What type of fossils were found near Macon Georgia?

The electric field strength at the midpoint of the rings is 0 N/C.

The electric field strength at the center  of the left ring is 2710.84 N/C.

The given parameters:

• Diameter of the rings, d = 10 cm
• Distance between the rings, r = 25 cm
• Charge of the rings, q = 20 nC

The electric field strength at the midpoint of the rings is calculated as follows;

The electric field strength at the center  of the left ring is calculated as follows;

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Final answer:

The electric field strength at the midpoint between the two rings is zero, and at the center of the left ring, it is 2.88 * 10^4 N/C.

Explanation:

The electric field strength at the

1. midpoint between the two rings is zero. The electric fields from each ring cancel each other out at this point because they are equal in magnitude and opposite in direction.
2. At the center of the left ring, the electric field strength can be calculated using the formula for the electric field due to a charged ring. The formula is E = k * (Q / r²), where E is the electric field strength, k is the Coulomb's constant, Q is the charge, and r is the distance from the center of the ring. Plugging in the values, we get:

E = (8.99 * 10^9 Nm²/C²) * (20.0 * 10^-9 C) / (0.05 m)² = 2.88 * 10^4 N/C

Answer: The correct answer is zero work done.

Explanation:

Work is said to be done when the object moves through a distance when the force is applied to the object.

If the object does not move a distance even the force is exerted on the object then the work done is zero in this case.

Therefore, when the force is exerted even when no work is done then this is called zero work done.

Final answer:

Force is experienced even when no work is done, such as when pushing against a wall. This is due to the fact that work in physics requires force to be applied over a distance. When no movement occurs, no work is done, yet a force was still exerted.

Explanation:

The concept you're referring to is known as force, a fundamental aspect in Newton's laws of motion. According to Newton's third law, every action has an equal and opposite reaction. So, when you push against a wall, it pushes back with an equal amount of force, even though no movement occurs, and therefore no work is done. This is due to the role distance plays in the calculation of work. In the physics sense, work is done when a force is applied over a certain distance.

This is also tied to the concept of potential energy. For example, when a force causes an object to deform, such as compressing a spring, the work done is stored as potential energy in the object until it is released. Yet, if the object does not move or deform, no work has been done, but a force was still exerted.

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Answer:

C. 2000kg rated sling

Explanation:

ensures better safety and can carry twice more mass than current mass.

Answer:

31.29 m/sec

Explanation:

We have given density of substance

We have convert this into

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram

We know that

So

So =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr

The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Given data :

separation between slits ( d ) = 1.5 x 10⁻³ m

wavelength of light ( λ ) =  514 * 10⁻⁹ m

Distance from narrow slit ( D ) = 5.0 m

Determine the distance between the adjacent bright fringes

we apply the formula below

w = D * λ  / d  ---- ( 1 )

where : w = distance between adjacent bright fringes

Back to equation ( 1 )

w =  ( 5 * 514 * 10⁻⁹ ) / 1.5 x 10⁻³

   = 1.7 * 10⁻³ M

Hence we can conclude that The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

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Answer:

m

Explanation:

d = separation between the two narrow slits = 1.5 mm = 1.5 x 10⁻³ m

λ = wavelength of the light = 514 nm = 514 x 10⁻⁹ m

D = Distance of the screen from the narrow slits = 5.0 m

w = Distance between the adjacent bright fringes on the screen

Distance between the adjacent bright fringes on the screen is given as

m

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

relative to the direction of the river.

b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

so, Dave's landing point is 540 m downstream.

d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).