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2 years ago

When air mass is caught between two cold fronts the result is a _______ front.

1 answer:

3 0

A. Occluded
Explanation- At an occluded front, the cold air mass from the cold front meets the cool air that was ahead of the warm front.

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A 30kg mass is brought from the earth to the moon what is the weight on the earth

ozzi

Answer:

Solution

verified

Verified by Toppr

Given:

Mass of body = 30 kg

gravitational acceleration on the moon = 1.62 m/s

Weight of the body on the moon = Mass of the body×gravitational acceleration on the moon=30×1.62=48 N

8 0

1 year ago

tom and ted are sitting on seprate chairs that have wheels. tom pushes ted and, in turn, he starts moving too. in which directio

IceJOKER [234]

In the direction opposite to his push.

5 0

3 years ago

he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee

Crazy boy [7]

Answer:

$F = -1.07 *10^{-8} \ N$

$F_r = 1.07 *10^{-8} \ N$

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

$F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}$

Here k is known as the proportionality constant with value $k = 2.31 * 10^ {-28} J \cdot m$

substituting -2 for $Q_{O}$ i.e the charge on oxygen , +1 for $Q_{Li}$ i.e the charge on Lithium and $[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}$ for r

$F = \frac{ 2.31 * 10^ {-28}* +1 * -2 }{ ( 0.208*10^{-9} )^2 }$

$F = -1.07 *10^{-8} \ N$

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

So Force of repulsionn is

$F_r = 1.07 *10^{-8} \ N$

6 0

3 years ago

When two magnets are brought close together their magnetic

Gnoma [55]

Answer:

You know the saying "Opposites attract" well that is how you can remember that South and North Magnetic Poles connect.

Explanation:

Hope this helps ya

6 0

3 years ago

While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 5.65 m/s. The stone subs

xxTIMURxx [149]

Answer:

1. 20.54m/s

2. 1.52s

Explanation:

QUESTION 1:

The speed the stone impact the ground is the final speed/velocity, which can be calculated using the formula:

v² = u² + 2as

Where;

v = final velocity (m/s)

u = initial velocity (m/s)

a = acceleration due to gravity (m/s²)

s = distance (m)

From the provided information, u = 5.65m/s, v = ?, s = 19.9m, a = 9.8m/s²

v² = 5.65² + 2 (9.8 × 19.9)

v² = 31.9225 + 2 (195.02)

v² = 31.9225 + 390.04

v² = 421.9625

v = √421.9625

v = 20.5417

v = 20.54m/s

QUESTION 2:

Using v = u + at

Where v = final velocity (m/s) = 20.54m/s

t = time (s)

u = initial velocity (m/s) = 5.65m/s

a = acceleration due to gravity (m/s²)

v = u + at

20.54 = 5.65 + 9.8t

20.54 - 5.65 = 9.8t

14.89 = 9.8t

t = 14.89/9.8

t = 1.519

t = 1.52s

3 0

3 years ago