Which is an example of a wedge?
A.
B
C
D

Answer:

Weight of car = 36034.88 lb.ft/s²

Explanation:

We are told the pressure of the four tires of your car is; P = 35.0 psi

Also, the surface area of contact is; A = 32 in²

Thus;

Weight of tires = Pressure × Area

W_tires = 35 × 32

Weight_tires = 1120lb.

To get the weight of the car, we will multiply the tire weight by acceleration due to gravity.

It's value in ft/s² is g = 32.174 ft/s²

Thus;

Weight of car = 1120 × 32.174

Weight of car = 36034.88 lb.ft/s²

Answer:

World War 1 was caused by entangled alliances, nationalism, imperialism, and major

advancements in military technology. Does the Treaty of Versaille address those issues?

Explain your answer using facts. (5 points)

Answer:

Wavelength

Explanation:

A vibration can cause a disturbance to travel through a medium, transporting energy without transporting matter. This is what a wave is.

Answer:

They each have one single proton (Z = 1), but differ in the number of their neutrons. Hydrogen has no neutron, deuterium has one, and tritium has two neutrons. The isotopes of hydrogen have, respectively, mass numbers of one, two, and three.

Explanation:

Answer:

Explanation:

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The principle of lateral continuity works, because they are continuous. Rocks that are alike but were separated by an erosional feature, can be originally continuous.

Answer: A. The reactants must collide in a certain way for the reaction to  occur.

Explanation: I did the quiz :)

Answer:

A) The reactants must collide in a certain way for the reaction to

occur.

Explanation:

A.P.E.X

Answer:

True.

Explanation:

Science can be defined as a branch of intellectual and practical study which systematically observe a body of fact in relation to the structure and behavior of non-living and living organisms (animals, plants and humans) in the natural world through experiments.

A concept is the general idea of objects, events, animals, or people based on common features, traits, or characteristics. It is typically considered to be an abstract idea or general notion that is formed by the mental combination of all the characteristics pertaining to the idea.

Generally, concepts are used in the problem-solving process as it serves as a benchmark or yardstick for the solutions. Some examples of concepts in physics are conservation of energy and momentum, electromagnetic waves, heat, motion and forces, electricity and magnetism, gravity, etc.

Answer:

TRUE

Explanation:

edge

Answer:

So an object with mass is attracted to another object with mass, and the gravitational force is directly proportional to the masses of the two objects, and inversely proportional to the square of the distance between the two objects.

If distance  were to increase, than the gravitational force would decrease. If mass were to increase, so would the gravitational force.

Explanation:

Explanation:

Lets take gravitational force(F) and mass(m) and distance (r)

now for a body in contact with the surface of the earth, its mass is also considered(m‘),now the mass of the earth(m") is also considered,then the distance from the body to the center of the earth (r).ie it r because its practically the radius of the earth. is also considered

So by using dimensional analysis ....

we get F a m'•m"/r² ,where a is proportional to.

now since F is directly increaseproportional to m ie. F a m, then an increase in mass of the body increases it's gravitational force(and clearly that makes sense because the bigger you are the stronger you get pulled to the ground)

then we also see that F is inversely proportional to r ie.F a 1/r ,then an increase in the distance between the ground an the object decrease it's gravitational force ( meaning as any object on earth keeps on moving away from the ground the gravitational force between the object and the center of the earth is weak, when it reaches space then the force becomes virtually negligible!)

So to answer the second question, we clearly see that doubling the mass of the body increases the gravitational force between it and the earth

and doubling the distance on the other hand will decrease the attraction between the body and the earth

So a body forcefully projected into the air fights against gravity but its easier as it keeps on getting higher, If it has a greater mass like that of a trail or truck , it will not even probably stay in the air for long , unless its projected with a very high velocity

I hope this helps, and you can ask me any question concerning this via the comments platform.

Answer:

link

Explanation:

ww.comhwhelp

Explanation:

Our solar system formed about 4.5 billion years ago from a dense cloud of interstellar gas and dust. The cloud collapsed, possibly due to the shockwave of a nearby exploding star, called a supernova. When this dust cloud collapsed, it formed a solar nebula—a spinning, swirling disk of material.

Answer:

29 m/s

Explanation:

Assuming the bridge and the platform are back to back,

Average speed = Total Distance/ Total Time

Avg S = 1,450m/50s = 29 m/s

The average speed of the train is 29 m/s

The average speed is defined as the ratio of total distance traveled and the total time taken.

