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NARA [144]
3 years ago
8

An electron is projected with horizontal speed 105m / s in a downwardly directed 304N / C electric field. Find the vertical posi

tion (in m) after 2.0 μs.
Physics
1 answer:
Vinil7 [7]3 years ago
4 0

Answer:

-107 m

Explanation:

Sum of forces in the y direction:

∑F = ma

-qE = ma

a = -qE/m

a = -(1.60×10⁻¹⁹ C) (304 N/C) / (9.11×10⁻³¹ kg)

a = -53.4×10¹² m/s²

Given in the y direction:

v₀ = 0 m/s

a = -53.4×10¹² m/s²

t = 2×10⁻⁶ s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (2×10⁻⁶ s) + ½ (-53.4×10¹² m/s²) (2×10⁻⁶ s)²

Δy = -107 m

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Gravitational potential energy is defined as the amount of energy an object possesses at a given height above the Earth's surface. The equation for gravitational energy is E = mgh. Here, m is the mass of the object, g is the acceleration due to gravity, and h is the height above the Earth's surface. Hence, the answer is E = 2000kg x 9.81m/s^2 x 2m = 39240 Joules. 
4 0
4 years ago
Count the dots to make sure that all of the valence electrons are represented 2
Semenov [28]

Do you have a diagram or anything?

5 0
4 years ago
At what distance is the electrostatic force between two protons equal to the weight of one?.
erik [133]

0.118 m is the distance between the two protons.

Mass of proton = 1.6726 × 10⁻²⁷ kg

Weight of proton= 1.6726 × 10⁻²⁷ x 9.81 N

                            = 1.6408 × 10⁻²⁶ N

Charge of proton = 1.602 × 10⁻²⁹ C

The force between two protons = kq²/r²     where, K is a proportionality    

                                                            constant, q is a charge of proton and  

                                                         r is the distance between two protons.          

                                                       = 9 × 10⁹ × (1.602×10⁻¹⁹)²/r²      

To calculate distance :

  Weight of proton= Force between protons

 ⇒   1.6408 × 10⁻²⁶ N = 9 × 10⁹ × (1.602×10⁻¹⁹)²/r²

 ⇒    r = 0.118m

Therefore, 0.118 m is the distance between the two protons.

Learn more about electrostatic force here:

brainly.com/question/18108470

#SPJ4

8 0
2 years ago
a quarterback throws a football with angle of elevation 40degrees and speed 60ft/s. Find the horizontal and vertical components
pshichka [43]
<span>
the horizontal velocity would be equal to
Vh = sin (40) /60
= 0.74 * 60
= 44
</span>
the Vertical velocity would be equal to
<span>Vv = cos(40) * 60
=40</span>
4 0
3 years ago
An airplane maintains a speed of 627 km/h relative to the air it is flying through as it makes a trip to a city 787 km away to t
amid [387]

Answer:1.33 hr

Explanation:

Given

Speed of airplane relative to air=627 km/h

Distance traveled =787 km

Speed of wind=38.4 km/h

Wind is blowing in opposite direction therefore net speed is

v_{net}=627-38.4=588.6 km/h

thus time taken is

t=\frac{distance}{speed}

t=\frac{787}{588.6}=1.33 hr

4 0
4 years ago
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