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earnstyle [38]
4 years ago
15

Please help on this one?

Physics
1 answer:
Vinvika [58]4 years ago
7 0

Answer:

C

Explanation:

Magnetic field lines go out of the north pole and into the south pole.

Thus, C is the south pole and A is the north pole.

The north pole of a bar magnet needs to be placed on the south pole of the magnet for it to be pulled.

Therefore, C is the answer

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How much GPE is stored when an 80kg astronaut climbs to the top of a 5m high lunar lander? The gravity strength on the moon is 1
8_murik_8 [283]

Answer:

The GPE, stored is 640 Joules

Explanation:

The given parameters are;

The given mass of the astronaut, m = 80 kg

The height of the top of the lunar lander to which the astronaut climbs, h = 5 m

The gravity strength on the moon, g = 1.6 N/kg

The Gravitational Potential Energy, GPE, stored is given according to the following equation;

GPE stored = m·g·h

Therefore, by substituting the known values, we have;

GPE Stored = 80 kg × 1.6 N/kg × 5 m = 640 Joules

The GPE, stored = 640 Joules.

6 0
3 years ago
What should you do immediately if a boat motor catches fire?
olganol [36]
 If a boat motor catches fire, you should immediately shut off the fuel supply, and try to put out the fire with an extinguisher. It is very important to have fully charged fire extinguishers on hand. As soon as you notice the fire activate the extinguisher and do not panic. In order to prevent fire remember to c<span>lean the bilges often and maintain proper gear stowage, make sure short-tie cables are properly connected...</span>
5 0
3 years ago
In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21
Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

6 0
4 years ago
100 Points
solniwko [45]

Answer:

4) Isaac Newton was the first scientist to propose the idea that gravity increases distance between objects decrease

Explanation:

1 - Kepler may have inspired most of Newton's propositions, but there is no record of collaborating on the theory of universal gravitation.

2 - The universal gravitation theory resemble the works of other great scientists, but Newton was indeed the first to propose this idea of gravitation

3 - On the contrary, these scientists have shared concepts, with differences in accuracy that supersede each other to be accepted by the community (i.e: Einstein being at the forefront despite sharing some ideas with Newton)

8 0
3 years ago
A leaky 10-kg bucket is lifted from the ground to a height of 14 m at a constant speed with a rope that weighs 0.5 kg/m. Initial
Lina20 [59]

solution:

Weight of bucket = 10kg

Length or distance =14m

Weight of rope=0.5kg/m

At any point x of the rope,

=(0.5)(14-x)

=(7-0.5x)

Since the water finishes draining at 14m level and total weight of water is 42kg

Total mass=(7-0.5x)+(42-3x)+10=(59-3.5x)kg

Force=(9.8)(59-3.5x)

work w =\lim_{n \to \infty }\sum_{i \to 1}^{n}(9.8)(59-3.5x)\Delta x\\=\int_{0}^{14}(9.8)(59-3.5x)dx\\=9.8\int_{0}^{14}(59-3.5x)dx\\9.8((59x-\frac{3.5x^2}{2})){_{0}}^{14}\\9.8(59(14)-\frac{3.5(14)^2}{2})\\=4733.4\\therefore,\\W=4733.4J\approx 4733J

5 0
3 years ago
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