Drag each tile to the correct box.
Arrange the events that are believed to have formed the hydrosphere in their order of occurrence.

Answer: [tex]1.875\ rad/s^2, 8\ s[/tex]

Explanation:

Given

Radius of disc [tex]r=10\ cm[/tex]

Angle turned [tex]\theta=60\ rad[/tex]

Initial angular velocity [tex]\omega_0=0[/tex]

Final angular velocity [tex]\omega_f=15\ rad/s[/tex]

using [tex]\omega^2-\omega_o^2=2\alpha \cdot \theta[/tex]

Substitute values

[tex]\Rightarrow 15^2-0=2\times \alpha \times 60\\\Rightarrow \alpha=1.875\ rad/s^2[/tex]

using [tex]\omega=\omega_o+\alpha t[/tex]

Substitute values

[tex]\Rightarrow 15=0+1.875\times t\\\Rightarrow t=8\ s[/tex]

Answer:

D. Pacific Plate

Explanation:

Answer:

d

Explanation:

i got it right

Answer:

C. The change of internal energy of a system is the sum of work and heat spent on it.

Explanation:

The law of conservation of Energy states that energy cannot be destroyed but can only be converted or transformed from one form to another. Therefore, the sum of the initial kinetic energy and potential energy is equal to the sum of the final kinetic energy and potential energy.

Mathematically, it is given by the formula;

Ki + Ui = Kf + Uf .......equation 1

Where;

Ki and Kf are the initial and final kinetic energy respectively.

Ui and Uf are the initial and final potential energy respectively.

The law of conservation of Energy is another way to describe the law of Thermodynamics. It states that the change of internal energy of a system is the sum of work and heat spent on it.

Mathematically, it is given by the formula;

ΔU = Q − W

Where;

ΔU represents the change in internal energy of a system.

Q represents the net heat transfer in and out of the system.

W represents the sum of work (net work) done on or by the system.

first of all the formula of force is F=ma,so we are searching for m,so we can divide a on both sides F/a=m, after this substitute the values given above 64N/4.51=14.2°Kg

Answer:

D

Explanation:

This will not change the weight and therefore not change the inertia

A dump truck contains a load of soil. The action that will leave the dump truck's inertia unchanged is that increase the force applied by the engine. Hence, option D is correct.

A body's ability to resist being propelled into motion or, if already moving, to modify the direction or magnitude of its velocity is known as inertia.

An object's lethargy is a passive quality that prevents it from doing anything other than obstructing active forces and torques. The only reason a moving body continues to move is the lack of a force that might slow it down, alter its trajectory, or accelerate it. This is not due to inertia.

A body's inertia moment about a certain axis and its mass, which determine how resistant it is to the application of forces to that axis, respectively, are two statistical measures of inertia.

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Answer:

[tex]4.61\times 10^{-5}\ \text{T}[/tex]

[tex]1.05\times 10^{-6}\ \text{Nm}[/tex]

Explanation:

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi10^{-7}\ \text{H/m}[/tex]

[tex]r_l[/tex] = Radius of loop = 15 cm

[tex]I_l[/tex] = Current in loop = 11 A

[tex]r_c[/tex] = Radius of coil = 0.76 cm

N = Number of turns of coil = 66

[tex]I_c[/tex] = Current in coil = 1.9 A

Magnetic field is given by

[tex]B=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}\times 11}{2\times 0.15}\\\Rightarrow B=4.61\times 10^{-5}\ \text{T}[/tex]

Magnitude of magnetic field produced by the loop at its center is [tex]4.61\times 10^{-5}\ \text{T}[/tex].

Torque is given by

[tex]\tau=BI_c\pi r_c^2N\sin90^{\circ}\\\Rightarrow \tau=4.61\times 10^{-5}\times 1.9\times \pi\times (0.76\times 10^{-2})^2\times 66\sin90^{\circ}\\\Rightarrow \tau=1.05\times 10^{-6}\ \text{Nm}[/tex]

Magnitude of torque on the coil due to the loop is [tex]1.05\times 10^{-6}\ \text{Nm}[/tex]

Answer: The bell has 8550 J energy.

