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MrRa [10]
2 years ago
6

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

Physics
1 answer:
Leokris [45]2 years ago
3 0

Answer:

she is 3 meters in north

Explanation:

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If you saw a waxing gibbous moon what phase would you expect one week later?
Leona [35]
Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.
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3 years ago
Define and Explain the Sovereignty
GarryVolchara [31]

Sovereignty is a political concept that refers to dominant power or supreme authority.

Explanation:

Sovereignty is an attribute of states that is both an idea and a reality of state power. It is one of the means, an important one, by which the government of a state seeks to ensure the best it possibly can for its people. As such, it also changes over time.

6 0
2 years ago
What is the definition of Cubit
disa [49]

Answer:

See below.

Explanation:

Cubit is a unit of length based on the length of the forearm from the elbow to the tip of the middle finger and usually equal to about 18 inches (46 centimeters).

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7 0
2 years ago
An electric charge q of mass m in an oscillating electric field Eosinot experiences force q Eosinot. Suppose it starts from rest
Masja [62]

Answer:

speed of the charge electric is  v = - (Eo q/m) cos t

Explanation:

The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,

          F = q Eo sint

a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity

        F = ma

        q Eo sint = ma

        a = Eo q / m sint

        a = dv / dt

        dv = adt

        ∫ dv = ∫ a dt

        v-vo = I (Eoq / m) sin  t dt

        v- vo = Eo q / m (-cos t)

We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0

        v = - (Eo q / m) cos t

b) Kinetic energy

       

         K = ½ m v2

          K = ½ m (Eoq / m)²2 (sint)²

         K = ¹/₂  Eo² q² / m sin² t

c) The average kinetic energy over a period

          K = ½ m v2

         <v2> = (Eoq / m) 2 <cos2 t>

The average of cos2 t = ½, substitute and calculate

          K = ½ m (Eoq / m)²  ½

          K = ¼ Eo² q² / m

7 0
3 years ago
An electron of mass 9.11×10−31 kgkg leaves one end of a TV picture tube with zero initial speed and travels in a straight line t
ki77a [65]

Explanation:

We will use the equations of constant acceleration to find out a_{x} and time t.

As we know that the initial speed is zero. So

(a)  

v_{0x} = 0

x - x_{o} = 1.25×10^{-2}m

v_{x} = 3.3×10^{6}m/s

v^{2} _{x} =  v^{2} _{x_{o} } + 2a_{x} (x - x_{o} )

a_{x} = \frac{v^{2} _{x} - v^{2} _{ox} }{2(x - x_{o}) }

   = \frac{(3.3 * 10^{6})^{2}  - 0 }{2(1.25 * 10^{-2}) }

   = 4.356×10^{14} m/s²

(b)

v_{x} = v_{ox} + a_{x}t

t = v_{x} - vo_{x}/a_{x}

t = \frac{3.00 * 10^{6} }{4.356*10^{14} } = 6.8870×10^{-9}s

(c)

ΣF_{x} = ma_{x}

       = (9.11×10^{-31})(4.356×10^{14}m/s²)

       = 3.968×10^{-16} N

6 0
3 years ago
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