Question 2 of 25
What is the term for a pair of forces described by Newton's third law?

A.
Normal-friction pair

B.
Action-reaction pair

C.
Kinetic-static pair

D.
Inertia-movement pair

The work function in eV for the given cutoff frequency is  5.05 eV.

The work function is related to the frequency as

Wo = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the cutoff frequency for a certain element is 1.22 x 10¹⁵ Hz

Wo = 6.626 x 10⁻³⁴ x  1.22 x 10¹⁵ Hz / 1.6 x 10⁻¹⁹

Wo = 5.05 eV

Thus, the work function is  5.05 eV

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The force the rod exerts on the ball at the given angle is determined as 3.94 N.

The force exerted is calculated as follows;

F = ma

F = mv²/r

F = mω²r

where;

r = 2cosθ

ωf² = ωi² + 2αθ

where;

ωf² = (1)² + 2(2)(15 x π/180)

ωf² = 2.04

ωf = √2.04

ωf = 1.428 rad/s

F = mω²(2cosθ)

F = (1)(1.428)²(2 x cos15)

F = 3.94 N

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Answer:

Explanation:

Comment

I think the best answer is probably A.  It's not the best reason for Earth loosing it's helium, but it is the only one that is close. The Earth derives some of its Helium (most) by the decomposition process of radio active high weight chemical that break down. Helium combines with practically nothing so once it gets moving, very little will stop it. That's the way we loose our Helium.

Still answer A.

Here is your answer mate,

Question,

[tex]Determine\: the\: amount\\ \: of\: power\:used\: in\\\: holding\: a\: 25\: kg\: box\:\\ , \: 1.5\: meters \: above\: the\: floor\\\: for\: 60\: seconds[/tex]

Answer,

[tex][/tex]

Solution,

[tex][/tex]

Given,

[tex]MASS \: IS\: 25\: KG\: \\ and \: HEIGHTIS\: 1.5m\: [/tex]

[tex][/tex]

WORK DONE (done against gravity) =

mass×acceleration due to gravity ×height

WORK = 25× 10× 1.5

[tex]\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: [/tex]= 375 Nm

[tex][/tex]

Now

POWER =

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{work}{time} [/tex]

POWER

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{375}{60} Watt [/tex]

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =6.25[/tex]

[tex]Therfore\: your \: answer\: is\: 6.25[/tex]

[tex][/tex]

Check this,

[tex]Acceleration\: due\: to \: gravity\\\: can\: be\: 9.8\: m/s²\: \\As\: nothing\: mentioned\\\: in\: question\: \\I \: took \: it \: as \: 10[/tex]

[tex][/tex]

Have a good day

Answer:

given -

initial velocity = 4.09 m/s

acceleration = 1.01 m/s²

distance = 23.4 m

time = ?

using second formula of motion,

s = ut + 1/2 at².

where, s = distance

u = initial velocity

t = time

a = acceleration

23.4 = 4.09(t) + 1/2(1.01)(t) ²

23.4 = 4.09t + 2.02t²

2.02t² + 4.09t - 23.4 = 0

solving the equation by using quadratic formula

Use the standard form, ax² + bx + c = 0 , to find the coefficients of our equation, :

a = 2.02

b = 4.09

c = -23.4

we get t=2.539 or t= -4.563

time cannot be negative so

t= 2.539 sec = 2.6 Sec is the answer

Hello!

We can use the following equation for magnetic force on a charged particle:
[tex]F_B = qv \times B[/tex]

[tex]F_B[/tex] = Magnetic force (N)

q = Charge of particle (1.6 × 10⁻¹⁹ C)
v = velocity of particle (5.2 × 10⁷ m/s)

B = Magnetic field strength (1.4 T)

This is a cross-product, so the equation can be rewritten to F = qvBsinφ where φ is the angle between the magnetic field and particle velocity vectors.  

Since the proton's velocity vector and the magnetic field vector are perpendicular, sin(90) = 1. We can reduce the equation to:

[tex]F_B = qvB[/tex]

Plug in the known values.