given, There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N

i.e., bell is located at the top of tower, h = 45m

weight of the bell, F = 190 N

workdone by the gravitational force = F.hcos180°

[ gravitational force (i.e., weight ) acting downward while body is located 45m above the ground. so, angle between force and h = 180° ]

workdone by the gravitational force = 190 × 45 × (-1)

= -8550 J

we know, potential energy = negative of workdone

= -(-8550 J) = 8550 J

Answer:

200,000J

Explanation:

KE= 0.5 x m x v^2

KE= 0.5x1000x20^2

KE= 0.5 x 1000 x 400

KE= 500 x 400

KE= 200,000

Answer:

B

Explanation:

To say that electric charge is conserved means that no case has ever been found where :_________

the total quantity of charge on an object has increased

According to conservation of energy, energy is neither created nor destroyed. But can only be converted into another form.

But in the case of charges, the number of charges can not be increased nor decreased it's only the rate of flow that can change due to the pressure supplied.

So therefore, the correct answer is B which say that:

To say that electric charge is conserved means that no case has ever been found where the total quantity of charge on an object has increased

Answer:

Sarah plans to use trickery and cunning to get back at Emma. Sarah thinks she is smarter than Emma.

Explanation:

i think this is right i dont know tho

Answer:

hey mate......looks like the question is incomplete

Answer:

    [tex]\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}[/tex]

Explanation:

For this interesting problem, we use the definition of centripetal acceleration  

      a = v² / r  

angular and linear velocity are related  

     v = w r  

we substitute  

    a = w² r

the rectangular body rotates at an angular velocity w  

We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A  

the distance OB = L₂  

the distance AB = L₁

the sides of the rectangle  

It is indicated that the acceleration in in A and B are related  

      [tex]a_A = n \ a_B[/tex]  

we substitute the value of the acceleration  

    w² r_A = n r_B  

the distance from the each corner is  

    r_B = L₂  

    r_A = [tex]\sqrt{L_1^2 + L_2^2}[/tex]  

we substitute  

   \sqrt{L_1^2 + L_2^2} = n L₂  

    L₁² + L₂² = n² L₂²  

    L₁² = (n²-1) L₂²  

Answer:

h = 5.38 10¹⁶ m

Explanation:

Let's start this exercise by assuming that all the potential energy of the electron is converted into kinetic energy, let's use the conservation of energy

starting point. Just before ionization

          Em₀ = U = qE

final point. Right after ionization

           Em_f = K = ½ m v²

Energy is conserved

           Em₀ = Em_f

           q E = ½ m v²

           v² = 2qE / m

Now we can use the relationship between net work and kinetic energy

           W_net = ΔK

net work is the work done by the electron minus the binding energy with the atom, called the work function, Ф = 3 10-19 J

           W - Ф = K_f - K₀

we assume that the electron converts all its initial initial kinetic energy to be zero

           W -Ф = ½ m v² - 0

            W = ½ m v² +Ф

we substitute

           W = 1/2 m 2qE/m + E

           W =  qE +Ф

           W = 1.6 10⁻¹⁹ 3 10⁶ + 3 10⁻¹⁹

            W = 4.8 10⁻¹³ + 3 10⁻¹⁹

           W = 4.8 10⁻¹³ J

When the electron is in air, its kinetic energy can be transformed into gravitational potential energy

As the electron is in the air, all work is transformed into scientific energy

           W = K

starting point Em₀ = K = W

end point Em_F = U = m g h

energy conservation Em₀ = Em_f

                       W = m g h

                       h = [tex]\frac{W}{mg}[/tex]

let's calculate

                       h = [tex]\frac{4.8 \ 10^{-13} x}{9.1 \ 10^{-31} \ 9.8 }[/tex]

                       h = 5.38 10¹⁶ m

Electron involved in making in spark travel through the air before it collides with an atom will be at the distance of 5.38 10¹⁶ m.

An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.

Let's begin this exercise by assuming that all of the electron's potential energy is turned into kinetic energy, and then we'll apply the law of conservation of energy.

Energy before ionization;

[tex]\rm Em_0 = U = qE[/tex]

Energy after ionization;

[tex]Em_f = K = \frac{1}{2} mv^2[/tex]

From the law of conservation of energy principle;

[tex]Em_0 = Em_f \\\\ q E =\frac{1}{2} m v^2\\\\ v^2 = \frac{2qE }{m}[/tex]

The relationship between net work and kinetic energy;

[tex]W_{net} = \triangle K[/tex]

The work function is defined as net work, which is the work done by the electron minus the binding energy with the atom.

