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Leokris [45]
4 years ago
13

List the names, charges and locations of the three kinds of particles that make up an atom

Physics
1 answer:
Marina86 [1]4 years ago
5 0

The atom contains three kinds of particles:

  1. <u>Electrons</u>: They are negative charged particles with charge of 1.6×10^{-19} C and they spin around the nucleus in specific orbits
  2. <u>Protons</u>: They are positive charged particles with the same charge of the electrons (but in positive sign) and they are located inside the nucleus
  3. <u>Neutrons</u>:  They are uncharged particles and they are located inside the nucleus with the protons

Then the atom contains positive particles and negative particles, it is electrically neutral.

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A box is 5 cm high, 4 cm wide, and 9 cm long. What is the
FromTheMoon [43]

Answer:

18

Explanation:

  1. 5+4=9+9=<em>18</em><em> </em><em>IS</em><em> </em><em>THE</em><em> </em><em>VOLUME</em><em> </em><em>OF</em><em> </em><em>CUBIC</em><em> </em><em>CENTIMETRES</em><em> </em>
6 0
3 years ago
The climber dropped her compass at the end of her 240-meter climb. How long did it take to strike bottom?
ANEK [815]

240 = 0+1/2 (-9.8t

240 = -4.9t

<span>240/-4.9 = t</span><span>
</span>49.0 = t

t= 7.0s

4 0
4 years ago
Read 2 more answers
A +13.4 nC charge is located at (0,9.4) cm and a -4.23 nC charge is located (4.99, 0) cm. Where would a -14.23 nC charge need to
antiseptic1488 [7]

In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge

E1 = (9x10^9)(13.4x10^-9)/(9.4x10^-2)^2 = 13648.7 V/m towards the negative y-axis

E2 = (9x10^9)(4.23x10^-9)/(4.99x10^-2)^2 = 15289.1 V/m towards the positive x-axis

The red arrow shows the direction of which the electric field points.

To make the electric field at the origin 0, we must find a location where q3 = the magnitude of q1 and q2

Etotal = sqrt(E1+E2) = 20494.97 V/m

E3 = 20494.97 = (9x10^9)(14.23x10^-9)/(d)^2

d = 0.079 m = 7.9 cm

4 0
1 year ago
You push a 85 kg shopping cart from rest with a net force of 250 n for 5 seconds,at which point it flies off a cliff that is 100
Vikki [24]

m = mass of the cart = 85 kg

F = net force on the cart = 250 N

a = acceleration of the cart

acceleration of the cart is given as

a = F/m

a = 250/85

a = 2.94 m/s²

t = time for which the force is applied = 5 sec

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart just before  it flies off the cliff = ?

using the equation

v = v₀ + a t

inserting the values

v = 0 + (2.94) (5)

v = 14.7 m/s

consider the motion of cart after it flies off the cliff in vertical direction :

v' = initial velocity in vertical direction = 0 m/s

a' = acceleration in vertical direction = g = acceleration due to gravity = 9.8 m/s²

t' = time taken for the cart to land = ?

Y' = vertical displacement of the cart = height of cliff = 100 m

using the kinematics equation

Y' = v' t' + (0.5) a' t'²

100 = (0) t' + (0.5) (9.8) t'²

t' = 4.52 sec


consider the motion of cart along the horizontal direction after it flies off the cliff

X = distance traveled from the base of cliff = ?

t' = time of travel = 4.52 sec

v = velocity along the horizontal direction = 14.7 m/s

distance traveled from the base of cliff is given as

X = v t'

X = 14.7 x 4.52

X  = 66.4 m


3 0
4 years ago
When light passes the
Leno4ka [110]

When light crosses the boundary between layers with different densities, the light is refracted. (A).

3 0
4 years ago
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