how much work is needed to move an object 20 meters with 40 newtons of force?

Answer:

inonic bond has greatest difference in electronegativiru

The type of bond that has elements with the greatest difference in electronegativity ionic bond. The correct option is is B.

Electronegativity is a chemical property that describes an atom's or functional group's tendency to attract electrons toward itself.

An atom's electronegativity is affected by both its atomic number and the distance between its valence electrons and the charged nuclei.

Electronegativity is a measurement of an atom's proclivity to attract electrons (or electron density) towards itself.

It governs the distribution of shared electrons between two atoms in a bond. The greater an atom's electronegativity, the more strongly it attracts electrons in its bonds.

Ionic bonds are formed by elements with large differences in electronegativity. Covalent bonds are formed between atoms of elements with similar electronegativity.

Thus, the correct option is B.

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Answer:

true

Explanation:

(sorry something wrong w my keyboard so write each line for the explnation!)

63.0 m

Explanation:

Acceleration of car

=

v

−

u

t

=

0 ms

−

1

−

21.0 ms

−

1

6.00 s

=

−

3.50 ms

−

2

S

=

v

2

−

u

2

2a

S

=

(

0 ms

−

1

)

2

−

(

21.0 ms

−

1

)

2

2

×

−

3.50 ms

−

2

S

=

63.0

m

Answer:

Object A

Explanation:

The object that would make you feel worse if you're hit by it is the object possessing the highest momentum. Thus, we need to find the momentum of the two objects.

Momentum of an object is the product of its mass and that of it's velocity. Momentum is given by the formula

P = M * V, where

P = momentum

M = mass of the object

V = velocity of the object

Now, solving for object A, we have

P(a) = 1.1 * 10.2

P(a) = 11.22 kgm/s

And then, solving for object B, we have

P(b) = 2 * 5

P(b) = 10 kgm/s

The object when the highest momentum is object A, and thus would make you feel worse when hit by it

Answer:

clouds are created when water vapor and invisible gas turns into water droplets and these water droplets form tiny particles like dust and float in the air.

you have cumulus

cirrus

cirrostraus  

Explanation:

Answer:

0.15 nm

Explanation:

d = Distance between the atomic planes

m = Order

[tex]\theta_{m}[/tex] = First angle = [tex]45.6^{\circ}[/tex]

[tex]\theta_{m+1}[/tex] = Adjacent angle = [tex]21^{\circ}[/tex]

[tex]\lambda[/tex] = Wavelength = 0.07 nm

From Bragg's relation we know

[tex]2d\cos\theta_{m}=m\lambda[/tex]

[tex]2d\cos45.6^{\circ}=m0.07[/tex]

[tex]2d\cos\theta_{m+1}=(m+1)\lambda[/tex]

[tex]2d\cos21^{\circ}=(m+1)0.07\\\Rightarrow 2d\cos21^{\circ}=m(0.07)+0.07[/tex]

So

[tex]2d\cos21^{\circ}=2d\cos45.6^{\circ}+0.07\\\Rightarrow 2d(\cos21^{\circ}-\cos45.6^{\circ})=0.07\\\Rightarrow d=\dfrac{0.07}{2(\cos21^{\circ}-\cos45.6^{\circ})}\\\Rightarrow d=0.14962\ \text{nm}[/tex]

The distance between the atomic planes is 0.15 nm.

Answer: the required height is 2.5 m

Explanation:

given the given data in question

Lime to come in rest;

using S = ((U1 + U2)/2) t

⇒ 0.5 = ((√(2gh) + 0)/2) t

t = (0.5 × 2) / √(2gh)

now fat = impulse = change in momentum dp

F = 5mg = dp/dt = [(m√(2gh))/(0.5 × 2)] × √(2gh)

5mg = 2mgh/(0.5 × 2)

5 = h / 0.5

h = 5 × 0.5

h = 2.5 m

Therefore the required height is 2.5 m

[tex] \small\bf \: let \: the \: distance \: of \: the \: man \: from \: the \: cliff \: be \: x[/tex]

