Complete the sentence below using one of the following
words: equilibrium, flux, motion.
If supported at its center of gravity, an object will remain in ______ any position.

Answer:

 v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

         Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

         Em_f = K = ½ m v_f²

         Emo = Em_g

         ½ m v₀² + m g y₀ = ½ m v_f²

        v_f² = v₀² + 2 g y₀

let's calculate

        v_f = √(10² + 2 9.8 25)

        v_f = 24.3 m / s

Answer:

Newton' second law  and third law describes the situation.

Explanation:

According to the Newton's second law, the force applied on a body is proportional to the rate of change of momentum of the body.

According to the Newton's third law, for every action there is an equal and opposite reaction.

When ice skater pushes harder means more force is applied so he moves fast and more be the action force more be the reaction force.

Thus, Newton' second law  and third law describes the situation.

Answer:

Data given.

focal length (f)=15m÷2=7.5m

Distance of the object(U)=10m

Image distance (v)=?

Magnification (M)=?

Solution:

From:

1/f=1/u+1/v

1/7.5=1/10+1/v=75

then v=75m

Magnification, M=u/v

=75/10=7.5

Then magnification=7.5

Answer:

v = 30 m and m = 3

Explanation:

Given that,

The radius of curvature of the mirror, R = 15 m

Focal length, f = 7.5 m

Object distance, u = -10 m

We need to find the image distance and the magnification of the object.

Using mirror's formula,

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(7.5)}+\dfrac{1}{(-10)}\\\\v=30\ m[/tex]

The magnification of the object in mirror is given by :

[tex]m=\dfrac{-v}{u}\\\\m=\dfrac{-30}{-10}\\\\m=3[/tex]

So, the distance of the image from the mirror and the magnification of the object are 30 m and 3 respectively.

Answer:

(a) t = 0.75 s

(b) d = 26 m

Explanation:

Height, h = 12 m

Angle, A = 30 degree

initial velocity, u = 40 m/s

(a) Let the time is t.

Use second equation of motion

 [tex]h = u t + 0.5 at^2\\\\12 = 40 sin 30 t - 0.5 \times10 \times t^2\\\\12 = 20 t - 5 t^2\\\\5 t^2 - 20 t + 12 =0 \\\\t = \frac{20\pm \sqrt{400-240}}{10}\\\\t = \frac{20\pm 12.65}{10}\\\\t = 3.3 s, 0.75 s[/tex]

(b) d = u cos A t

d = 40 x cos 30 x 0.75 = 26 m

Since fundamental units are elementary in nature and cannot be reduced further, so it can not be reduced further, so it cannot be expressed in other units

Answer:

The magnetic field through the wire must be changing

Explanation:

According to Faraday's law, the induced emf, ε in a metallic conductor is directly proportional to the rate of change of magnetic flux,Φ  through it. This is stated mathematically as ε = dΦ/dt.

Now for the wire, the magnetic flux through it is given by Φ = ABcosθ where A = cross-sectional area of wire, B = magnetic field and θ = angle between A and B.

So, dΦ/dt = dABcosθ/dt

Since A and B are constant,

dΦ/dt = ABdcosθ/dt = -(dθ/dt)ABsinθ

Since dθ/dt implies a change in the angle between A and B, since A is constant, it implies that B must be rotating.

So, for an electric current (or voltage) to be produced in the wire, the magnetic field must be rotating or changing.

Answer:

b

Explanation:

bc

Answer:

C voltage

Explanation:

Voltage is the change in electric potential so basically current flows from high potential to low potential due to voltage.

Answer:

the answer to this would be D.

Answer: Its D but it might swap to another letter so keep your eye out bud pass this test good luck

Explanation:

Answer:

Explanation:

This is a First Law of thermodynamics problem. We have to remember that the total energy available to a system is constant throughout the whole problem and that energy cannot be created or destroyed. So we need to find the total energy available right at the start. Well it just so happens that we are told that the total energy is 1000J and that it is all potential energy when the sphere is at rest and is 25 m off the ground. If the object isn't moving, all the energy is potential until it starts moving and the energy begins to convert from potential to kinetic a little bit at a time. The thing that we don't know is the mass of the shpere. Begin with the fact that the PE = 1000 (I'm going to se 2 sig fig's since there's only 1 in 1000). If

PE = 1000 and PE = mgh, then

1000 = m(9.8)(25) so

m = 4.1 kg

We also need the height at which this sphere has a PE of 600. Again, if

PE = 600 and PE = mgh, then

600 = (4.1)(9.8)h so

h = 15 Filling in the total energy equation now, using the fact that the total energy available to the system is 1000J:

TE = PE + KE and

1000 = (4.1)(9.8)(15) + [tex]\frac{1}{2}(4.1)v^2[/tex] and we are looking for v.

1000 = 6.0 × 10² + 2.1v² so

[tex]v=\sqrt{\frac{1000-6.0*10^2}{2.1} }[/tex] and

[tex]v=\sqrt{\frac{4.0*10^2}{2.1} }[/tex] gives us

v = 14 m/s

Answer:

[tex]f = m \frac{v1 - v2}{t} \\ = 120 \times \frac{25 - 5}{2} \\ = 120 \times \frac{20}{2} \\ = 120 \times 10 \\ = 1200N \\ thank \: you[/tex]

Answer:

The sun uses hydrogen for nuclear fusion.

Answer:

Hydrogen

Explanation:

After half of a revolution ...

