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2 years ago

Ummmm ya girl needs help

Physics

2 answers:

Novosadov [1.4K]2 years ago

8 0

Hydrogen 1
Nitrogen 1
Oxygen 3

trasher [3.6K]2 years ago

5 0

hydrogen: 4

nitrogen: 4

oxygen: 12

this should be correct

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I need help with a physics worksheet about Newtons Laws.

Viktor [21]

The diagram of the object on the inclined plane is shown below

The mass of 7kg is exerted on the plane surface. It is acted upon by gravity. The force exerted on the surface is calculated by applying the formula,

F = mass x acceleration due to gravity

Assumme acceleration due to gravity is 9.8m/s^2, then

Force = 7 x 9.8 = 68.6N

The surface exerts an opposite force of the same magnitude with the force of the object but in opposite direction. Since it is inclined at an angle of 36.9 degrees,

Normal force = mgCosθ = 68.6Cos36.9 = 54.86N

Recall, frictional force = normal reaction x coefficient of friction

Given that coefficient of kinetic friction = 0.35,

Frictional force = 54.86 x 0.35 = 19.201N

This is the force that must be overcome to keep the object moving.

The force pulling the object upwards along the inclined plane is

mgSinθ = 7 x 9.8 x Sin36.9 = 41.19N

Since the velocity is constant, it means that there is no acceleration. The net force is zero. The force required to pull the mass and make it move at constant velocity must be equal to the sum of the exerted force and the the frictional force(Since it involves coeffricient of kinetic friction, it would cause the object to keep moving)

Thus,

Required force = 19.2 + 41.19 = 60.4N

Option E is correct

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1 year ago

Removing an electron from a neutral atom will result in an atom that?

Alex73 [517]

Removing an electron from a neutral atom will result in an atom that is positive.

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2 years ago

Read 2 more answers

An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.

I am Lyosha [343]

Answer:

The density of this object is approximately $1.36\; {\rm kg \cdot L^{-1}}$ .

The density of the oil in this question is approximately $0.600\; {\rm kg \cdot L^{-1}}$ .

(Assumption: the gravitational field strength is $g =9.806\; {\rm N \cdot kg^{-1}}$ )

Explanation:

When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

$\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}$ .

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was $(63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}$ . Hence, the weight of water that this object displaced would be $47.0 \; {\rm N}$ .

The mass of water displaced would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}$ .

The volume of that much water (which this object had displaced) would be:

$\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}$ .

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately $4.793\; {\rm L}$ .

The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately $4.793\; {\rm L}$ .

The weight of oil displaced would be equal to the magnitude of the buoyancy force: $63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}$ .

The mass of that much oil would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}$ .

Hence, the density of the oil in this question would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}$ .