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bija089 [108]
3 years ago
8

A stationary boat bobs up and down with a period of 2.4 s when it encounters the waves from a moving boat.

Physics
1 answer:
lapo4ka [179]3 years ago
8 0
It seems like the question is asking for the frequency.
Given: 
Time period (T) = 2.4 sec

Frequency (f) =? 

We know that the formula for frequency is:

Frequency (f) = 1/time period (T) 

= 1 / 2.4 s 

= 0.42 Hz. is the frequency for this problem.
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A 50.0 ohm and a 30.0 ohm resistor are connected in parallel. What is their equivalent resistance? Unit=Ohms
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R(parallel) = product/ sum

50×30/50+30

1500/80

18,75 ohms

4 0
3 years ago
What evidence can you cite for the particle nature of light? 1. Refraction phenomenon of light 2. The many colors of light 3. Di
GalinKa [24]
Evidence for the particle nature of light are not: 1. refraction, 2.  many colors of light, 3. diffraction. These are all phenomenon that support wave theory of light. Evidence for particle nature of light is photoelectric effect. Because it was discovered that you need discrete energies of light to eject electrons from a metal surface and not continuous as the wave theory of light suggests. 
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3 years ago
After creating a prototype, which of the following would not be and appropriate next step in the design process of a new technol
Dafna11 [192]

The answer would be to research the need.  This should have been done before the project began.

6 0
3 years ago
Read 2 more answers
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
2 years ago
An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the polic
Montano1993 [528]

Answer:

Speed of the speeder will be 28 m/sec

Explanation:

In first case police car is traveling with a speed of 90 km/hr

We can change 90 km/hr in m/sec

So 90km/hr=\frac{90\times 1000m}{3600sec}=25m/sec

Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m

After that car is accelerating with a=2m/sec^2 for 7 sec

So distance traveled by car in these 7 sec

S=ut+\frac{1}{2}at^2=25\times 7+\frac{1}{2}\times 2\times 7^2=224m

So total distance traveled by police car = 224 m

This distance is also same for speeder

Now let speeder is moving with constant velocity v

so 224=\left ( 1+7 \right )v

v = 28 m/sec

4 0
3 years ago
Read 2 more answers
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