[tex]F_B = (1.6*10^{-19})(5.2*10^7)(1.4) = \boxed{1.1648 *10^{-11} N}[/tex]

The velocity of the air relative to the runner is 5 m/s.

We must recall that velocity is a vector quantity and the relative velocity must be obtained vectorially. Thus we know that;

Velocity of the runner = 4m/s. due west

Velocity of the wind =  3m/s due south

The relative velocity is;

Vr = √(4)^2 + (3)^2

Vr = 5 m/s

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The strength of the electric field on the point charge at this distance will be 4000 V/m.

The strength of the electric field is the ratio of electric force per unit charge.

The given data in the problem is;

Qis the unit charge = 4.0 × 10⁻⁶ C

E is the strength of the electric field

R is the distance from point charge = 3 m

The strength of the electric field is;

[tex]\rm E = \frac{KQ}{R^2} \\\\ \rm E = \frac{9 \times 10^9 \times 4 \times 10^{-6} \ C}{3^2} \\\\ E= 4000 V/m[/tex]

Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.

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Answer:

0.5 m/s²

Explanation:

according to Newton's second law, we are goven a relationship between force, mass and acceleration, with the formula:

F = m×a

F for force

m for mass

a for acceleration

we use the given data and get:

20 = 40×a

we find a=20/40=0.5m/s²

acceleration of the box is 20m/s^2

given

mass (m) = 40 kg

force (f) = 20n

acceleration (a) =?

we know,

f = m × a

20 = 40 × a

40 - 20 = a

20m/s^2 = a

Explanation:

This case is inertia.

Mass is directly proportional to inertia

The resultant of the two forces is about 170 N.

The term resultant force has to do with the single force that has the same magnitude and direction as two or more forces acting together.

In this case, the both forces are acting in the forward direction. This implies that the resultant force is 90 N  + 80 N = 170 N.

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Answer:

35 m/s^2

Explanation:

Decceleration =   change in velocity / change in time

                        =  ( 70 m/s ) / 2 seconds = 35 m/s^2

   DEcceleration = 35 m/s^2

       Jason's speed changes by  - 35 m/s^2

Explanation:

⠀

⠀

(a) The segment A shows acceleration as velocity increases with the increase in time.

⠀

⠀

⠀

(b) The segment C shows the object is slowing down as the time increases in segment C, the velocity decreases and afterwards it comes to rest.

⠀

⠀

⠀

(c) The velocity is segment B is 40m/s. And in the diagram there is no change in velocity.

⠀

⠀

⠀

(d) The acceleration of segment B is zero. As there in no change in curve and it is moving with uniform velocity.

⠀

⠀

⠀

[tex] \: [/tex]

Answer:

1 A and C

2 C

3 40m/s

4 zero

Explanation:

The mass and tension due to the system are as follows:

Tension is a type of pulling force due transmitted by means of a string or cable.

a) The mass of the child and scale = 104/9.81 = 10.6 kg

b) The tension T, in the cord attaching the child to the scale = scale reading = 104N

c) The tension T', in the cord attaching the scale to the ceiling = scale reading + weight of scale

T' = 104 + (0.5 * 9.81)

T' = 108.9 N

d) The sketch is attached in the picture

In conclusion, the tension is force exerted on the cord due to the weight of the scale and the baby.

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(a)  The charge stored in the 5.00-μF capacitor is 37.2  μC.

(b) The potential difference across the 10.0-μF capacitor is 3.72 V.

The capacitance of a capacitor is defined as the ratio of the charge stored and the potential difference between the capacitor.

The capacitance of a capacitor is denoted by C and expressed as

C = Q/V

Given, 5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.

(a) The equivalent capacitance is

1 / Ceq = 1 / C₁ +1 / C₂ + 1/ C₃

Substitute the values, we get

Ceq = 3.1  μF

The charge stored in 5.00-μF capacitor is

Q  = Ceq x V

Q = 3.1  μF x 12 V

Q = 37.2  μC

Thus, the charge stored in the 5.00-μF capacitor is 37.2  μC

(b) The potential difference across the 10.0-μF capacitor is given by

V = Q/C₂

Put the values, we get

V = 37.2 / 10

V = 3.72 V

Thus, the potential difference across the 10.0-μF capacitor is 3.72 V.

