. What is a normal force? It acts against an object slowing it down. It acts perpendicular to an object. It acts with an object adding to its speed. It acts parallel to an object.

oh this question seems to be difficult I don't think I can answer this

Answer:

The fundamental frequency of the stretched string is:

[tex]f=[/tex] [tex]\frac{1}{2} \sqrt{\frac{T}{L} }[/tex] [ T = Tension and μ = mass per unit length]

Here,

 μ = [tex]\frac{m}{L} = \frac{Vp}{L} = Ap[/tex]

[tex]f= \frac{1}{2} \sqrt{\frac{T}{Ap} }[/tex]

If T is halved and A is doubled,

[tex]f= \frac{1}{2} \sqrt{\frac{T'}{A'p} } = \sqrt{\frac{1}{2* 2* A* p} } = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{T}{Ap} } = \frac{1}{2} f[/tex]

Thus, the frequency is reduced to half if its tension is halved and the area of cross-section of the string is doubled.

When the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.

The frequency of transverse vibration of a stretched string is calculated as follows;

[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]

where;

[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }\\\\f = \frac{1}{2l} \sqrt{\frac{Tl}{m} } \\\\f = \frac{l^2}{2l} \sqrt{\frac{T}{m} } \\\\f = \frac{A}{2l} \sqrt{\frac{T}{m}}[/tex]

when the tension is halved and the area of cross section is doubled, the frequency is calculated as;

[tex]\frac{f_1}{A_1 \sqrt{T_1} } = \frac{f_2}{A_2\sqrt{T_2} } \\\\f_2 = \frac{f_1 A_2\sqrt{T_2}}{A_1 \sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1\sqrt{0.5T_1} }{\sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1 \times 0.7071\sqrt{T_1} }{\sqrt{T_1} }\\\\f_2 = 1.414 f_1[/tex]

Thus, when the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.

Learn more about tension in a string here: https://brainly.com/question/25743940

The plane has a centripetal acceleration a of

a = v ²/r

where v is the plane's tangential speed and r is the radius of the circle. By Newton's second law,

F = mv ²/r

Solve for the mass m :

m = Fr/v ² = (3000 N) (18.3 m) / (55.0 m/s)² ≈ 18.1 kg

hope it is helpful to you

Answer:

English only

Explanation:

When solving problems related to Electric Fields, care must be taken about symmetries. In our particular case when we take a look to at the drawings of the attached file, we realize:

1.-By symmetry each dx associated at a, has an opposite dx with point b as reference. The respective dE ( the charge is uniform ) is the same, as the charge of the wire is positive the force and the Field on a test charge (+) located at h will be upward, therefore the components dEx will cancel each other and the Electric Field becomes E = Ey = ∫ 2×dE× cosθ

The solutions:

A) Ey = 4623 N/C

B) Ey = 19.34 N/C

E = Ey = ∫ 2×dE× cosθ

Here     cosθ   = h/ d   ⇒  cosθ = h/√h² + x²      dE = K× dQ / d²

d² = h² + x²

k = 8.9 ×10⁹ Nm²C⁻²  ;   dQ = λ×dx     λ = 150×10⁻⁹ C    h = 0.08 m

Then by substitution

Ey =  2 ∫[K× λ×dx/ (h² + x²) ] × h / √h² + x²

reordering that equation:

Ey = 2×K×λ×h ∫ dx / [√ ( h² + x² ) ]³          (2)

To solve the integral we make use of a change of variables

x = h × tanα     then   x² = h² ×tan²α   and  dx = h× sec²α dα

plugging that values in equation (2)

Ey  =  2×K×λ×h ∫  h× sec²α× dα / [√ ( h² + h²tan²α)]³

Ey  = 2×K×λ×h² ∫ sec²α× dα / [ h × √ (1 + tan²α)]³            1 + tan²α = sec²α

Ey = 2×K×λ×h²× ∫ (sec²α / h³× sec³α )×dα

Ey = 2×K×λ/h × ∫ ( 1 / secα dα

Ey = 2×K×λ/h × sinα             now we αneed to come back to our original variables:

as   x = h × tanα         tanα = x/h   then x is the opposite leg in a right triangle  and h the adjacent one then the hypothenuse is √ (h² + x²)         then    sin α = x/ √ (h² + x²)      

