There are two main types of collisions that you will study: elastic and perfectly inelastic. In an elastic collision, kinetic energy is conserved. In a perfectly inelastic collision, the particles stick together and thus retain the same velocity after the collision.

a. True
b. False

Answer:

0.116 N/m

Explanation:

Since the net force acting on the ring must be greater than 2.00 × 10⁻² N, and the surface tension T = F/L where F = net force = 2.00 × 10⁻² N and L = circumference of ring = 2πr where r = radius of ring = 2.75 cm = 2.75 × 10⁻² m.

So, T = F/L

= F/2πr

= 2.00 × 10⁻² N ÷ 2π(2.75 × 10⁻² m)

= 1/2.75π  N/m

= 1/8.64 N/m

= 0.116 N/m

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:

The recipe for a hurricane is a combination of warm, humid wind over tropical waters. The temperature of tropical waters must be at least 80 degrees F for up to 165 feet below the ocean’s surface. As this warm water meets the wind that blows west from Africa across the ocean, it causes the water to vaporize. The water vapor then rises into the atmosphere, where it cools and liquefies.

Explanation:

I hope this helps

Answer:

Volume of water displaced = 450 - 375 = 75 ml

Vr = volume of rock = 75 ml

Wr = 9.22 - 7.60 = 1.62 N  weight of 75 ml of rock

Density of rock = 1.62 N / 75 ml = .0216 N / ml

Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml

Density of rock / density of water = .0216 / .0098 = 2.20

The specific gravity of the rock in the given water volume is 0.2.

The given parameters;

The specific gravity of the rock is calculated as follows;

[tex]S.G = \frac{weight \ in \ air}{Weight \ in \ water} \\\\S.G = \frac{450 - 375}{375} \\\\S.G = 0.2[/tex]

Thus, the specific gravity of the rock in the given water volume is 0.2.

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This problem involved half projectile.

initial velocity, vo = 20 m/s

time of flight, t = 7 s

(a) Simply use the formula to get the height, h:

h = vo*t - (1/2)gt^2

(b) To get the final vertical velocity or terminal velocity (vf), use the formula:

(vf)^2 - (vo)^2 = 2gh

(c) Use the formula find the horizontal distance traveled, R:

R = vo * cos(θ) * t

But since the angle involved with respect to horizontal is zero, and cos(0) = 1, we have

R = vo * t

Hope this helps~ `u`

Jai

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:

[tex]f = g \times \frac{m1 \times m2}{ {d}^{2} } [/tex]

[tex]f = 6.67 \times {10}^{ - 11} \times \frac{40 \times 40}{9} [/tex]

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.

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Answer:

The motion of an object is accelerated when its speed increases.

Answer:

igneous = melted rocks formed by cooled magma

sedimentary is brken rocks, kayers with fossil...

metamophic is rocks formed by pressure and heat

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:

E =  1.64 x 10⁶ V/m

Explanation:

The electric field in the region between the plates can be given by the following formula:

[tex]E = \frac{\Delta V}{d}[/tex]

where,

E = Electric Field = ?

ΔV = Poetential Difference across the plates = 23 KV = 23000 V

d = distance between plates = 1.4 cm = 0.014 m

Therefore, using these values in the equation, we get:

[tex]E = \frac{23000\ V}{0.014\ m}[/tex]

E =  1.64 x 10⁶ V/m

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:

3.39 rad / s.

Explanation:

Given data:

mass of disk = 4.50 Kg

radius of wheel = 0.650 m

mass of the clay = 0.870 kg

The moment of inertial of the wheel = I = 4.5 kg x ( 0.65 m )2 / 2 = 0.95 kg . m2.

Now, applying the principle of angular momentum conservation :

Iω_i = ( I + mr2 )ω_f.

where ω_i = initial angular speed= 4.70 rad/s, ω_f = final angular speed

Hence, ω_f = Iω_i / ( I + mr2 )

= ( 0.95 kg . m2 x 4.7 rad / s ) / [ 0.95 kg . m2 + 0.87 kg x ( 0.65 m )2 ]

= 3.39 rad / s.

Hence, correct answer is  : 3.39 rad / s.

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

Answer:

Drought

Explanation:

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:

v = 9.04 m / s

Explanation:

For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.

          W = Em_f - Em₀         (1)

Starting point. Lower slope

        Em₀ = K = ½ m v²

highest point. Where is the skier at a height h

        Em_f = U = m g h

The work of rubbing

        W = -fr x

the negative sign is because the friction force opposes the movement.

Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular

let's use trigonometry to break down the weight

        cos θ = W_y / W

        sin θ = Wₓ / W

        W_y = W cos θ

        Wₓ = W sin θ

Y axis

        N - Wₓ = 0

        N = mg sin  θ

X axis

         fr = m a

the friction force has the expression

         fr = μ N

         fr = μ mg sin θ

we look for the job

         W = - μ mg sin θ  x

where x is the distance along the slope

we substitute in 1

         -μ mg sin θ x = mg h - ½ m v²

let's use trigonometry to find the distance x

        tan 30 = h / x

        x = h / tan 30

we substitute

          -   [tex]\mu \ mg \ sin \theta \ \frac{h}{tan 30} \ x[/tex] = m gh - ½ m v²

we use  

          tan 30 = sin30 / cos30

          v² = 2g h + 2 μ g h cos 30

          v = [tex]\sqrt{ 2gh \ (1+ cos 30}[/tex]

let's calculate

          v = [tex]\sqrt{ 2 \ 9.8 \ 4 \ (1 + 0.05 \ cos \ 30)}[/tex]

          v = 9.04 m / s

Answer:

option 5

Explanation:

because all u do is have to add them up

Answer:

The answer is "892.90 N"

Explanation:

Following are the solution to these question:

Calculating the vertical force of the summation that is equal to zero:

[tex]\to TL \cos 65 + TR \cos 80 -520 = 0\\\\\to 0.4226\ TL + 0.1736\ TR = 520\\[/tex]

Calculating the sum of horizontal forces that is equal to zero:

[tex]\to TL\sin 65 - TR \sin 80 = 0 \\\\\to 0.9063TL - 0.9846TR = 0\\\\\to TL = (\frac{0.9846}{0.9063})\ TR \ \ = 1.0866\ TR\\\\\to 0.4226(1.0866) \ TR +0.1736\ TR =520 \ N\\\\\to 0.6328 \ TR = 520 \\\\\to TR = 821.74 \ N \\\\\to TL = 1.0866 \times 821.74 = 892.90\ N[/tex]

Answer:

c is trueeeeeeeeeeeeeee

The growth of algae in ocean water is limited by their need for dissolved oxygen. Thus, the correct answer is option C.

The term "algae" refers to a large and diverse group of photosynthetic eukaryotic organisms. It is a polyphyletic grouping of species from several distinct clades. Most are aquatic and autotrophic, lacking many of the distinct cell and tissue types found in land plants, such as stomata, xylem, and phloem.

Algae, like all organisms, normally grow in balance with their ecosystems, with the amount of nutrients in the water limiting their growth. Deep ocean water contains fewer algae because algae require sunlight and carbon dioxide to thrive.

Therefore, due to need for dissolved oxygen the growth of algae in ocean water is limited.

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Answer:

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

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

Explanation:

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Explanation:

force=35

distance=18

time=30

power=f×d/t

p=35×18/30

p=21