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Darina [25.2K]
3 years ago
11

A particle (q = +5.0 µC) is released from rest when it is 2.0 m from a charged particle which is held at

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
3 0
The answer to your question is A. Have a great day
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Heterotrophs convert solar energy into chemical energy.<br> a. True<br> b. False
jonny [76]

The answer would be false

3 0
4 years ago
My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin
Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0  ⇒  K₀

and  h₁ = 0   ⇒    U₁ = 0

we get

U₀ = K₁    

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0
3 years ago
Cancer causing agents are also called
BARSIC [14]
Cancer-causing agents are also called <span>carcinogens</span>
5 0
4 years ago
Can someone please help me with this
Mekhanik [1.2K]

Answer:

B because the earth rotational axis tilt away from the sun.

3 0
3 years ago
What is the final velocity of a car that is originally traveling 12 m/s and then undergoes an acceleration of 2.3m/s squared for
Katarina [22]

To solve this problem we use the general kinetic equations.

We need to know the time it takes for the car to reach 130 meters.

In this way we have to:

x(t) = x_0 + v_0t + 0.5at ^ 2

Where

x_0 = initial position

v_0 = initial velocity

a = acceleration

t = time

x(t) = position as a function of time

130 = 0 + 12(t) + 0.5(2.3)t ^ 2

1.15t ^ 2 + 12t - 130.

We use the quadratic formula to solve the equation.

t = \frac{-12 \± \sqrt {(12) ^ 2-4(1.15)(- 130)}}{2 (1.15)}

t = 6.63 s and t = -17.1 s

We take the positive solution. This means that the car takes 6.63 s to reach 130 meters.

Then we use the following equation to find the final velocity:

v_f = v_0 + at

Where:

v_f = final speed

v_f = 12 +2.23(6.63)

The final speed of the car is 27.25 m/s

3 0
3 years ago
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