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Misha Larkins [42]
3 years ago
6

Write the electric field at point p in component form. assume that x-axis is horizontal and points to the right, and y-axis poin

ts upward.
Physics
1 answer:
salantis [7]3 years ago
5 0
E = q/r^2

where r is perpendicular distance between the point and source charge

Ex = E cos theta

Ey = E sin theta

take theta angle between horizontal and line joining the point charge and the point

 
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Is light fastest than speed and why.
Genrish500 [490]

Answer:

No

Explanation:

There is no limit to how fast the universe can expand, says physicist Charles Bennett of Johns Hopkins University. Einstein's theory that nothing can travel faster than the speed of light in a vacuum still holds true, because space itself is stretching, and space is nothing.

7 0
3 years ago
A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant
mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

3 0
2 years ago
A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration to a velocity of 13 m/s. What is the magn
olga55 [171]

Answer:

The acceleration is a = 2.75 [m/s^2]

Explanation:

In order to solve this problem we must use kinematics equations.

v_{f} = v_{i} + a*t\\

where:

Vf = final velocity = 13 [m/s]

Vi = initial velocity = 2 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Now replacing:

13 = 2 + (4*a)

(13 - 2) = 4*a

a = 2.75 [m/s^2]

5 0
3 years ago
Mention the name of the following frequencies depending upon their values a) 12Hz b) 20100Hz
Kaylis [27]
It can be a) 12Hz.................
5 0
3 years ago
The car's initial speed was 15 m / s and the distance the car travels before it comes to a complete stop after the driver applie
pentagon [3]

Initial speed of the car (u) = 15 m/s

Final speed of the car (v) = 0 m/s (Car comes to a complete stop after driver applies the brake)

Distance travelled by the car before it comes to halt (s) = 63 m

By using equation of motion, we get:

\bf \longrightarrow  {v}^{2}  =  {u}^{2}  + 2as \\  \\ \rm \longrightarrow  {0}^{2}  =  {15}^{2}  + 2 \times a \times 63 \\  \\ \rm \longrightarrow 0 = 225 + 126a \\  \\ \rm \longrightarrow 126a =  - 225 \\  \\ \rm \longrightarrow a =  -  \dfrac{225}{126}  \\  \\ \rm \longrightarrow a =  - 1.78 \: m {s}^{ - 2}

\therefore Acceleration of the car (a) = -1.78 m/s²

Magnitude of the car's acceleration (|a|) = 1.78 m/s²

5 0
3 years ago
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