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dimulka [17.4K]
3 years ago
14

A student is given an assignment to demonstrate diffraction. He takes a photograph of a straw in a glass of water. The straw app

ears bent at the water level. Which best describes this example?
A) This is a good example of diffraction.
B) This is an example of dispersion and not diffraction.
C) This is an example of refraction and not diffraction.
D) This is an example of reflection and not diffraction.
Physics
2 answers:
arlik [135]3 years ago
8 0

The answer is "c". because the straw the straw moves from air to water and appears refracted at the water level

wel3 years ago
7 0

The answer is C, This is an example of refraction and not diffraction.

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Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
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4 years ago
Convert: 8 mm: ____________ cm
vfiekz [6]

-1^10

centi=10 mm and mm=1/10=10^-1 cm

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3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subj
zloy xaker [14]

Answer:

T = 0.00889 N*m

Explanation:

Given the initial speed

Vo = 9.0rev/s.

V = 65 rev/10s

V = 6.5 rev/s.

V = Vo + a*t

Solve to acceleration knowing the initial velocity and the velocity at 10 s

6.5 rev/s - 9 rev/s = a*10s

a = -0.25 rev/s^2.

Now the solve the time at stop time so V=0

V = Vo + a*t

0 = 9.0 - 0.25 rev/s *t,

t = 36 s The Stopping time.

36s -  10s = 26s

The torque can be find using the acceleration using the equation

T = I*a

I = 1/2*m*r^2

I = 1/2*0.725kg*(0.315m)^2= 0.0359kg*m^2

T = 0.0359kg*m^2*-0.25rev/s^2

T = 0.00889 N*m

7 0
3 years ago
33] You have long hair, you should:
Sveta_85 [38]

Answer:

B

Explanation:

hair can be a safety hazard

4 0
3 years ago
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