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3 years ago

What will the weather most likely be like the day after a cold front?

1 answer:

8 0

<span>As a cold front arrives, you are, most likely in the warm sector of the frontal low pressure. You could say that the air will be warmer than after the passing of the front.</span>

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9) An electrical appliance has a resistance of 25 N. When this electrical ap-

Neko [114]

(C)

Explanation:

From Ohm's law,

V = IR

Solving for I,

I = V/R

= (230 V)/(25 ohms)

= 9.2 A

6 0

3 years ago

12. A frain moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?

ziro4ka [17]

initial velocity=u=0m/s
Final velocity=v=25m/s
Time=t=30s

$\\ \tt\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \tt\longmapsto Acceleration=\dfrac{25-0}{30}$

$\\ \tt\longmapsto Acceleration=\dfrac{25}{30}$

$\\ \tt\longmapsto Acceleration=0.8m/s^2$

6 0

3 years ago

While filming an intense action sequence for the next James Bond movie, a controlled explosion detonates 1.3 km away from the ac

brilliants [131]

To solve this problem we must basically resort to the kinematic equations of movement. For which speed is defined as the distance traveled in a given time. Mathematically this can be expressed as

$v = \frac{d}{t}$

Where

d = Distance

t = time

For which clearing the time we will have the expression

$t = \frac{d}{v}$

Since we have two 'fluids' in which the sound travels at different speeds we will have that for the rock the time elapsed to feel the explosion will be:

$t = \frac{1300m}{3000m/s}$

$t = 0.433s$

In the case of the atmosphere -composite of air- the average speed of sound is 343m / s, therefore it will take

$t = \frac{1300m}{343m/s}$

$t = 3.79s$

The total difference between the two times would be

$\Delta t = 3.79s-0.433s$

$\Delta t = 3.357s$

Therefore 3.357s will pass between when they feel the explosion and when they hear it

8 0

4 years ago

Compare and contrast rutherford's "planetary model" of the atom with our current understanding of an atom's internal structure.

andrew11 [14]

<span>Rutherford felt that the atom had an internal structure much like our solar system, with a nucleus surrounded by orbiting electrons. Today, we know that the electrons inhabit specific shells and sub-shells, and there is an order to which these electrons are located.</span>

5 0

4 years ago

A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simp

Wittaler [7]

Answer:

l = 10.16 m

Explanation:

In this case, we have the period of oscillation of the pendulum is 6.4 s. It is required to find the height of the tower.

We know that the pendulum executes SHM. Let l is the height of the tower. The time period of simple pendulum is given by :

$T=2\pi \sqrt{\dfrac{l}{g}}$

g is acceleration due to gravity

We need to rearrange the above equation such that,

$l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(6.4)^2\times 9.8}{4\pi ^2}\\\\l=10.16\ m$

So, the height of the tower is 10.16 m.

4 0

4 years ago