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bearhunter [10]
1 year ago
7

Identify the following terms:

Physics
1 answer:
anastassius [24]1 year ago
4 0

Answer:

atom -
the smallest particle of a chemical element that can exist.

atomic mass-
the quantity of matter contained in an atom of an element

atomic weight -
ratio of the average mass of a chemical element's atoms to some standard

protons-
stable subatomic particle that has a positive charge equal in magnitude to a unit of electron charge and a rest mass of 1.67262 × 10−27 kg

electrons-
a stable subatomic particle with a charge of negative electricity, found in all atoms and acting as the primary carrier of electricity in solids

neutrons-
a subatomic particle of about the same mass as a proton but without an electric charge, present in all atomic nuclei except those of ordinary hydrogen.

energy levels-
one of the stable states of constant energy that may be assumed by a physical system  
[used especially of the quantum states of electrons in atoms and of nuclei. — called also energy state.]

Covalent bonds

the interatomic linkage that results from the sharing of an electron pair between two atoms.

ionic bonds
type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.

Valence electrons
a single electron or one of two or more electrons in the outer shell of an atom that is responsible for the chemical properties of the atom.

Lewis Dot Diagram
A way of representing atoms or molecules by showing electrons as dots surrounding the element symbol. One bond is represented as two electrons.

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A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
ikadub [295]

Answer:

f_{fr}=1590.85 N

Explanation:

Here the total force at the horizontal components will be equal to the centripetal force on the car. So we will have:

f_{fr}cos(25)+Nsin(25)=m\frac{v^{2}}{r} (1)

  • f(fr) is the friction force
  • N is the normal force

Now, the sum of forces at the vertical direction is equal to 0.

Ncos(25)-mg-f_{fr}sin(25)=0 (2)          

Let's combine (1) and (2) to find f(fr)

f_{fr}=\frac{(mv^{2}/r)-mgtan(25)}{cos(25)+tan(25)sin(25)}

f_{fr}=\frac{(600*30^{2}/120)-600*9.81*tan(25)}{cos(25)+tan(25)sin(25)}  

f_{fr}=1590.85 N

I hope it helps you!

5 0
3 years ago
Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is r
katen-ka-za [31]

Answer:

<em>171.5m</em>

Explanation:

The velocity of sound in water = 343m/s

Time taken = 1.00secs

using the formula to calculate the distance

2x = vt

x is the distance

v is the speed of sound

t is the time

x = vt/2

x = 343(1)/2

x = 171.5m

<em>hence their separation 1.00 s after the second object is released is 171.5m</em>

4 0
2 years ago
in a Mercury thermometer the level of Mercury Rises when its bulb comes in contact with a hot object what is the reason for this
olasank [31]

Answer:

because thermometric liquid readily expands on heating or contracts on cooling even for a small difference in the temperature of the body.

3 0
2 years ago
A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is sp
Lorico [155]

Answer:

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2

b) Assume the grind stone is a solid disk, its moment of inertia is

I = mR^2/2

Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.

I = 28*0.15^2/2 = 0.315 kgm^2

So the friction torque is

T_f = I\alpha = 0.315*0.21 = 0.06615 Nm

The friction force is

F_f = T_f/R = 0.06615 / 0.15 = 0.441 N

Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone

N = F_f/\mu = 0.441/0.2 = 2.205 N

7 0
3 years ago
A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied o
Masja [62]

Answers is  F=7.84 N

Friction force resists the effect of horizontal force and trying to approch to a limiting force.

we have formula for limiting friction force between block and floor

               F=Ц N

where N=mg

putting values we get answer.

Download pdf
3 0
2 years ago
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