All categories

3 years ago

It's a cloudy and rainy day. The air pressure is most likely _____.

1 answer:

kramer3 years ago

5 0

The air pressure ( or atmospheric pressure ) is the force of air over a unit of area. Changes in the air pressure causes the weather changes. High pressure usually brings good weather with dry and cool air. But in a low pressure zone warm air is rising up. This vertical movements are caused by winds high in the troposphere. Water molecules stay as a gas in warmer air. After the vertical movement they condense and bring steady continuous rain. Therefore the low pressure brings cloudly and rainy weather. Answer: The air pressure is most likely low<span>. </span>

You might be interested in

Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil

laila [671]

Answer:

$W_n_e_t=7.648512 \approx 7.6J$

$K.E=0.8J$

$v=0.7844645406 \approx 0.78m/s$

Explanation:

From the question we are told that

Mass of pitcher $M= 2.6kg$

Force on pitcher $f=8.8N$

Distance traveled $48cm=>0.48m$

Coefficient of friction $\mu=0.28$

a)Generally frictional force is mathematically given by

$F=\mu N$

$F=0.28*2.6*9.8$

$F=7.1344N$

Generally work done on the pitcher is mathematically given as

$W_n_e_t=W_f+W_F$

$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

$W_n_e_t=4.224-3.424512$

$W_n_e_t=0.799488\approx 0.8J$

b)Generally K.E can be given mathematically as

$K.E= W_n_e_t$

Therefore

$K.E=0.8J$

c)Generally the equation for kinetic energy is mathematically represented by

$K.E=1/2mv^2$

$0.8=1/2mv^2$

Velocity as subject

$v=\sqrt{\frac{K.E*2}{m} }$

$v=\sqrt{\frac{0.8*2}{2.6} }$

$v=0.7844645406 \approx 0.78m/s$

6 0

3 years ago

two large boxes sit side by side on a sidewalk. the box on the left has a mass of 80kg and the box on the right has a mass of 50

garri49 [273]

The force that prevents motion when the surfaces of two objects come into contact is known as friction. Friction decreases a machine's mechanical advantage, or, to put it another way, reduces the output to input ratio.

<h3>How can I figure out the frictional force?</h3>

The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value.

The formula fr = Fr/N serves as a representation of it.

Therefore, 100N of force is needed to move an item with a mass of 50 kg.

It will accelerate by 10 m/s2.

If a substance's mass does not change over time, friction cannot affect it. Instead, friction can be affected in a variety of ways by an object's mass.

To Learn more About Friction, Refer:

brainly.com/question/24338873

#SPJ13

8 0

1 year ago

An object accelerating at 16 m/s/s doubles its mass and triples its net force acting on it. What will the new acceleration be? (

nataly862011 [7]

Answer:

24 m/s²

Explanation:

The given parameters are;

The initial acceleration of the object, a = 16 m/s²

Let 'm' represent the initial mass of the object

The initial force acting on the object, F = m × a

∴ F = 16 × m = 16·m

When the mass is doubled, we have;

The new mass of the object, m₂ = 2 × m = 2·m

When the net force acting on the object triples, we have;

The new net force acting on the object, F₂ = 3 × F = 3 × 16·m = 48·m

From F = m × a, we have;

a = F/m

∴ The new acceleration of the object, a₂ = F₂/m₂

From which, by plugging in the values, we have;

a₂ = 48·m/(2·m) = 24

The new acceleration of the object, a₂ = 24 m/s².

6 0

3 years ago

A wave traveling in water has a frequency of 250 Hz and a wavelength of 6.0m. What is the speed of the wave?

vladimir2022 [97]

Since you are looking for the speed, you need to rearrange the formula which is f = speed / wavelength. That should give you speed = f (wavelength.) All you need to do next is to substitute the value to the following equation. speed = 250 Hz (6.0m) that should leave you with 1500 m/s which is very fast.

6 0

3 years ago

Ron is on a Ferris wheel of radius 30 ft that turns counterclockwise at a rate of one revolution every 12 seconds. The lowest po

alexgriva [62]

Answer:

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

Explanation:

1 full revolution is $2\pi.$ let \theta be the angle of Ron's position.

At t = 0. $\theta = 0$

one full revolution occurs in 12 sec, so his angle at t time is

$\theta =2\pi \frac{t}{12} = \frac{\pi}{6}t$

r is radius of circle and it is given as

$x = rcos\theta$

$y = rsin\theta$

for r = 30 sec

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t$

however, that is centered at (0,0) and the positioned at time t = 0 is (30,0). it is need to shift so that the start position is (30,45). it can be done by adding to y

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

7 0

2 years ago