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2 years ago

What is an neutral atom?

2 answers:

7 0

Atoms are neutral; they contain the same number of protons as electrons. By definition, an ion is an electrically charged particle produced by either removing electrons from a neutral atom to give a positive ion or adding electrons to a neutral atom to give a negative ion.

Alex2 years ago

5 0

It is one that the overall charge is zero

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A baseball is hit into the air with a velocity of 27 m/s. How high does it go?

Marizza181 [45]

Answer:

Explanation:

If a baseball is hit into the air with a velocity of 27 m/s, we want to determine the maximum height of the ball. Using the projecile formula;

Max height H = u²/2g

u is the initial velocity of the body = 27m/s

g is the acceleration due to gravity = 9.81m/s²

H = 27²/2(9.81)

H = 729/19.62

H = 37.16m

Hence the ball went 37.16m high

8 0

2 years ago

TRUE OR FALSE? The genetic code is determined by the SIZES of the nitrogen bases.

lidiya [134]

False it is not determined by size

5 0

2 years ago

Which object has a net force of -68 N?

Neko [114]

Answer:

5 or 6

Explanation:you

8 0

2 years ago

A soccer ball is kicked from the ground with an initial speed of 20.5 m/s at an upward angle of 45°. A player 55 m away in the d

Tanzania [10]

Answer:

S = 4.08 m/s

Explanation:

given,

initial speed, v = 20.5 m/s

angle made with horizontal, θ = 45°

Distance of player = 55 m

Average speed of the player to catch the ball = ?

vertical velocity of the ball =

v_y = v sin θ = 20.5 x sin 45° = 14.5 m/s

horizontal velocity of the ball

v_x = v cos θ = 20.5 x cos 45° = 14.5 m/s

time taken by the ball to reach at the highest point

v = u + g t

v = V_y - g t

0 = 14.5 - 9.8 x t

t = 1.48 s

total time of flight = 2 t = 1.48 x 2 = 2.96 s

Horizontal distance travel by the ball

s = vₓ x t

s = 14.5 x 2.96

s = 42.96 m

distance player has to run

D = 55 - 42.96 = 12.08 m

Average speed of the player

$S = \dfrac{D}{t}$

$S = \dfrac{12.08}{2.96}$

S = 4.08 m/s

Speed of the player is equal to 4.08 m/s

4 0

2 years ago

You want to get from a point a on the straight shore of the beach to a buoy which is 60 meters out in the water from a point b o

loris [4]

The time t for a constant speed is given by:
t = distance/ velocity

Let's call the distance to run on the beach x₁ and the remaining distance to swim x₂, then the equation becomes:
$t = \frac{x_1}{5} + \frac{x_2}{3}$

The distance x₂ depends on x₁:
$x_2 = \sqrt{(70 - x_1)^2 + 60^2}$

So: $t = \frac{x_1}{5} + \frac{ \sqrt{(70 - x_1)^2 + 60^2} }{3}$

Find the minimum by setting the derivative to zero: $\frac{dt}{dx_1} = 0$
$\frac{dt}{dx_1} = \frac{1}{5} - \frac{1}{3} \frac{70 - x_1} {\sqrt{70 - x_1)^2 + 60^2} } = 0$
$70 - x_1 = \frac{3}{5} \sqrt{(70 - x_1)^2 + 60^2} \\ \\ (70 - x_1)^2 = \frac{9}{25}((70 - x_1)^2 + 60^2) \\ \\ \frac{16}{25}(70 - x_1)^2 = \frac{9}{25}60^2 \\ \\ \frac{4}{5}(70 - x_1) = \frac{3}{5}60 \\ \\ x_1 = 25$

You should stop running 70m - 25m = 45m from b.

4 0

2 years ago