Which equation represents a neutralization reaction?

An ocean liner leaves New York City and travels 18.0o north of east for 155 km. How far east and how far north has it gone? In o

Monica [59]

I’m sorry if i took up a lot of space, hope this is a valid approximate answer

6 0

1 year ago

A wave is a disturbance that carries from one place to another through matter of space

krok68 [10]

There's supposed to be a blank in the statement, where the answer is supposed to be inserted.

The question is supposed to say: "A wave is a disturbance that carries ______ from one place to another through matter or space".

To answer the question, write the word "<em>energy</em>" in the blank.

4 0

2 years ago

Tires of a Bigfoot truck has a diameter of 2.2 m. If it rotates 60 revolutions find distance travel on the road.

Firlakuza [10]

Answer:

$s = 414.7 m\\$

Explanation:

The relationship between the linear distance covered by an object and its angular displacement is given by the following formula:

s = rθ

where,

s = distance traveled on road = ?

r radius of tires = diameter/2 = 2.2 m/2 = 1.1 m

θ = angular displacement = (60 rev)(2π rad/1 rev) = 377 rad

Therefore,

$s = (1.1 m)(377 rad)\\s = 414.7 m$

8 0

2 years ago

A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net

Schach [20]

Explanation:

Let us first calculate long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

negative sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0

2 years ago

Two balls are thrown against a wall. Ball 1 has a much higher speed than ball 2.

Sunny_sXe [5.5K]

Let both the balls have the same mass equals to m.

Let $v_1$ and $v_2$ be the speed of the ball1 and the ball2 respectively, such that

$v_1>v_2\;\cdots(i)$

Assuming that both the balls are at the same level with respect to the ground, so let h be the height from the ground.

The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, $v$ , is $\frac 12 m v^2$

and the potential energy is due to the change in height is $mgh$ [where $g$ is the acceleration due to gravity]

So, the total energy of ball1,

$=\frac 12 m v_1^2 + mgh\;\cdots(ii)$

and the total energy of ball1,

$=\frac 12 m v_2^2 + mgh\;\cdots(iii)$ .

Here, the potential energy for both the balls are the same, but the kinetic energy of the ball1 is higher the ball2 as the ball1 have the higher speed, refer equation (i)

So, $\frac 12 m v_1^2 >\frac 12 m v_2^2$

Now, from equations (ii) and (iii)

The total energy of ball1 hi higher than the total energy of ball2.

6 0

2 years ago