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dlinn [17]
2 years ago
13

The initial velocity of a truck is 10 m/s. How long must it accelerate at a constant acceleration of 2m/s2 before its average ve

locity is equal to three times its initial velocity
Physics
1 answer:
Yuliya22 [10]2 years ago
7 0

Answer:

10s

Explanation:

vf = vi+at

30=10+2t

20= 2t

t = 10 s

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A combination of two identical resistors connected in series has an equivalent resistance of 12. ohms. What is the equivalent re
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Answer:

R1 + R2 = R = 12 for resistors in series - so R1 = R2 if they are identical

2 R1 = 12         and R1 = R2 = 6 ohms

1 / R = 1 / R1 + 1 / R2     for resistors in parallel

R = R1 * R2 / (R1 + R2) = 6 * 6 / (6 + 6) = 3

The equivalent resistance would be 3 ohms if connected in parallel

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Does a compressed spring transfer elastic energy to its surroundings?
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Yes

Explanation:

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What is meant by sundial???​
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Sundial is an instrument showing the time by the shadow of a pointer cast by the sun on to a plate marked with the hours of the day.

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An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

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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