[tex] \small\bf \: thus \: time \: taken \: by \: sound \: to \: hit \: the \: cilff \: and \: return = \frac{2x}{v} = 1[/tex]

[tex] \bf \to \: x = \frac{320}{2} m = 160m[/tex]

[tex] \small \bf \: thus \: the \: distance \: between \: the \: cliffs \: = 160m \times 2 = 320m[/tex]

Answer:

Since binary is only 1 and 0, you can use a flashlight to display something similar to Morse code (see explanation below)

Explanation:

In binary, 1 means "on" and 0 means "off". A way you can use visible light is through turning on and off a flashlight. If the flashlight is turned on, it would represent a 1. If the flashlight is turned off, it would represent a 0. To make the message easier and more accurately understood for the receiver make sure to flash the lights in a consistent pattern (ex. each flash lasts no longer than half a second, one second between each digit, etc.)

For example, let's say you're trying to send the message "11001"

  on     on    off     off     on

0       1       2       3       4       5      Numbers represent seconds

As you can see above the message starts at 0 seconds. Between 0 and 1 seconds the flashlight is turned on once. Between 1 and 2 seconds the flashlight is turned on again, Between 2 and 3 seconds as well as 3 and 4 seconds the flashlight is not turned on at all. And finally between 4 and 5 seconds the flashlight is turned on.

Answer:

The value of the inductance is 1.364 mH.

Explanation:

Given;

amplitude current, I₀ = 200 mA = 0.2 A

amplitude voltage, V₀ = 2.4 V

frequency of the wave, f = 1400 Hz

The inductive reactance is calculated;

[tex]X_l = \frac{V_o}{I_o} \\\\X_l = \frac{2.4}{0.2} \\\\X_l =12 \ ohms[/tex]

The inductive reactance is calculated as;

[tex]X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2 \pi f}[/tex]

where;

L is the inductance

[tex]L = \frac{12}{2 \pi \times \ 1400} \\\\L = 1.364 \times \ 10^{-3} \ H\\\\L = 1.364 \ mH[/tex]

Therefore, the value of the inductance is 1.364 mH.

Given values are:

As we know the formula,

→ [tex]X_l = \frac{V_0}{I_0}[/tex]

By substituting the values, we get

       [tex]= \frac{2.4}{0.2}[/tex]

       [tex]= 12 \ ohms[/tex]

Now,

The inductive reactance will be:

→ [tex]X_l = \omega L[/tex]

or,

→ [tex]X_l = 2 \pi f L[/tex]

Hence,

The Inductance will be:

→ [tex]L = \frac{X_i}{2 \pi f}[/tex]

By putting the values, we get

      [tex]= \frac{12}{2 \pi\times 1400}[/tex]

      [tex]= 1.364\times 10^{-3} \ H[/tex]

      [tex]= 1.36 \ mH[/tex]

Thus the above answer is right.

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Answer: the required mass is 1.7628 × 10²⁵ μ

Explanation:

Given that;

energy released in the fission of one ²³⁵U₉₂ is 206.6 MeV

power p = 28.7 Mega Watts = 28.7 × 10⁶ W = 28.7 × 10⁶/ 1.6× 10⁻¹³ = 17.9375 × 10¹⁹ MeV/s

now fission need per second will be;

⇒ power / energy released i  fission

= 17.9375 × 10¹⁹ MeV /  206.6 MeV = 8.68 × 10¹⁷ per second

now fission need per day will be;

⇒ ( 8.68 × 10¹⁷ × 360 × 24 ) = 7.5 × 10²² per day

hence mass of ²³⁵U₉₂ that is consumed in one day in this reactor will be;

⇒ (235 × 7.5 × 10²²)μ

= 1.7628 × 10²⁵ μ

Therefore the required mass is 1.7628 × 10²⁵ μ

Answer:

The options are

a. Sally is a very creative kind of person who likes to build things.

b. Jerry only works because he receives a very large income.

c. Rikki is usually shy, but at work she appears to be quite outgoing.

d. Maury gives money to charities because he wants other people to think he is very generous.

The answer is c. Rikki is usually shy, but at work she appears to be quite outgoing.

