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3 years ago

11

How do constellations differ from other groupings of stars such as galaxies?

1 answer:

SashulF [63]3 years ago

3 0

I think a constellation is an area of sky that contains up a picture, object, animal, ect. - a galaxy is a grouping of several million to billion stars, gas and dust, that is gravitationally bound.

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3. What distance will a car, traveling 65 km/hr, cover in 3.0 hrs? (195 km)

qwelly [4]

Answer:

$\boxed{\sf Distance \ travelled \ by \ car = 195 \ km}$

Given:

Speed = 65 km/hr

Time = 3.0 hrs

To Find:

Distance

Explanation:

Distance = Speed × Time

= 65 × 3

= 195 km

7 0

4 years ago

1. Which of the following air pollutants combines with haemoglobin in our blood and renders it incapable to

enyata [817]

B, Carbon monoxide is the answer :)

7 0

3 years ago

n astronaut who weighs 800 N on the surface of the earth lifts off from planet Zuton in a space ship. The free-fall acceleration

ANTONII [103]

Answer: 0.29 kN

Explanation:

We have the following data:

$W_{E}=800 N$ is the weight of the astronaut on Earth

$g_{E}=9.8 m/s^{2}$ is the free fall acceleration due gravity on Earth (directed downwards)

$g_{Z}=3 m/s^{2}$ is the free fall acceleration due gravity on Zuton (directed downwards)

$a=0.5 m/s^{2}$ is the acceleration of the spaceship at litoff (directed upwards)

We have to find the <u>magnitude of the force</u> $F$ the space ship exerts on the astronaut.

Firstly, we have to know weight has a direct relation with the mass and the acceleration due gravity. In the case of Earth is:

$W_{E}=mg_{E}$ (1)

Where $m$ is the mass of the atronaut.

Isolating $m$ :

$m=\frac{W_{E}}{g_{E}}$ (2)

$m=\frac{800 N}{9.8 m/s^{2}}$ (3)

$m=81.63 kg$ (4)

Now that we know the mass of the astronaut, we can find its weight on Zuton:

$W_{Z}=mg_{Z}$ (5)

$W_{Z}=(81.63 kg)(3 m/s^{2})$ (6)

$W_{Z}=244.89 N$ (7)

Then, we can calculate the force the space ship exerts on the astronaut by the following equation:

$F-W_{Z}=m.a$ (8)

Isolating $F$ :

$F=m.a+W_{Z}$ (9)

$F=(81.63 kg)(0.5 m/s^{2})+244.89 N$ (10)

$F=285.7 N \frac{1 kN}{1000 N}=0.285 kN$ (11)

Finally:

$F=0.285 kN \approx 0.29 kN$

5 0

4 years ago

Which of these experiments would make use of qualitative data?

Artyom0805 [142]

The answer is A study of different surfaces to compare ability to repel water. Hope this helps!

8 0

4 years ago

Read 2 more answers

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

4 years ago