All categories

3 years ago

Which of an atom's electrons do you think are most likely to interact with other atoms?

Physics

2 answers:

max2010maxim [7]3 years ago

7 0

A. The elections neartest to tye nuclear

leva [86]3 years ago

6 0

A) The electrons nearest to the nucleus
correct me if i’m wrong

You might be interested in

The answer and how to do it

Arturiano [62]

Current = charge per second
2 Coulombs per second = 2 Amperes

Potential difference = (current)x(resistance) in volts.

That's (2 Amperes) x (2 ohms).

That's how to do it.
I think you can find the answer now.

8 0

4 years ago

two ships leave a port at the same time. The first ship sails on a bearing of 40 degrees at 18 knots and the second at a bearing

Yuliya22 [10]

Answer:

The distance between the ships is 87.84 km.

Explanation:

Given that,

Angle of first ship= 40°

Speed of first ship = 18 knots

Angle of second ship= 130°

Speed of second ship = 26 knots

We need to calculate the resultant velocity

Using cosine rule

$v=\sqrt{v_{1}^2+v_{2}^2-2v_{1}v_{2}\cos\theta}$

Put the value into the formula

$v=\sqrt{18^2+26^2-2\times18\times26\times\cos90}$

$v=\sqrt{18^2+26^2}$

$v=\sqrt{324+676}$

$v=10\sqrt{10}$

We need to calculate the distance between the ships

$d =v\times t$

Put the value into the formula

$d=10\sqrt{10}\times1.5\times1.852$

$d=87.84\ km$

Hence, The distance between the ships is 87.84 km.

7 0

3 years ago

1) Which of the following is not a dwarf planet?

vodomira [7]

1 io is a moon not a planet.
2)C,The Milky Way is a spiral galaxy
3)D,this is called the particle wave duality,stuff can act like particles and waves at the same time.
4)the white dwarfs
5)using distance and compass direction(I’m not sure about this one though)

5 0

3 years ago

Read 2 more answers

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

4 0

3 years ago

A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration?

Scrat [10]

Answer:

${ \boxed{ \bold{ \sf{Acceleration \: ( \: a) = 8 \: m/ {s \: }^{2} }}}}$

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

♨ Question :

A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?

♨ $\underbrace{ \sf{Required \: Answer : }}$

☄ Given :

Initial velocity ( u ) = 0
Final velocity ( v ) = 60.0 m/s
Time ( t ) = 7.50 s

☄ To find :

Acceleration ( a )

✒ We know ,

$\boxed{ \underline{ \bold{ \sf{Acceleration \: ( \: a) = \frac{Final velocity ( v ) - Initial velocity ( u)}{t} }}}}$

Substitute the values and solve for a.

➛ $\sf{a = \frac{60.0 - 0}{7.50}}$

➛ $\sf{a = \frac{60.0}{7.50}}$

➛ $\boxed{ \boxed{ \sf{a = 8 \: m/ {s \: }^{2} }}}$

---------------------------------------------------------------

✑ Additional Info :

When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
When a moving object comes in rest , in the case , final velocity ( v ) = 0
If the object is moving with uniform velocity , in the case , u = v.
If any object is thrown vertically upwards in the case , a = -g
When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.

Hope I helped!

Have a wonderful time ツ

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

4 0

3 years ago