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Answer:

Given - Force = 21.4 N

Work done = 1974 J

To find - Distance

Solution -

Work done = Force * displacement

1974 j = 21.4 N * displacement

1974/21.4 = displacement

92.24

[tex]\\ \rm\Rrightarrow Work=Force\times Displacement [/tex]

[tex]\\ \rm\Rrightarrow 1974=21.4×Displacement[/tex]

[tex]\\ \rm\Rrightarrow Displacement=92.24m[/tex]

The I3 will be 158 A.

To calculate I3 firstly, V4 has to be calculated,

[tex]V_{4} =I_{4} R_{4}[/tex]

[tex]V_{4} = V_{2} / R_{4} + R_{5} * R_{4}[/tex]

[tex]V_{4} = 12 * 135 / 135+61[/tex]

[tex]V_{4} = 8.26V[/tex]

For I3,

[tex]I_{3} = R_{1} /(R1+R3 + (R1+R3)(R2+R6) * (V2 - V1 (R1+R2+R6/R1)[/tex]

[tex]I3=(61)/((61)(50)+(61+50)(141+141)) (12 -18 (1+(141+141)/61)) = -.158 A[/tex]

Hence, the current through I3 will be 158 A.

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Answer:

CABD

.....................

The image is cut and jumbled and placed into different places, so the correct order of the image that it can be back together is CABD.

A specific piece of something is an image. It can provide information to the optical system in two dimensions, three components, or in another way. An item that mimics a subject, such as a picture or other two-dimensional image, might be considered an image. An image in the context of signal analysis is a dispersed color amplitude.

A graphical image need not make use of the complete visual system. A common illustration of this is a grayscale image, which does not employ color, but instead relies on the visual game's sensitivity to brightness across all wavelengths. Even if it doesn't make proper use of the visual system, a black-and-white and white visual depiction of something is nonetheless an image.

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The equivalent capacitance of these capacitors as they are arranged in series and parallel will be 0.093 μF and 71.1 μF.

A capacitor is a device that can store electrical energy. It is a two-conductor configuration separated by an insulating medium that carries charges of equal size and opposite sign.

If the capacitor is connected in the series the equivalent capacitance is found as;

[tex]\rm \frac{1}{c_{eq}} = \frac{1}{c_1} +\frac{1}{c_2} +\frac{1}{c_3} +\frac{1}{c_4} +\frac{1}{c_5} \\\\\ \rm \frac{1}{c_{eq}} = \frac{1}{10} +\frac{1}{50} +\frac{1}{9} +\frac{1}{2} +\frac{1}{0.1} \\\\\ \frac{1}{c_{eq}} = 0.1+0.02+0.11+0.5+10 \\\\ \frac{1}{c_{eq}} = 10.7 \\\\ c_{eq} = 0.093 \ \mu F[/tex]

If the capacitor is connected in parallel the equivalent capacitance,cₐ is found as;

cₐ= c₁+c₂+c₃+c₄+c₅

cₐ = 10+50+9+2+0.1 μF

cₐ =71.1 μF

Hence, the equivalent capacitance of these capacitors as they are arranged in series and parallel will be 0.093 μF and 71.1 μF.

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The force is reasonable for making fusion possible in the Sun is heat energy.

When the slow moving neutrons are bombarded with the heavy radioactive nuclei, the product is the more number of neutrons are produced with the large amount of energy. This multiplying process is called nuclear fusion.

The amount of energy produced in such a reaction can be calculated using the equivalence of mass and energy relationship.

E = mc²

The same happens in nuclear fusion where large amount of energy is needed to make more heavy nuclei.

Thus, fusion requires heat energy to continue the reaction.

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The average velocity is 4i m/sec

The average acceleration is 4i

Average velocity is defined as the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs. T

According to the question,

x=at²i + btj ( t is in second and x is in meter).