Ey = 2×K×λ/h × x/ √ (h² + x²) |₀⁰°⁰⁵

Ey  = 2×8.9×10⁹× 150×10⁻⁹× 5×10⁻²/8× 10⁻²× √ 10⁻² ( 8 + 5 )   N/C

Ey = 4623 N/C

To answer the second question again we will make use of symmetries if you look at drawing ( Figure 2 ) you see that again the components in direction of x-axis cancel each other and the components in y-axis direction will add. Then

Ey = ∫ dE× cosθ

following the same procedure  we will find:

Ey = ∫ [K×λ × dl/d²] × h/ d

The importan point here is that the radius of the circle is

2×π×r = 0.01      ( the length of the wire)  ⇒  r = 0.16×10⁻² m

And we need to take into account that the integration is over the circle and the length of the circle is 0.01 m or ××2×π×r. All other factors are constant. Then by substitution

Ey = [K×λ ×h×  / ( √ r² + h²)³ ] × 10⁻²    N/C

Ey = 8.9 × 10⁹ × 150× 10⁻⁹ × 6× 10⁻² × 10⁻² / √ 10⁻² ( 0.16 + 6)

Ey = 0.8 × 10² / 6

Ey = 19.34 N/C

Answer:

We know that for a pendulum of length L, the period  (time for a complete swing) is defined as:

T = 2*pi*√(L/g)

where:

pi = 3.14

L = length of the pendulum

g = gravitational acceleration = 9.8 m/s^2

Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.

Then the period is independent of:

The mass of the child

The initial angle

Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.

Answer:

The number of turns in the second coil is more than the coil 1.

Explanation:

The magnetic field lines are the imaginary path on which an isolated north pole moves if it is free to do so.

The tangent at any point to the magnetic field line, gives the direction of magnetic field at that point.

More be the crowd ness of magnetic field lines more is the strength of magnetic field.

Here the crowd ness of magnetic field lines is more in figure 2 , so the magnetic filed in figure 2 is more than 1. It shows that the number of turns in the second coil is more than the 1 and also the current in the coil 2 is more than 1 .

Answer:

Explanation:

[tex]KE=\frac{1}{2}mv^2[/tex] so filling in:

[tex]KE=\frac{1}{2}(1000)(30)^2[/tex] so to 2 sig figs (which is actually not accurate, but oh well...)

KE = 450000 J

If we want to find out how high it will have to travel up a hill so that its PE is the same as the KE at this speed, we set the value for KE = to PE:

450000 = (1000)(9.8)h so

[tex]h=\frac{450000}{(1000)(9.8)}=45.9m[/tex]

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:

there is no context??????

Answer:

[tex]H_{max}=3.4m[/tex]

Explanation:

From the question we are told that:

Mass 1 [tex]m_1=5kg[/tex]

Mass  [tex]m_2=2kg[/tex]

Distance above floor [tex]d=2.5m[/tex]

Generally the equation for Conservation of energy is mathematically given by

[tex]0.5m_1v^2+0.5m_2v^2=2mg[/tex]

[tex]0.5m_1v^2+0.5m_2v^2=2g(m_1-m_2)[/tex]

[tex]v^2(0.5*m_1+0.5*m_2)=2*g(m_1-m_2)[/tex]

[tex]v^2(0.5*5+0.5*2) = 2 * 9.8 * (5 - 2)[/tex]

[tex]v^2=\frac{58.8}{3.5}[/tex]

[tex]v=4.1m/s[/tex]

Generally the equation for The maximum height of lighter block  is mathematically given by

[tex]H_{max}=d+\frac{v^2}{2g}[/tex]

[tex]H_{max}=2.5+\frac{16.2}{2*9.81}[/tex]

[tex]H_{max}=3.4m[/tex]

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:

2 cm.

Explanation:

Please see attached photo for diagram.

In the attached photo, y is the distance from the pivot to which the 25 g is placed.

The value of y can be obtained as follow:

Clockwise moment = 5 × 10

Anticlock wise moment = y × 25

Anticlock wise moment = Clockwise moment

y × 25 = 5 × 10

y × 25 = 50

Divide both side by 25

y = 50/25

y = 2 cm

Thus, the distance is 2 cm

Answer:

10 Joule

Explanation:

The solution and answer are well written in the Pic above.