[tex]W - \phi = K_f - K_0[/tex]

[tex]W = K_f+ \phi[/tex]

[tex]W = \frac{1}{2} m \times \frac{2qE}{m} + E\\ \\W = qE + \phi \\\\ \rm W = 1.6 \times 10^{-19}\times 3 \tims 10^6 3 10⁶ +3 \times 10^{-19} \\\\ W = 4.8 \times 10^{-13}+ 3 \times 10^{-19}\\\\ W = 4.8 \times 10^{-13} J[/tex]

EMF at starting point;

[tex]\rm Em_0 = K = W[/tex]

EMF at the endpoint;

[tex]\rm Em_F = U = m g h[/tex]

From the law of conservation of energy principle;

[tex]Em_0 = Em_f \\\\ W = m g \\\\ h = \frac{W}{mg}\\\\\ h = \frac{4.8 \timjes 10^{-13}}{9.1 \times 10^{-31} \times 9.81 }\\\\ \rm h= 5.38 \times 10^{16}[/tex]

Hence electron involved in making in spark travel through the air before it collides with an atom will be at a distance of 5.38 10¹⁶ m.

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Answer:

Answer:

0 J

Explanation:

Work equals force times distance, but the force is zero because the crate being dragged will have zero acceleration. Force equals mass times acceleration and since acceleration is zero, force has to equal zero as well. Since the force is zero, the work required also has to be zero.

Answer:

α = 2,857 10⁻⁵ ºC⁻¹

Explanation:

The thermal expansion of materials is described by the expression

          ΔL = α Lo ΔT

          α = [tex]\frac{\Delta L}{L_o \ \Delta T}[/tex]

in the case of the bar the expansion is

         ΔL = L_f - L₀

         ΔL=   1.002 -1

         ΔL = 0.002 m

the temperature variation is

         ΔT = 100 - 30

         ΔT = 70º C

we calculate

         α = 0.002 / 1 70

         α = 2,857 10⁻⁵ ºC⁻¹

Answer:

the induced electric field is 9.95 V/m

Explanation:

Given the data in the question;

Number of turns N = 52

Diameter of coil D = 17 cm = 0.17 m

Radius r = D/2 = 0.17/2 = 0.085 m

Now,

cross-section area A of the coil  = πr²

A = π × ( 0.085 m )²

A = 0.0227 m²

Also given that;

Initial magnetic field B₁ = 0.45 T

Final magnetic field B₂ = 0

∴ change in magnetic field ΔB = B₁ - B₂ = 0.45 T - 0 = 0.45 T

Time taken dT = 0.10 seconds

Now, we know that;

Induced emf ∈ = N[tex](\frac{d\eta }{dt} )[/tex]

where η = BAcosθ

We know that, magnetic field is cylindrically symmetric, coil is also perpendicular to magnetic field.

Hence, the angle between B & A is 0°

∴ θ = 0°

Induced emf ε = N[tex](\frac{d }{dt} )BAcos\theta[/tex]

we substitute

ε = N[tex](\frac{d }{dt} )[/tex] (BAcos0°)

A is constant and cos0° = 1

so

ε = NA[tex](\frac{dB }{dt} )[/tex]  

We now substitute in our values;

ε = 52 × 0.0227 m² ×  [tex](\frac{0.45T }{0.10s} )[/tex]

ε  =  5.3118 V

we know that, from the relation between electric and emf

ε = ∫∈.dl or ε = ∈∫dl         { for coil; ∫dl = πD }

so we have;

ε = ∈πD

solve for ∈

∈ = ε/πD

we substitute

∈ = 5.3118 V / ( π × 0.17 m )

∈ = 9.95 V/m

Therefore, the induced electric field is 9.95 V/m

Answer:

The aswer is D

Explanation:

Answer:

d

Explanation:

cumulus clouds that are tall with a flat top

Answer: the answer is B

Explanation: 80 is not the same as 150 so it will go the way 150 units of force is pulling.

