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Consider the element carbon. How many electrons can be found in a carbon atom? How many are valence electrons?

Finger [1]

The answer is B: 6; 4. Carbon has four electrons in its outermost shell, which are its valence electrons.

8 0

2 years ago

Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec

SOVA2 [1]

Answer:

$F_{net} = 6.879\times 10^{- 7}\ N$

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

$F_{net} = F + F'$

$F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N$

6 0

2 years ago

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

$F=mrw^2$

$m=$ mass of the corresponding bugs

$r=$ corresponding radial distance of each bug

$w=$ angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

$F$ ∝ $mr$

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0

2 years ago

Read 2 more answers

If an object is thrown in an upward direction from the top of a building 1.6 x 102 ft. high at an initial velocity of 21.82 mi/h

stiv31 [10]

In this case, the object is thrown upwards from the building. Therefore, it first achieves some height before its starts dropping.

Now, when going upwards
v^2 = u^2 - 2gs

Where,
v = final velocity
u = initial velocity
g = gravitational acceleration
s = height achieved from the top of he bulding

Using the values given;
v = 0 (comes into rest before it starts dropping)
u = 21.82 mi/h = 32 ft/s
g = 9.81 m/s^2 = 32.174 ft/s^2

Then,
0^2 = 32^2 - 2*32.174*s
32^2 = 2*32.174*s
s = (32^2)/(2*32.174) = 15.91 ft

After achieving that height, it starts to drop from rest to maximum velocity when it hits the ground.
Applying the same formula;
v^2 = u^2 + 2gs

Where;
v = velocity when it hits the ground
u = initial velocity, 0 ft/s as it starts from rest
s = 15.91+1.6*10^2 = 15.91+160 = 175.91 ft

Therefore,
v^2 = 0^2 + 2*32.174*175.91
v^2 = 11319.68
v = Sqrt (11319.68) = 106.39 ft/s ≈ 32.43 m/s moving downwards.

5 0

2 years ago

Read 2 more answers

What instrument is normally used to measure atmospheric pressure?

Anuta_ua [19.1K]

B. Barometer is used to measure atmospheric pressure.

6 0

2 years ago