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3 years ago

Please help on this one someone

Physics

1 answer:

Nonamiya [84]3 years ago

6 0

Valence electrons are the electrons in the outermost energy level of an atom — in the energy level that is farthest away from the nucleus.

I think it's A.

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On Earth, the gravitational Field strength is much than on the moon. If a piece of rock was taken from the Moon to the Earth, st

nikklg [1K]

Ok so I’m pretty sure the answer would be 2 because the mass of the rock would have the same mass on Earth as it has on the moon. Also the Density of a solid object remains constant meaning it doesn’t change. But the weight would change because the Earths gravitational pull is more than that of the moon. I hope this helped!

7 0

2 years ago

the apparent weight of a body wholly immersed in water is 32N and its weight in 96N and calculate volume of the body

ArbitrLikvidat [17]

Answer:

0.0065 m³

Explanation:

Apparent weight = weight − buoyancy

32 N = 96 N − (1000 kg/m³) (9.8 m/s²) V

V = 0.0065 m³

3 0

2 years ago

Help please!

JulsSmile [24]

3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:

(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v

where v is the velocity of the combined players. Solve for v :

450 kg•m/s - 320 kg•m/s = (155 kg) v

v = (130 kg•m/s) / (155 kg)

v ≈ 0.84 m/s

4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that

(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v

where v is the new velocity of the 4-kg ball. Solve for v :

30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v

v = (16.4 kg•m/s) / (4 kg)

v = 4.1 m/s

7 0

2 years ago

You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r

Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

$t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s$

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

$d=ut-\dfrac{1}{2}gt^2$

Putting all values

$d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m$

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

8 0

2 years ago

What is the magnitude of the momentum of a 3400 kg airplane traveling at a speed of 450 miles per hour?

Vitek1552 [10]

Answer:

The magnitude of momentum of the airplane is $6.83\times 10^5\ kg-m/s$ .

Explanation:

Given that,

Mass of the airplane, m = 3400 kg

Speed of the airplane, v = 450 miles per hour

Since, 1 mile per hour = 0.44704 m/s

v = 201.16 m/s

We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,

$p=mv$

$p=3400\ kg\times 201.16 \ m/s$

$p=683944\ kg-m/s$

$p=6.83\times 10^5\ kg-m/s$

So, the magnitude of momentum of the airplane is $6.83\times 10^5\ kg-m/s$ . Hence, this is the required solution.

7 0

3 years ago

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