Average velocity =  [tex]\frac{dx}{dt}[/tex]

[tex]\frac{dx}{dt} = 2at i + b j\\\\\frac{dx}{dt} = 2 * 2 * t i +1 j\\\\\frac{dx}{dt} = 4t i +j[/tex]

[tex]\frac{dx}{dt}[/tex] in the time interval of 2 to 3 sec is

at 2 sec,

[tex]\frac{dx}{dt} = 4 * 2 i +j = 8i +j[/tex]

at 3 sec ,

[tex]\frac{dx}{dt}[/tex] = 4 * 3 i + j = 12i+j

Therefore the average velocity is

[tex]\frac{(12i +j - 8i -j)}{(3-2)} = 4i[/tex]

The velocity at any given time is   [tex]\frac{dx}{dt} = 4t i +j[/tex]

The acceleration at any given time is [tex]\frac{dx^2}{dt^2}[/tex]

[tex]\frac{dx^2}{dt^2} = 4 i[/tex]

The average acceleration in the time interval  t₁=2sec and t₂=3sec

Average Acceleration =  4i

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a. The magnitude of the impulse which the player must impart to the ball is 17.25 kg.m/s.

b. The magnitude of average acceleration of the player's head during the impact is   195.66 m/s².

The change in momentum is equal to the product of impact force applied while colliding and time for that impact.

Impulse F. t = m (Vf -Vi)

where, Vf is the final velocity and Vi is the initial velocity.

Given, a soccer player 'heads' a size 3 ball deflecting it by 40.0°, and keeps its speed of 13.20 m/s constant. A size 3 ball has a mass of approximately 2.000 kg.

a. Substitute the values into the expression, we get

Impulse (x) = 2 x (13.20 - 13.20cos40°)

Impulse  (x) = 3.088 kg.m/s

Impulse in y direction, is

Impulse (y) = 2 x  13.20sin40°

Impulse (y)  = 16.97 kg.m/s

so, the magnitude of impulse is

I = sq rt(Ix² + Iy²)

Put the values, we get

I = 17.25 kg.m/s

Thus, the magnitude of the impulse which the player must impart to the ball is 17.25 kg.m/s

b. The player's head has a mass of 2.90 kg. The brief time of the impact, 30.40 ms,

F.t = I

Put the given values, we have

F = 17.25 / 30.40 x 10⁻³

F = 567.434 N

From the Newtons second law of motion,

F = ma

Plug the values from the question, we have

567.434 N =  2.90 x a

a = 195.66 m/s²

Thus the magnitude of average acceleration of the player's head during the impact is  195.66 m/s².

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Answer:

See the answer Explain why aircraft are carefully designed so that parts do not resonate. Expert Answer This virtually takes place, however maximum usually in small piston-engined airplanes, in particular dual-engined airplanes. The resonant frequency of the fuselage of a small plane goes to have numerous nodes, withinside the low loads of hertz.

Answer:

false

Explanation:

it cant defined the messy and clean states

A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.

In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.

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The difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl are HCl is a polar covalent compound because the chloride ion is extra electronegative than the hydrogen ion.

HCl is a polar molecule. This is due to the fact the Chlorine (Cl) atom withinside the HCl molecule is extra electronegative and does now no longer proportion the bonding electrons similarly with Hydrogen (H). But H2 And Cl2 are nonpolar because of comparable electronegativity of each the atoms withinside the molecule H2 And Cl2 .

Hydrogen chloride is a diatomic molecule, such as a hydrogen atom H and a chlorine atom Cl related with the aid of using a polar covalent bond. The chlorine atom is an awful lot extra electronegative than the hydrogen atom, which makes this bond polar.

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Answer:

Answer is C because light travels in a sight line but when light pass through a refractor the light from the source changes direction when passes through a refractor

Answer:

He has a hangover.

Explanation:

Just something I know.

You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, the average force on the jet is due to the catapult is mathematically given as

What is the average force on the jet is due to the catapult?

Generally, the equation for acceleration is mathematically given as

[tex]a=\frac{vf^2-vi^2}{2s}\\\\\Therefore\\\\a=\frac{69.29^2-0^2}{2(90}\\\\a=26.673m/s^2[/tex]

In conclusion, The force

F=m*a

F=15100*26.673

F=40272.3N

F 402 KN

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