Answer:D

Explanation:

Answer:

true it all changes

Explanation:

________________________

Answer:

2.75 N

Explanation:

Given that,

Box A has a mass 21.0 kg and box B has a mass 8.0 kg.

A horizontal force of 100N is exerted on box A.

Let a be the acceleration of the system. Using second law of motion,

[tex]F=(m_A+m_B)a\\\\a=\dfrac{F}{(m_A+m_B)}\\\\a=\dfrac{10}{(21+8)}\\\\a=0.344\ m/s^2[/tex]

Now applying Newton's second law to box B. So,

[tex]F_A=m_Ba\\\\=8\times 0.344\\\\=2.75\ N[/tex]

So, 2.75 N is the force that box A exerts on box B.

The key concept here is that when two objects interact, the forces they exert on each other are equal in magnitude but opposite in direction. The magnitude of the force that box A exerts on box B is also 100N.

Newton's third law of motion states that for every action, there is an equal and opposite reaction. It is one of the three fundamental principles described by Sir Isaac Newton in his Laws of Motion.

According to Newton's third law, when an object exerts a force on another object, the second object simultaneously exerts a force of equal magnitude but in the opposite direction on the first object. In simpler terms, if object A applies a force on object B, then object B applies an equal force in the opposite direction on object A.

Since boxes A and B are in contact on a frictionless surface, the force exerted on box A will be transmitted to box B. According to Newton's third law of motion, the magnitude of the force that box A exerts on box B will be equal in magnitude but opposite in direction to the force exerted on box A.

Therefore, the magnitude of the force that box A exerts on box B is also 100N.

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This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.

The minimum horizontal velocity required is "2.6 m/s".

First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.

[tex]h=v_it+\frac{1}{2}gt^2[/tex]

where,

h = height = 3 m

vi = initial vertical speed = 0 m/s

t = time interval = ?

g = acceleration due to gravity = 9.81 m/s²

therefore,

[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]

t = 0.78 s

Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:

[tex]s = vt\\\\v = \frac{s}{t}[/tex]

where,

s = horizontal distance = 2 m

t =0.78 s

v = minimum horizontal velocity = ?

Therefore,

[tex]v = \frac{2\ m}{0.78\ s}[/tex]

v = 2.6 m/s

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The attached picture shows the equations of motion in the horizontal and vertical directions.

Answer:

I think so but i could be wrong..

Explanation:

Answer: yes it does

Explanation: Yes, you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.

The height of the tower is 20.41 m.

To determine the height of the tower, we need to understand the concept of the energy conservation principle since the speed and acceleration due to gravity are involved in the system.

The principle of energy conservation lets us know that in an isolated system, energy can neither be created nor destroyed.

[tex]\mathbf{mgh = \dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2}[/tex]

By applying the energy conservation principle, we have:

[tex]\mathbf{gh +\dfrac{1}{2}v_1^2 = \dfrac{1}{2}v_2^2}[/tex]

[tex]\mathbf{h = \dfrac{v_2^2- v_1^2 }{2 \times g}}[/tex]

[tex]\mathbf{h = \dfrac{25^2-15^2 }{2 \times 9.8}}[/tex]

h = 20.41 m

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Answer:

6 oxygen

Explanation:

3 multiply O2 and 6 Oxygen

Answer:

The "butterfly Effect"

Explanation:

The "butterfly effect" will probably have big changes in the future.

Answer:

Both Magnetic

Explanation:

Answer: Newtons third law

Explanation: According to Newton's third law of motion, forces always act in equal but opposite pairs. Another way of saying this is for every action, there is an equal but opposite reaction. This means that when you push on a wall, the wall pushes back on you with a force equal in strength to the force you exerted.

Newton's third law we find that the reaction to the force is an out applied by the putting door towards to the person.

given parameters

to find

Newton's third law states that the forces between bodies always occur in pairs, that is, the two bodies interacting with each other, therefore these forces are of equal magnitude, but opposite direction. Consequently each force acts on one of the bodies, these forces are called action and reaction.

By applying Newton's third law to the open door; we exert a force on the door therefore the door exerts a force of equal magnitude, but opposite direction on us.

Using Newton's third law we find that the reaction to the force is an out applied to the person.

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Answer:

hi

Explanation:

false because the light of some stars can be very bright, but it can also be very far from the earth.

have a nice day