Lewin's interactionist perspective explains that an individual’s behavior is usually dependent on his personal behavior/ trait and the environment. The best option is that Rikki is usually shy which is her personal behavior but at work she appears to be quite outgoing due to her environment.

Answer:

The force of friction required to keep the car from sliding down the hill is 898.054 newtons and since [tex]f<f_{max}[/tex], the car will not slide down the hill.

Explanation:

The free body diagram is included below and now we prepare each equation of equilibrium for respective orthogonal axis:

[tex]\Sigma F_{x'} = f-m\cdot g \cdot \sin \theta = 0[/tex] (1)

[tex]\Sigma F_{y'} = N-m\cdot g \cdot \cos \theta = 0[/tex] (2)

Where:

[tex]f[/tex] - Static friction force, measured in newtons.

[tex]N[/tex] - Normal force from the inclined hill to the car, measured in newtons.

[tex]m[/tex] - Mass of the car, measured in newtons.

[tex]g[/tex] - Gravitational constant, measured in meters per square second.

[tex]\theta[/tex] - Angle of inclination of the hill, measured in sexagesimal degrees.

If we know that [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 3.5^{\circ}[/tex], then the friction force required to keep the car from sliding down the hill is:

[tex]f = m\cdot g \cdot \sin \theta[/tex]

[tex]f = (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \sin 3.5^{\circ}[/tex]

[tex]f = 898.054\,N[/tex]

The force of friction required to keep the car from sliding down the hill is 898.054 newtons.

If the car does not slide down the hill, then the following condition must be observed:

[tex]f\le \mu_{s}\cdot m\cdot g \cdot \cos \theta[/tex] (2)

Where [tex]\mu_{s}[/tex] is the static coefficient of friction, dimensionless.

If we know that [tex]\mu_{s} = 0.45[/tex], [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 3.5^{\circ}[/tex], then the maximum allowable static friction force is:

[tex]f_{max} = \mu_{s}\cdot m \cdot g \cdot \cos \theta[/tex]

[tex]f_{max} = (0.45)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 3.5^{\circ}[/tex]

[tex]f_{max} = 6607.378\,N[/tex]

Since [tex]f<f_{max}[/tex], the car will not slide down the hill.

Answer:

   ω = ω₀ + α t

      ω² = ω₀² + 2 α  θ

      θ = θ₀ + ω₀ t + ½ α t²

Explanation:

Rotational kinematics can be treated as equivalent to linear kinematics, for this change the displacement will change to the angular displacement, the velocity to the angular velocity and the acceleration to the angular relation, that is

     x → θ

     v → ω

     a → α

with these changes the three linear kinematics relations change to

      ω = ω₀ + α t

      ω² = ω₀² + 2 α  θ

      θ = θ₀ + ω₀ t + ½ α t²

where it should be clarified that to use these equations the angles must be measured in radians

Answer:

[tex]I_{rms}=2.12\ A[/tex]

Explanation:

Given that,

The instantaneous current in the circuit is giveen by :

[tex]I=3\sin\theta\ A[/tex]

We need to find the rms value of the current.

The general equation of current is given by :

[tex]I=I_o\sin\theta[/tex]

It means, [tex]I_o=3\ A[/tex]

We know that,

[tex]I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A[/tex]

So, the rms value of current is 2.12 A.

Answer: The height of the cliff is 104.59 m

Explanation:

The horizontal speed of the car when it leaves the cliff is 13 m/s, and it hits the ground 60m from the shoreline.

Here we can use the relationship:

Time*Speed = Distance.

To find the time that the car is in the air, we know that:

speed = 13m/s

distance = 60m

time = T

13m/s*T = 60m

T = (60m)/13m/s = 4.62 s

This means that the car is falling for 4.62 seconds.

Now let's analyze the vertical problem.

As the car leaves the cliff, it only has horizontal velocity, this means that the vertical initial velocity will be zero

The only force acting in the vertical axis is the gravitational force, this means that the acceleration will be equal to the gravitational acceleration, which is:

g = 9.8m/s^2

then:

a = -9.8m/s^2

Where the negative sign is because the acceleration is pulling the car downwards.

