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Katarina [22]
2 years ago
7

WEATHER: What kind of weather occurs along this type of front?

Physics
2 answers:
Radda [10]2 years ago
4 0
Cold front because the warm front or hit front is lower on the right and goes downwards and cold front is in the middle and goes to Minneapolis
konstantin123 [22]2 years ago
3 0

Answer:

Cold front because the warm front or hit front is lower on the right and goes downwards and cold front is in the middle and goes to Minneapolis

Explanation:

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One way to describe a wave crest is the amount of ___________ applied over an area.
Pani-rosa [81]
I believe it’s force but i’m not really sure.
8 0
3 years ago
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Why are there two types of core?
dlinn [17]

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because the dog cat moster ate a bean.  the answer is 68

Explanation:

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3 years ago
The equation of a transverse wave traveling along a string is 1 1 y (2.00 mm)sin[(20 m )x (600 s )t] − − = − Find the (a) amplit
Lelu [443]

Answer:

a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V=  30 m/s  ( + x direction )

d)  λ = 0.31 m

e)Umax= 1.2 m/s

Explanation:

Given that

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

As we know that standard form of wave equation given as

y=A sin(\phi -\omega t)

A= Amplitude

ω=Frequency (rad /s)

t=Time

Φ = Phase difference

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

So from above equation we can say that

Amplitude ,A = 2 mm

Frequency ,ω= 600 rad/s                     (2πf=ω)

ω= 2πf

f= ω /2π

f= 300/π = 95.49 Hz

K= 20 rad/m

So velocity,V

V= ω /K

V= 600 /20 = 30 m/s  ( + x direction )

V = f λ

30 = 95.49 x  λ

 λ = 0.31 m

We know that speed is the rate of displacement

U=\dfrac{dy}{dt}

U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s

The maximum velocity

Umax = 1200 mm/s

Umax= 1.2 m/s

8 0
3 years ago
A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo
m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

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As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

7 0
3 years ago
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Arlecino [84]
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4 0
3 years ago
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