Answer:

The standard quantity with which we carry out the measurement of any physical quantity of the same kind is called a unit.

Answer:

All the sides and inner angles of a regular form must be equal. The sides and angles of an irregular form aren't the same. An equilateral triangle, for example, is a regular form because all of its sides and angles are the same length.

OAmalOHopeO

Answer:

I don't really know this one trust me am not lieing

The answer would be B..

Since sand can heat up quickly, it will also cool off quickly. But water takes a long time to heat up and cool down.

Answer: The HUMAN EYE

Explanation:

The human eye is made up of different parts which ranges from controlling the amount of light that enters the eye to the focusing of the image that is formed. The camera is a device which is both mechanically and electronically operated which shares a number of similarities with the eye.

In the human eye, the IRIS helps to regulate the amount of rays passing through the pupil to the lens by either contracting or dilating in light or dark environment respectively. While in the camera, the DIAPHRAGM controls the amount of light entering the camera.

The PUPIL serves as the passage for light into the eye while in the camera, the APERTURE does the same.

The photosensitive surface in the eye is the YELLOW SPOT while in the camera, the photosensitive surface is the PHOTOGRAPHIC FILM.

Answer:

To explore the other planets, the satellite must have the velocity more than the escape velocity.

Explanation:

The minimum velocity required by any object to escape from the earth gravitational pull is called the escape velocity.

The escape velocity for any planet depends on the mass of planet and radius of planet. It does not depends on the mass of object. The escape velocity is same for any mass for a particular planet.

So, to explore the other planets, the satellite must have the velocity more than the escape velocity.

Answer:

1. 100 J

2. 225 J

Explanation:

We'll begin by calculating the mass of the object. This can be obtained as follow:

Velocity (v) = 5 ms¯¹

Kinetic energy (KE) = 25 J

Mass (m) =?

KE = ½mv²

25 = ½ × m × 5²

25 = ½ × m × 25

25 = 25m / 2

Cross multiply

25m = 25 × 2

25m = 50

Divide both side by 25

m = 50 / 25

m = 2 Kg

1. Determination of the kinetic energy when the velocity is doubled.

Mass (m) = 2 Kg

Velocity (v) = double the initial velocity

= 2 × 5 ms¯¹

= 10 ms¯¹

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 2 × 10²

KE = ½ × 2 × 100

KE = 100 J

2. Determination of the kinetic energy when the velocity increased three times.

Mass (m) = 2 Kg

Velocity (v) = three times the initial velocity

= 3 × 5 ms¯¹

= 15 ms¯¹

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 2 × 15²

KE = ½ × 2 × 225

KE = 225 J

Answer: See explanation

Explanation:

From the information given, we are informed that an instrument rated pilot is planning a flight under IFR on July 10, this year.

It should be noted that before conducting the flight, the pilot must have performed and logged the prescribed tasks and repetitions that are required for instrument currency no earlier than January, 10 for the year.

Answer:

Explanation:

Una magnitud fundamental es aquella que se define por si misma y es independiente de las demás (masa, tiempo, longitud, etc.). magnitud derivada. Una magnitud derivada es aquella que se obtiene mediante expresiones matemáticas a partir de las magnitudes fundamentales (densidad, superficie, velocidad).

Answer:

01.

Explanation:

Half the acceleration. Its heavier and moves slower. If it moved the same acceleration, the forces would also have to be doubled since the mass was.

Answer:

 u = - 40 ft / s

Explanation:

The Galilean relation for the relative velocity is

         v ’= v + u

where u is the speed between the two reference frames, v is the speed of the fixed system and v 'the speed of the mobile system.

In this case the truck has a speed with respect to the ground (fixed system) 0 m / s (it is stopped), the car has a speed with respect to the ground of v = 40 ft / s,

        u = v'- v

        u = 0 - 40

        u = - 40 ft / s

the speed perceived by the car if the system is fixed on it is -40 ft / s

Answer:

a. i. 6.608 V ii. 5.507 V iii. 1.89 V iv. 1.89 V

b. i. 0.22 mJ ii. 0.182 mJ iii. 0.027 mJ iv. 0.036 mJ

Explanation:

a. The voltage across each capacitor

Since the 15 μf and 20 μf capacitors are in parallel, their total capacitance is C = 15 μf + 20 μf = 35 μf.

