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3 years ago

Question 13 of 20

Physics

2 answers:

Dominik [7]3 years ago

6 0

Answer:

B. It has a central nucleus composed of 29 protons and 35 neutrons,

surrounded by an electron cloud containing 29 electrons.

Explanation:

Protons and neutrons are the only subatomic particles with mass, and they are located in the nucleus. If this atom has an atomic number of 29 and is copper, it must have 29 protons (protons define which element is being observed). This means all remaining mass is from neutrons. 64-29 = 35.

Electrons have no mass and orbit the nucleus in the electron cloud. Since this copper atom is neutral (we are not told it has a charge), there must be an equal number of protons and electrons.

sp2606 [1]3 years ago

6 0

Answer:

B. It has a central nucleus composed of 29 protons and 35 neutrons,

surrounded by an electron cloud containing 29 electrons.

Explanation:

Hope this helps

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1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.

harkovskaia [24]

Answer:

(a) The net force is 80.394 N

The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

Apply Newton's second law of motion;

F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$

7 0

3 years ago

PLEASEEE HELPPPP

solniwko [45]

Answer:

Explanation:

average speed more than 25.0m/s.

5 0

3 years ago

Read 2 more answers

Gold has a density of 19300 kg/m³. Calculate the mass of 0.02 m³ of gold in kilograms.

hjlf

Answer:

Mass = 386 kg

Explanation:

<u><em>Density = Mass / Volume</em></u>

Mass = Density × Volume

Where D = 19300 kg/m³ , V = 0.02 m³

<em>Putting the given in the above formula</em>

Mass = 19300 × 0.02

Mass = 386 kg

7 0

3 years ago

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A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has

LekaFEV [45]

The distance D where the object comes to rest is 1.08.m.

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$W_{total} =$ ΔKE ⇒ -0.25*1*9.8*D = 0-1/2*1* $2.3^{2}$