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nika2105 [10]
3 years ago
10

A cube of copper was found to have a mass of 630 g. What are the dimensions of the cube? (The density of copper is 8.94 g/cm3.)​

Physics
1 answer:
raketka [301]3 years ago
3 0

Answer:

Density = mass/volume

Volume of cube = mass/density

630g/8.94 = 70.5cm^3 (3s.f.)

Since it is a cube, its dimensions of Length x breadth x height=volume

Take 70.5 amd cube root it

\sqrt[3]{70.4697986577)

=4.13 (3s.f)

hence , it would be 4.13 x 4.13 x 4.13 to find the volume

Hence, the answer would be 4.13cm(3s.f.)

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Examine the total number of orbitals in the n = 1 through 4 shells. what is the relationship between the n value and the total n
Veseljchak [2.6K]

The total number of orbitals in the n = 1 through 4 shells is only 1.

The relationship between the n value and the total number of orbitals is given as

Total number of orbitals = n²

Because there is only one way a sphere can be orientated in space, there is only one(1)  orbital in the n = 1 shell.

An orbital has n² total orbitals for a given value of n.

Each atom's electron is given one of four quantum numbers, the primary quantum number (symbolized n), to describe the state of that electron. It is a discrete variable since the values are all natural numbers starting at 1.

The orbital's size is defined by the primary quantum number (n). For example, orbitals with n = 2 are larger than those with n = 1.

Electrons are drawn to the atom's nucleus because their electrical charges are in opposition to one another.

In order to excite an electron from an orbital where it is close to the nucleus (n = 1) to an orbital where it is distant from the nucleus (n = 2), energy must be absorbed.

Thus, the energy of an orbital is indirectly described by the primary quantum number.

Learn more about Quantum numbers here brainly.com/question/5927165

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4 0
2 years ago
A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T
Nady [450]
The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse
6 0
4 years ago
A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where
const2013 [10]

Answer:

The work done by the force on the particle is 29.85 J.

Explanation:

The work is given by:  

W = ^{x_{2}}_{x_{1}}\int F_{x} dx

Where:

x₁: is the lower limit = 0 m    

x₂: is the upper limit = 1.9 m

Fₓ: is the force in the horizontal direction =  (4.5 + 13.7x - 1.5x²)N

W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx  

W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0}  

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}    

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}      

W = 29.85 J

Therefore, the work done by the force on the particle is 29.85 J.

I hope it helps you!                                

6 0
3 years ago
A load is to be pulled up an inclined plane using a pulley system. The inclined plane is 25.0 ft long and 5.00 ft high. The pull
tangare [24]
Advantage equals 0.1, mate
1) advantage of plane: 5 divided by 25 = 0.2
2) advantage of fixed pulleys = 0
3) advantage of movable pulley = 0.5
4) and you multiply 0.2 and 0.5 = 0.1

to lift 1kg you don't need force 10N, you will just need 1N
4 0
4 years ago
A 3.0N force to the right acts on a 0.5kg object at rest during a time interval of 4.0 seconds .What is the velocity of the obje
monitta

Answer:

24 m/s

Explanation:

Applying Newton's second law of motion,

F = ma................ Equation 1

Where F = force acting on the object, m = mass of the object a = acceleration of the object.

First we calculate for a.

make a the subject of the formula in equation 1 above

a = F/m............... Equation 2

Given: F = 3 N, m = 0.5 kg.

Substitute into equation 2

a = 3/0.5

a = 6 m/s²

Secondly we calculate the velocity by using,

v = u+at................... Equation 3

Where v = Final velocity of the object, u = initial velocity of the object, t = time interval

Given: u = 0 m/s (at rest), a = 6 m/s², t = 4 s

Substitute into equation 3

v = 0+6(4)

v = 24 m/s

5 0
3 years ago
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