the work done to cover the distance of

d = 5.08 cm = 0.0508 m

Therefore

W_f = -(0.0323)(0.0508)=-1.64\cdot 10^{-3} J

So, the kinetic energy of the ball at the point of maximum speed is

K_f = E+W_f = 0.0102 - 1.64\cdot 10^{-3}=0.00856 J

And so the final speed is

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00856)}{0.00538}}=1.78 m/s

7 0
3 years ago
The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Figure below.. Det
Savatey [412]

Answer:

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Patrick: (clearly triggered) <em>Why'd</em><em> </em><em>you</em><em> </em><em>say</em><em> </em><em>"</em><em>bye</em><em> </em><em>squidward</em><em>"</em><em> </em><em>twice</em><em>?</em>

Spongebob: <em>I</em><em> </em><em>LiKe</em><em> </em><em>SqUiDwArD</em>

5 0
3 years ago
In a single-slit diffraction experiment, a coherent light source illuminates a slit in a barrier, and the resulting pattern is p
UkoKoshka [18]

Answer:

hsjdsdddwqdqwdd

Explanation:

4 0
3 years ago
A Boeing 737—a small, short-range jet with a mass of 51,000 kg— sits at rest at the start of a runway. The pilot turns the pair
lara31 [8.8K]

Answer:

67000N

Explanation:

We solve for the acceleration using the the 3rd constant-acceleration equation.

(Vx)f² = (Vx)i² + 2ax∆x

We have the displacement to be

∆x = Xf - Xi = 940m

Vx = 70m/s

The acceleration = (70m/s)²/2(940m)

= 4900/1880

= 2.61m/s²

From isaac newton's second law,

51000kg x 2.61m/s²

= 133,000N

The engines thrust is half of this value

Therefore thrust = 67000N or 67kN

8 0
3 years ago
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