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maksim [4K]
3 years ago
14

A skater is using very low friction rollerblades. A friend throws a Frisbee at her, on the straight line along which she is coas

ting. Describe each of the following events as an elastic, an inelastic, or a perfectly inelastic collision between the skater and the Frisbee.
Required:
a. She catches the Frisbee and holds it.
b. She tries to catch the Frisbee, but it bounces off her hands and falls to the ground in front of her.
c. She catches the Frisbee and immediately throws it back with the same speed (relative to the ground) to her friend.
Physics
1 answer:
kupik [55]3 years ago
5 0

Answer:

a)  perfectly inelastic,  b)  collision is inelastic,  c)   elastic  

Explanation:

In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved

Initial instant. Before the skater touches the frisbee

    p₀ = M v₁ + m v₂

where M and m are the masses of the skater and frisbee, respectively

for the final moment they give us several possibilities, in all case the moment is conserved

       p₀ = p_{f}

case a)

Final instant. grabs the frisbee and holds it

    p_{f} = (M + m) v '

     p₀ = p_{f}

We can see that this shock is perfectly inelastic, it holds the fressbee

case b)

final instant.

This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.

case c)

final instant. Grab the fressbee and resend it

      p_{f} = M v_{1f} + m v_{2f}

this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.

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A 1.20 m wire has a mass of 6.80 g and is under a tension of 120 N. The wire is held rigidly at both ends and set into oscillati
swat32

Answer:

145.52137 m/s

1.4 m

0.7 m

60.6339 Hz

121.2678 Hz

Explanation:

T = Tension = 120 N

\mu = Linear density  = \frac{m}{L}

m = Mass of wire = 6.8 g

L = Length of wire = 1.2 m

n = Number of loops

Velocity is given by

v=\sqrt{\frac{T}{\mu}}\\\Rightarrow v=\sqrt{\frac{T}{\frac{m}{L}}}\\\Rightarrow v=\sqrt{\frac{120}{\frac{6.8\times 10^{-3}}{1.2}}}\\\Rightarrow v=145.52137\ m/s

The speed of waves on the wire is 145.52137 m/s

Wavelength is given by

\lambda=\frac{2L}{n}\\\Rightarrow \lambda=\frac{2\times 1.2}{1}\\\Rightarrow \lambda=1.4\ m

The wavelength of the waves that produces one-loop standing waves is 1.4 m

\lambda=\frac{2L}{n}\\\Rightarrow \lambda=\frac{2\times 1.2}{2}\\\Rightarrow \lambda=0.7\ m

The wavelength of the waves that produces two-loop standing waves is 0.7 m

Frequency is given by

f=\frac{nv}{2L}\\\Rightarrow f=\frac{1\times 145.52137}{2\times 1.2}\\\Rightarrow f=60.6339\ Hz

The frequency of the waves that produces one-loop standing waves is 60.6339 Hz

f=\frac{nv}{2L}\\\Rightarrow f=\frac{2\times 145.52137}{2\times 1.2}\\\Rightarrow f=121.2678\ Hz

The frequency of the waves that produces two-loop standing waves is 121.2678 Hz

4 0
3 years ago
A single slit of width 0.50 mm is illuminated with light of wavelength 500 nm, and a screen is placed 120 cm in front of the sli
Vaselesa [24]

Answer:

a) 2.4 mm

b) 1.2 mm

c) 1.2 mm

Explanation:

To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:

m\lambda=asin\theta

a: width of the slit

λ: wavelength

m: order of the minimum

for little angles you have:

y=\frac{m\lambda D}{a}

y: height of the mth minimum

a) the width of the central maximum is 2*y for m=1:

w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm

b) the width of first maximum is y2-y1:

w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm

c) and for the second maximum:

w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm

4 0
3 years ago
A positive charge of 6.0 × 10-4 C is in an electric field that exerts a force of 4.5 × 10-4 N on it. What is the strength of the
VLD [36.1K]

Answer:

0.75 NC⁻¹

Explanation:

Electric field intensity ( or strength of the electric field ) is the force per a 1 C charge,

So, Force (F) = Electric field intensity(E) × Charge (q)

                 F = E×q ⇒ q = F/E

                                      = 4.5×10⁻⁴/6×10⁻⁴ = 0.75 NC⁻¹

According to cool om's law electric fields are generated due to charges. When charges are same there is a repulsive force acted on both charges. When charges are opposite there is a attraction force acted  on both charges.

According to cool om's law,

F =G×q1×q2 / r²

F = force exerted of two charges

q1 , q2 = charges

r = distance between two charges

And also Electric field intensity is a vector which has a magnitude and direction both. Direction is depending on a charge and the sign of the charge

6 0
3 years ago
Read 2 more answers
When a plane flies with the​ wind, it can travel 1575 miles in 3.5 hours. when the plane flies in the opposite​ direction, again
Zepler [3.9K]

When plane is flying along the wind then we can say

V_{plane} + v_{wind} = \frac{distance}{time}

V_{plane} + v_{wind} = \frac{1575}{3.5}

V_{plane} + v_{wind} = 450 mph

Now when its going against the wind the speed is given by

V_{plane} - v_{wind} = \frac{distance}{time}

V_{plane} - v_{wind} = \frac{1575}{4.5}

V_{plane} - v_{wind} = 350 mph

Now by the above two equations we will have

V_{plane} = 400 mph

v_{wind} = 50 mph

3 0
3 years ago
The bond enthalpy value for a carbon to hydrogen bond is 413 kJ. What does this mean?
AlladinOne [14]
Breaking bond requires energy. The bond between the carbon and hydrogen is broken when the energy is absorbed. The enthalpy is defined to be the energy taken to break the one mole of the stated carbon and hydrogen bond. Thus a should be the correct answer
6 0
4 years ago
Read 2 more answers
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