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Elenna [48]
3 years ago
6

Plzz answer this questions will mark as brainlist ​

Physics
1 answer:
EastWind [94]3 years ago
4 0

Answer:

1. 19.28 secs

2. 154.22 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

Force (F) = 1000 N

Mass (m) = 1200 Kg

Time (t) =..?

Distance (s) =...?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Force (F) = 1000 N

Mass (m) = 1200 Kg

Acceleration (a) =.?

Force (F) = mass (m) x acceleration (a)

F = ma

1000 = 1200 x a

Divide both side by 1200

a = 1000/1200

a = 0.83 m/s²

Since the car is coming to rest, it means it is decelerating. Therefore, the acceleration is – 0.83 m/s²

1. Determination of time taken for the car to halt i.e stop. This can be obtained as follow:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

acceleration (a) = – 0.83 m/s²

Time (t) =.?

v = u + at

0 = 16 + (–0.83 x t)

0 = 16 – 0.83t

Rearrange

0.83t = 16

Divide both side by 0.83

t = 16/0.83

t = 19.28 secs.

Therefore, the time taken for the car to halt is 19.28 secs.

2. Determination of the distance travelled by the car before coming to rest. This can be obtained as follow:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

acceleration (a) = – 0.83 m/s²

Distance (s) =..?

v² = u² + 2as

0 = 16² + (2 x –0.83 x s)

0 = 256 – 1.66s

Rearrange

1.66s = 256

Divide both side by 1.66

s = 256/1.66

s = 154.22 m

Therefore, the distance travelled by the car before coming to rest is 154.22 m.

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Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

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Answer:

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Answer:

a) 141.6m

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Explanation:

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