To get the vertical velocity, we could integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the constant of integration and the initial vertical velocity, that we already know that is equal to zero, then the vertical velocity as a function of time can be written as:

v(t) = (-9.8m/s^2)*t

To get the vertical position equation, we need to integrate again over the time:

P(t) = (1/2)*(-9.8m/s^2)*t^2 + H

Where H is the constant of integration and the initial vertical position, then H will be the height of the cliff.

We know that the car needs 4.62 seconds to hit the ground, this means that:

P(4.6s) = 0m

Then:

P(t) = (1/2)*(-9.8m/s^2)*(4.62s)^2 + H = 0

            (-4.9m/s^2)*(4.62s)^2 + H = 0

          H =  (4.9m/s^2)*(4.62s)^2 = 104.59 m

This means that the cliff is 104.59 meters high

Answer:

No they just orbit the sun

Explanation:

Answer:

155.17N.

Explanation:

The magnitude of the net force is expressed as;

F = mv²/r where;

m is the mass

v is the velocity of the airplane

r is the radius of the loop

Given

m = 95kg

v = 70m/s

r = 3km = 3000m

Required

Magnitude of the net force

F = 95*70²/3000

F = 95*4900/3000

F = 95*49/30

F = 4655/30

F = 155.17N

Hence the magnitude of the net force on the 95 kg pilot at the bottom of this loop is 155.17N.

The net force on the pilot is 1551.67 N.

Force can be defined as the product of mass and acceleration.

To calculate the magnitude of the net force at the bottom of the loop, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the net force on the pilot is 1551.67 N.

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Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

Answer:

hurricanes,Typhoons and cyclones

Answer:

B) In the near-ultraviolet

Explanation:

The best option to this problem is : In the near-ultraviolet

Answer:

the second pendulum ( also called the royal pendulum) o.994m (39.1 in) long, in which each swing takes one second, became widely used in quality clocks.

Answer:

λ = 698 nm (Approx)

Explanation:

Given:

Width d = 0.002 cm = 0.002 x 10⁻² nm

θ = 2°

Computation:

d sinθ = mλ

(0.002 x 10⁻²) sin2° = mλ

λ = 698 nm (Approx)

Answer:

I think C

Explanation:

if im wrong then what is it im taking the exam too

Answer: C. She can increase the number of wire loops.

Explanation:

Answer:

ok fine I'll answer u on comment

Answer:

4.4 min

Explanation:

To solve this, we use the law of conservation of momentum

0 = -m(a).v(a) + m(c).v(c), where

m(a) = mass of the astronaut

m(c) = mass of the camera

v(a) = velocity of the astronaut

v(c) = velocity of the camera

Making v(a) subject of the formula, we have

v(a) = m(c).v(c)/m(a)

Now, we make one last assumption, that the astronaut is moving at constant speed and time, thus the time needed will be

t = s/v -> since s = vt

Now, we already have our v from above, so we substitute

t = s/[m(c).v(c)/m(a)]

t = s.m(a)/m(c).v(c)

Applying the values, we have

t = (41.4 * 70.1) / (0.916 * 12)

t = 2902.14 / 10.992

t = 264 seconds or 4.4 minutes

Answer:

1832

Explanation:

From;

Δp Δx = h/4π

Δp = uncertainty in momentum

Δx = uncertainty in position

h= Plank's constant

But p =mv hence, Δp= Δmv

m= mass, v= velocity

mass of electron = 9.11 * 10^-31 Kg

Mass of proton = 1.67 * 10^-27 Kg

since m is a constant,

Δv = h/Δxm4π

For proton;

Δv = 6.6 * 10^-34/4 * 3.14 * 1.67 * 10^-27 * 1 * 10^-10

Δv = 315 ms-1

For electron;

Δv = 6.6 * 10^-34/4 * 3.14 * 9.11 * 10^-31 * 1 * 10^-10

Δv = 577000 ms-1

Ratio of uncertainty of electron to that of proton = 577000 ms-1/315 ms-1= 1832

Answer:

basketball

Explanation: cause it is

Answer:

It honestly depends on which one you're better at.

Explanation: I mean that's what I think.