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Natasha_Volkova [10]
3 years ago
6

Why does it take more force to get a resting object to move than it does to keep

Physics
1 answer:
Triss [41]3 years ago
6 0

Answer:

Objects have a natural tendency to resist change. ... Heavier objects (greater mass) resist change more than lighter objects. Example: Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop.

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A wave has a wavelength of 7 mm and a frequency of 16 hertz. What is its speed?
Leokris [45]
 0.000112m/s hope this helps


3 0
3 years ago
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do constructive inference occur when the compression of one wave meets up with the compression of a second wave
Ugo [173]

Answer:

Yes

Explanation:

There are two types of interference possible when two waves meet at the same point:

- Constructive interference: this occurs when the two waves meet in phase, i.e. the crest (or the compression, in case of a longitudinale wave) meets with the crest (compression) of the other wave. In such a case, the amplitude of the resultant wave is twice that of the original wave.

- Destructive interferece: this occurs when the two waves meet in anti-phase, i.e. the crest (or the compression, in case of a longitudinal wave) meets with the trough (rarefaction) of the other wave. In this case, the amplitude of the resultant wave is zero, since the amplitudes of the two waves cancel out.

In this problem, we have a situation where the compression of one wave meets with the compression of the second wave, so we have constructive interference.

6 0
3 years ago
In an emergency situation, firemen need to respond as quickly as possible. If a fireman is responding from the second floor, how
rjkz [21]

Answer:

4m/s

Explanation:

May be different considering how long the pole is and how heavy the firefighter is.

8 0
3 years ago
What is the pendulum length whose period is 2.0s ?
Mashutka [201]
Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24mT=2 \pi   \sqrt{\frac{L}{g}} \\
 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi }  = \frac{L}{g} \\
 \frac{T^2}{2 \pi }*g=L\\
L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m

7 0
3 years ago
A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the
Elena L [17]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

8 0
3 years ago
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