Also, since C is in series with the 10 μf and 12 μf which are in series, their total capacitance, C' is gotten from 1/C' = 1/10 μf + 1/12 μf + 1/35 μf  

1/C' = (12 + 42 + 35)/420 /μf

1/C' = 89/420 /μf

C' = 420/89 μf

C' = 4.72 μf

The total charge in the circuit' is thus Q = C'V where V = voltage = 14 V

So, Q = C'V =  4.72 μf × 14 V = 66.08 μC

Since the 10 μf and 12 μf are in series, Q is the charge flowing through them.

Since Q = CV and V = Q/C

i. The voltage across the 10 capacitor is

V = 66.08 μC/10 μF = 6.608 V

ii. The voltage across the 12 capacitor is

V = 66.08 μC/12 μF = 5.507 V

The voltage across the 15 μF and 20 μF capacitors.

Since the capacitors are in parallel, the voltage across them is the voltage across their combined capacitance, C

So, V = Q/C = 66.08 μC/35 μF = 1.89 V

iii. The voltage across the 15 μF capacitor is 1.89 V

iv. The voltage across the 20 μF capacitor is 1.89 V

b. The potential energy of each capacitor

i. The potential energy of the 10 μF capacitor

E = 1/2CV² where C = Capacitance = 10 μF = 10 × 10⁻⁶ F and V = voltage across capacitor = 6.608 V

E = 1/2CV²

E = 1/2 × 10 × 10⁻⁶ F(6.608 V)²

E = 5 × 10⁻⁶ F(43.666) V²

E = 218.33 × 10⁻⁶ J

E = 0.21833 × 10⁻³ J

E = 0.21833 mJ

E ≅ 0.22 mJ

ii. The potential energy of the 12 μF capacitor

E = 1/2CV² where C = Capacitance = 12 μF = 12 × 10⁻⁶ F and V = voltage across capacitor = 5.507 V

E = 1/2CV²

E = 1/2 × 12 × 10⁻⁶ F(5.507 V)²

E = 6 × 10⁻⁶ F(30.327) V²

E = 181.96 × 10⁻⁶ J

E = 0.18196 × 10⁻³ J

E = 0.18196 mJ

E ≅ 0.182 mJ

iii. The potential energy of the 15 μF capacitor

E = 1/2CV² where C = Capacitance = 15 μF = 15 × 10⁻⁶ F and V = voltage across capacitor = 1.89 V

E = 1/2CV²

E = 1/2 × 15 × 10⁻⁶ F(1.89 V)²

E = 7.5 × 10⁻⁶ F(3.5721) V²

E = 26.79 × 10⁻⁶ J

E = 0.02679 × 10⁻³ J

E = 0.02679 mJ

E ≅ 0.027 mJ

iv. The potential energy of the 15 μF capacitor

E = 1/2CV² where C = Capacitance = 20 μF = 15 × 10⁻⁶ F and V = voltage across capacitor = 1.89 V

E = 1/2CV²

E = 1/2 × 20 × 10⁻⁶ F(1.89 V)²

E = 10 × 10⁻⁶ F(3.5721) V²

E = 35.721 × 10⁻⁶ J

E = 0.035721 × 10⁻³ J

E = 0.035721 mJ

E ≅ 0.036 mJ

Answer:

Blood alcohol content (BAC) is a measure of the amount of alcohol that is present in the blood.

A heavier person will have a lower blood alcohol level due to a greater amount of water in their body. The more a person weighs, the more water he/she tends to have in his/her body. The water has a diluting effect on the alcohol so the BAC will be lower in comparison with a person who is slimmer  and has less water in the body.

Explanation:

Answer:

horizontal velocity remaing constan thorough out the motion but the vertical motion's velocity changes due to the gravity acting on it.

for everl 1 second the velocity decreases by 9.8 that is the gravity

The molecules making up object collide with the molecules of the other and some of the kinetic energy from a warmer object is transferred to the cooler one.