==> Distance = π•r

==> Displacement = 2•r

Answer:

answer is E

Explanation:

Answer:

900C

Explanation:

Change 2.5 mins to secs by multiplying by 60 which is 150secs

hope this helps please like and mark as brainliest

26. D. crushing the sugar cube and dissolving it in water.

27. A. atom

28. B. molecule

29. B. plum pudding model of Joseph John Thomson

30. B. He used cathode ray tubes which showed that all atoms contain tiny negatively charged subatomic particles or electrons.

31. D. protons and neutrons are relatively heavier than electrons.

Answer:

As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.

Formula: g = GM/r2

Dimensional Formula: M0L1T-2

Values of g in SI: 9.806 ms-2

Explanation:

Please Mark me brainliest

Answer:

♡

Explanation:

As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.

Answer:

12m/s

Explanation:

Using the equation of  motion

v = u + at

Given that

u = 0m/s

a = 2m/s²

time t = 6s

Substitute the given values into the expression

v = 0 + 2(6)

v = 0+12

v = 12m/s

Hence the required velocity is 12m/s

Answer:

[tex]Q=50.3[/tex]

Explanation:

From the question we are told that:

Frequency [tex]F=330Hz[/tex]

Sound intensity drop [tex]I_d=2[/tex]

Time [tex]T=4s[/tex]

Therefore

Sound intensity Ratio

 [tex]\frac{I}{I_x}=\frac{1}{2}[/tex]

Generally the equation for Sound intensity is mathematically given by

 [tex]\frac{I}{I_x}=e^{-4\ \=t}[/tex]

 [tex]\frac{1}{2}=e^{-4\ \=t}[/tex]

 [tex]\=t =5.8s[/tex]

Generally the equation for Quality Factor is mathematically given by

 [tex]Q=2 \pi\frac{E}{\triangle E}[/tex]

 [tex]Q=2 \pi\frac{E}{\frac{E}{2*4}}[/tex]

 [tex]Q=50.3[/tex]

Answer:

0.25s

Explanation:

5/20 = 0.25s

This might be correct

Answer:

7.35kg

Explanation:

Hope this is helpful :)

Answer:

[tex]F=-8.2*10^{-8}N[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=5.3*10^{-11}[/tex]

Where

Proton Charge [tex]q_1=1.6*10^{-19}C[/tex]

Charge of Electron [tex]q_2=-1.6*10^{-19}C[/tex]

Generally the equation for Coulomb's Law is mathematically given by

 [tex]F=\frac{9*10^9(q_1)(q_2)}{r^2}[/tex]

 [tex]F=\frac{9*10^9(1.6*10^{-19}C)(-1.6*10^{-19}C)}{5.3*10^{-11}}[/tex]

 [tex]F=-8.2*10^{-8}N[/tex]

Answer:

Explanation:

As far as the displacement goes, we have 2 displacement vectors. If we didn't have the angles to deal with, this would be a much simpler process, but then that wouldn't be any fun at all, would it? I'll deal with the average speed first, then the displacement, which is a vector addition problem.

The average speed is found by adding together the distances the student traveled and then dividing this sum by the total time he spent traveling. If we are told that the student runs at 4.5 m/s for 3.0 minutes, we can use this to find out the distance he ran during that time interval. However, the units are not the same. We will find the distance the student traveled by convering the time to seconds.

3.0 minutes = 180 seconds, and

4.1 minutes = 246 seconds.

That means that the distance he ran in 180 seconds is found by multiplying this time be the speed at which he ran:

4.5 m/s(180 s) = 810 m and

3.5 m/s(246 s) = 860 m (rounded to follow the rules of sig dig).

This makes the speed equation look like this:

[tex]s=\frac{810+861}{180+246}=\frac{1671}{426}=3.9\frac{m}{s}[/tex] That's the average speed, which is NOT at all the same as the displacement. Displacement is where he ended up in reference to where he started. The angles play a huge part in this math (that is very involved, to say the least). We begin by restating the displacement of each "leg" of this journey.

The first leg took him 810 m at 207 degrees and

the second leg took him 860 m at 325 degrees

To find the x and y components of these 2 legs, or parts, we have to use the cos and sin formulas. We will call the first leg A and the second leg B. First the x components of both A and B:

[tex]A_x=810cos207[/tex] and

[tex]A_x=-720[/tex]

[tex]B_x=860cos325[/tex] and

[tex]B_x=704[/tex] and we add these to get the x-component of the resultant vector, C:

  -720

+  704

   -10 (rounded, as needed, to the tens place).

Now for the y-components of the resultant vector:

[tex]A_y=810sin207[/tex] and

[tex]A_y=-370[/tex]

[tex]B_y=860sin325[/tex] and

[tex]B_y=-490[/tex] and we add these to get the y-component of the resultant vector, C:

  -370

+ -490

 -860

Since the x component is negative and so is the y, we are in QIII, so when we finally find our angle, we will have to add 180 to it.

For the magnitude of the displacement vector, in m:

[tex]C_{mag}=\sqrt{(-10)^2+(-860)^2}[/tex] which gives us

[tex]C_{mag}=860m[/tex]

Now, because displacement is vector, we also need the angle. We find that is the formula

[tex]\theta=tan^{-1}(\frac{C_y}{C_x})[/tex] and filling in:

[tex]\theta=tan^{-1}(\frac{-860}{-10})=90[/tex] (rounded correctly), and then we add 180 to give us a final direction of 270 degrees.

So the final displacement of the student is 860 m at 270 degrees

26. D. crushing the sugar cube and dissolving it in water.

27. A. atom

28. B. molecule

29. B. plum pudding model of Joseph John Thomson

30. B. He used cathode ray tubes which showed that all atoms contain tiny negatively charged subatomic particles or electrons.

31. D. protons and neutrons are relatively heavier than electrons.

Answer:

the answer is well known as gravity