    Total distance = 650 m + 800 m = 1450 m

     Total Time      = 20 s + 30 s = 50 s

    Average speed = total distance/total time

                               = 1450 / 50

                               = 29 m/s is the average speed of the train.

Learn more about average speed:

https://brainly.com/question/11753352

Answer:

.lyrjhg.

Explanation:

qqlqfgtjnfh

Answer:

C

Explanation:

The initial velocity of the projectile is 19.8m/s

Explanation:

First, find time.

From our kinematics equations:

delta y = Vi•t + (1/2)at^2

rearrange,

t = sqrt[(2•delta y)/a]

t = sqrt[(2•20m)/9.8m/s^2]

t = 2.02s

Next, plug time into new kinematics equation to solve for the Vi in the x direction (horizontal)

delta x = Vi•t + (1/2)at^2

delta x = Vi•t

Rearrange:

Vi = delta x/t

Vi = 40m/2.02s

Vix = 19.8m/s

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}[/tex]

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = [tex]k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}[/tex]

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = [tex]k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )[/tex]

            E_ {total} = [tex]k \frac{Q}{x_1^2}[/tex]   [tex]( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )[/tex]

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = [tex]-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}[/tex]

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}[/tex]

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = [tex]+ k \frac{Q}{(d-x_1)^2}[/tex]+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

5) Right side point

            E_ {total} = [tex]k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}[/tex]

             E_ {total} = [tex]- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )[/tex]- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

Answer:

ferromagnetic materials are materials that are highly attracted to magnets while antiferromagnetic materials are materials that are not attracted by magnets

Explanation:

force=35

distance=18

time=30

power=f×d/t

p=35×18/30

p=21

Answer:

A.

Explanation:

Electrons

Answer:2.23 m/s

Explanation:

Given

Mass of person and vehicle is [tex]m=1600\ kg[/tex]

Total work done is [tex]W_t=6000\ J[/tex]

Friction consumes one-third of the energy

The remaining two-third is consumed to increase the kinetic energy

[tex]\Rightarrow \dfrac{2}{3}\times 6000=\dfrac{1}{2}\times 1600\times v^2\\\\\Rightarrow 4000\times 2=1600\times v^2\\\\\Rightarrow v=\sqrt{5}\ \approx 2.23\ m/s[/tex]

Answer:

Light Winds

Explanation:

Answer:

[tex]1.554\times 10^{32}\ \text{kg}[/tex]

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

Radius of orbit is given by

[tex]R=\dfrac{vT}{2\pi}[/tex]

We have the relation

[tex]\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}[/tex]

The mass of each star is [tex]1.554\times 10^{32}\ \text{kg}[/tex]

Answer:

B

Explanation:

Hi there! :)

[tex]\large\boxed{17.32 m/s}[/tex]

Use the following equation in solving for kinetic energy:

KE = 1/2mv² where:

KE = kinetic energy (J)

m = mass (kg)

v = velocity (m/s)

Plug in the given values:

12,000 = 40v²

Divide both sides by 40:

12,000 / 40 = v²

300 = v²

Take the square root of both sides:

√300 = v

v ≈ 17.32 m/s

Solution :

Let [tex]$m_1=m_2=4$[/tex] kg

[tex]$u_1 = 5$[/tex] m/s

Let [tex]$v_1$[/tex] and [tex]$v_2$[/tex] are the speeds of the disk [tex]$m_1$[/tex] and [tex]$m_2$[/tex]  after the collision.

So applying conservation of momentum in the y-direction,

[tex]$0=m_1 .v_1_y -m_2 .v_2_y $[/tex]

[tex]$v_1_y = v_2_y$[/tex]

[tex]$v_1 . \sin 60=v_2. \sin 30$[/tex]

[tex]$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$[/tex]

[tex]$v_2=1.732 \times v_1$[/tex]

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

[tex]$m_1.u_1=m_1.v_1_x+m_2.v_2_x$[/tex]

[tex]$u_1=v_1_x+v_2_x$[/tex]

[tex]$5=v_1. \cos 60 + v_2 . \cos 30$[/tex]

[tex]$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$[/tex]

[tex]$v_1 = 2.50$[/tex] m/s

So, [tex]$v_2 = 1.732 \times 2.5$[/tex]

          = 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s