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3 years ago

Proof of conservation of energy

1 answer:

6 0

If you compare the energy at two different times in an equation, you'll see no difference, because it's been conserved. The total energy of a system is the sum of its energy in motion - its kinetic energy - and its energy due to its position, which is its potential energy

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A trip is taken that passes through the following points in order

riadik2000 [5.3K]

Answer:

35, I got you bro, i got you

8 0

3 years ago

Suppose that a certain battery produces a voltage of 1.55V without a load connected (open circuit) and a current of 500mA when s

lubasha [3.4K]

Answer:

Explanation:

Let the internal resistance be r .

Since in open circuit the volt is 1.55 V , this will be the source voltage .

Source voltage = 1.55

If external resistance be R .

1.55 / (R + r ) = .500

R + r = 3.1 ohm

So sum of internal resistance and external resistance will be 3.1 ohm.

7 0

3 years ago

Daffy Duck is standing 6.8 m away from Minnie Duck. The attractive gravitational force between them is 5.4x10-8 N. If Daffy Duck

artcher [175]

Answer:

432.78 Kg

Explanation:

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

Distance apart (r) = 6.8 m

Force of attraction (F) = 5.4×10¯⁸ N

Mass of Daffy Duck (M₁) = 86.5 kg

Mass of Minnie Duck (M₂) =?

NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

The mass of Minnie Duck can be obtained as follow:

F = GM₁M₂ / r²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24

Cross multiply

6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24

Divide both side by 6.67×10¯¹¹ × 86.5

M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5

M₂ = 432.78 Kg

Therefore, the mass of Minnie Duck is 432.78 Kg

8 0

3 years ago

An airplane traveling 919 m above the ocean at 505 m/s is to drop a case of twinkies to the victims below. How much time before

Elena L [17]

Answer:

a. 13.7 s b. 6913.5 m

Explanation:

a. How much time before being directly overhead should the box be dropped?

Since the box falls under gravity we use the equation

y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.

So,

y = ut - 1/2gt²

y = 0 × t - 1/2gt²

y = 0 - 1/2gt²

y = - 1/2gt²

t² = -2y/g

t = √(-2y/g)

So, t = √(-2 × 919 m/-9.8 m/s²)

t = √(-1838 m/-9.8 m/s²)

t = √(187.551 m²/s²)

t = 13.69 s

t ≅ 13.7 s

So, the box should be dropped 13.69 s before being directly overhead.

b. What is the horizontal distance between the plane and the victims when the box is dropped?

The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m

6 0

3 years ago

What accounts for the two precipitation peaks in mbandaka?

slava [35]

The two precipitation peaks in Mbandaka during March to April and September to November is due to the intertropical convergence zone.

Intertropical convergence zone is a narrow zone located near the equator. It is where the northern and southern air masses intersect which results to low atmospheric pressure. Due to the intertropical convergence zone’s meeting of air masses, often times the air pressure are lower which will results to colder air, or even rainfall during the period of March to April, and most especially September to November in Mbandaka.

<span>Since Mbandaka is located at the cented of Tumba-Ngiri-Maindombe area, which is named as a Wetland of International importance, there is really a bigger chance that this place experience above 60mm precipitation in a year, temperatures averaging from 23 – 26 degrees Celsius.</span>

7 0

3 years ago