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vladimir2022 [97]
3 years ago
15

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

face which causes the block to come to a complete stop. The coefficient of kinetic friction between the block and the surface is 0.5.
Calculate the distance the block slides while coming to a stop. Show all work.
Physics
1 answer:
emmasim [6.3K]3 years ago
8 0

• Net vertical force on the block:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

(<em>n</em> = magnitude of normal force, <em>w</em> = weight)

<em>n</em> = <em>w</em> = <em>m g</em>

(<em>m</em> = mass, <em>g</em> = 9.8 m/s²)

<em>n</em> = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ <em>F</em> = -<em>f</em> = <em>m a</em>

(<em>f</em> = mag. of friction, <em>a</em> = acceleration)

We have <em>f</em> = <em>µ</em> <em>n</em> = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) <em>a</em>

<em>a</em> = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆<em>x</em> in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆<em>x</em>

∆<em>x</em> = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

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Answer:

\lambda=3.99*10^{-7}m

Explanation:

According to the law of conservation of energy, the energy of the absorbed photon must be equal to the binding energy of the electron plus the energy of the released electron:

E_p=E_b+E_r\\\frac{hc}{\lambda}=\frac{13.6eV}{n^2}+\frac{m_ev^2}{2}

1 eV is equal to 1.6*10^{-19}J, so:

13.6eV*\frac{1.6*10^{-19}J}{1eV}=2.18*10^{-18}J

Solving for \lambda and replacing the given values:

\lambda=\frac{hc}{\frac{2.18*10^{-18}J}{n^2}+\frac{m_ev^2}{2}}\\\lambda=\frac{6.63*10^{-134}J\cdot s(3*10^8\frac{m}{s})}{\frac{2.18*10^{-18}J}{3^2}+\frac{(9.11*10^{-31}kg)(7.5*10^5\frac{m}{s})^2}{2}}\\\lambda=3.99*10^{-7}m

4 0
3 years ago
Use the accompanying seismogram to answer which of the three types of seismic waves reached the seismograph first.
UkoKoshka [18]

Answer:

Primary waves (P-waves)

Explanation:

Due to excess of the energy inside the earth when the tectonic plates begin to slide or fracture then the energy is released in the form of seismic waves, this causes the earthquake.

<u>Two types of seismic waves are generally responsible for the earth quakes:</u>

  1. body waves
  2. surface waves

Body waves are of two types:

Primary waves (P-waves)

These are the fastest of all the waves involved in the earth-quake which travel at a speed of 1.6 km to 8 km per second.

They can pass trough solids, liquids and gases. They arrive at the surface as an instant thud.

Secondary waves (S-waves)

They can only pass through the solids and they move slower than the P-waves.

As S-waves move, they displace the rock particles, pushing them outwards perpendicular to the wave-path that leads to the earthquake-related first rolling period.

Surface waves (L-waves/ long waves)

  • These waves move along the surface of the earth. They are responsible for the earthquake's carnage.
  • They move up and down the Earth's surface, rocking the foundations of man-made structures.
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3 years ago
If a FM radio station broadcasts at 80. 3 MHz (megahertz), what is its wavelength in m (speed of light 3. 0 x 108 m/s)
hammer [34]

Answer:

Wavelength = 3.74 m

Explanation:

In order to find wavelength in "metres", we must first convert megahertz to hertz.

1 MHz = 1 × 10⁶ Hz

80.3 Mhz = <em>x</em>

<em>x </em>= 80.3 × 1 × 10⁶ = 8.03 × 10⁷ Hz

The formula between wave speed, frequency and wavelength is:

v = fλ  [where v is wave speed, f is frequency and λ is wavelength]

Reorganise the equation and make λ the subject.

λ = v ÷ f

λ = (3 × 10⁸) ÷ (8.03 × 10⁷)

λ = 3.74 m [rounded to 3 significant figures]

8 0
2 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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