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2 years ago

10

Which things determine an element's electronegativity? Check all that apply. A. The size of the atom of that element B. Whether

the element is a gas or a liquid at room temperature C. The number of neutrons D. The number of electrons in the valence shell Physics

1 answer:

KengaRu [80]2 years ago

6 0

D. The number of electrons

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Sarah walks WEST a distance of 5 meters then turns NORTH and walks 8

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Answer:

she is 3 meters in north

Explanation:

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2 years ago

An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0

Reptile [31]

Answer:

28 m/s^2

Explanation:

distance, s = 14 m

time, t = 2 - 1 = 1 s

initial velocity, u = 0 m/s

Let a be the acceleration.

Use third equation of motion

$s = ut + \frac{1}{2}at^{2}$

$14 = 0 + \frac{1}{2}a\times 1^{2}$

a = 28 m/s^2

Thus, the acceleration is 28 m/s^2.

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3 years ago

A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseba

Marina CMI [18]

Answer:

4.5m/s

Explanation:

Linear speed (v) = 42.5m/s

Distance(x) = 16.5m

θ= 49.0 rad

radius (r) = 3.67 cm

= 0.0367m

The time taken to travel = t

Recall that speed = distance / time

Time = distance / speed

t = x/v

t = 16.5/42.5

t = 0.4 secs

tangential velocity is proportional to the radius and angular velocity ω

Vt = rω

Angular velocity (ω) = θ/t

ω = 49/0.4

ω = 122.5 rad/s

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Vt = 0.0367 * 122.5

Vt =4.5 m/s

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2 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

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We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

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The electric field outside the solid conducting sphere and direction is away of solid sphere.

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2 years ago

What is the weight of a basketball with a mass of 0.5 kg

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3 years ago