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dmitriy555 [2]
3 years ago
11

How is a warm front differnt from a cold front

Physics
1 answer:
NISA [10]3 years ago
8 0
Bruh... but a warm front is different because it brings warm ness and sun, a cold front brings clouds , cloudy weather, and cold weather
You might be interested in
What is the amplitude and wavelength of the wave shown below?
masha68 [24]

Answer:

Option B

Explanation:

I did a question like this a long time ago so i'm not sure if its right so sorry if its wrong

3 0
2 years ago
A cheetah runs at a speed of 27.6 m/s. If the
aleksklad [387]

Answer:

256.68m

Explanation:

that is the procedure above

8 0
2 years ago
A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans
alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

                 y = (0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)

Now, we will compare above equation with the standard form of transeverse wave equation,

                 y = A sin(kx + \omega t)

where,    A is the amplitude = 0.09 m

              k is the wave vector = \frac{\pi}{11}

              \omega is the angular frequency = 4\pi

              x is displacement = 1.40 m

              t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave  will be:

                   v(t) = \frac{dy}{dt}

                v(t) = A \omega cos(kx + \omega t)

        v(t) = (0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)

          v(t) = -0.84 m/s

The acceleration of the particle in the location is

            a(t) = \frac{dv}{dt}

           a(t) = -A \omega 2sin(kx + \omega t)

           a(t) = -(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)

           a(t) = -9.49 m/s^{2}

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 m/s^{2} .

(b)  Wavelength of the wave is given as follows.

               \lambda = \frac{2\pi}{k}

              \lambda = (frac{2\pi}{\frac{\pi}{11})


              \lambda = 22 m

The period of the wave is

             T = \frac{2 \pi}{\omega}

             T = \frac{2 \pi}{4 \pi}

                = 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

                    v = \frac{\lambda}{T}

                       = \frac{22 m}{0.5 s}

                       = 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0
3 years ago
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
Maksim231197 [3]

Answer:

Tt = 70 + 135e^-0.031t

13 minutes

Explanation:

Given that :

Initial temperature, Ti = 205°

Temperature after 2.5 minutes = 195°

Temperature of room, Ts= 70

Using the relation :

Tt = Ts + Ce^-kt

Temperature after time, t

When freshly poured, t = 0

205 = 70 + Ce^-0k

205 = 70 + C

C = 205 - 70 = 135°

T after 2.5 minutes to find proportionality constant, k

Tt = Ts + Ce^-kt

195 = 70 + 135e^-2.5k

125 = 135e^-2.5k

125 / 135 = e^-2.5k

0.9259 = e^-2.5k

Take In of both sides :

−0.076989 = - 2.5k

k = −0.076989 / - 2.5

k = 0.031

Equation becomes :

Tt = 70 + 135e^-0.031t

t when Tt = 160

160 = 70 + 135e^-0.031k

90 = 135e^-0.031t

90/135 = e^-0.031t

0.6667 = e^-0.031t

In(0.6667) = - 0.031t

−0.405465 = - 0.031t

t = 0.405465/ 0.031

t = 13.071

t = 13 minutes

8 0
2 years ago
How can two strengths be added together
Ket [755]
By using mind and dna
6 0
3 years ago
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