How far would you push a car if you did 28,000J of work, exerting a force of 825N?

The radius of the Sun is approximately 654497.9 kilometers.

A subtended angle in geometry is an angle created by a common point (in this case, the Earth) that crosses two points of a nearby circular arc (the Sun in this case). Below is a visual illustration of the subtended angle.

The law of cosine can be used to calculate the sun's radius, which is 654497.950 kilometers in kilometers.

The Sun's radius is nearly 109 times bigger than the Earth's. The Earth's radius is 6378 kilometers. Or to put it another way, both dimensions are 1: 109 ratios.

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The value of parameter C for the function in the figure is 2.

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

f(x) = Acos(x - C)

where;

Angular frequency is  the angular displacement of any element of the wave per unit time.

From the blue colored graph; at y = 1, x = -2 cm

1 = cos(2 - C)

(2 - C) = cos^(1)

(2 - C) = 0

C = 2

Thus, the value of parameter C for the function in the figure is 2.

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The glass is classified the way it is because of option(b)i.e, Free electrons can carry a charge.

An amorphous solid is a glass. Glass shows all the mechanical properties of a solid even though its atomic-scale structure resembles that of a supercooled liquid. Transparency, heat resistance, pressure and breakage resistance, and chemical resistance are among glass' primary properties. 90% of all manufactured glass is soda-lime glass, making it the most prevalent and least priced type.

High electronegativity and tightly bonded electrons are found in the oxygen atoms that make up glass. This is so that glass can conduct electricity, which requires a significant amount of energy to release the electrons from the oxygen atom. As a result, the glass serves as an insulator at room temperature.

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1. The mass of the air, given the data is 85.88 Kg

2. The weight of the air is 841.624 N

3. The mass of an equal volume of water is 67338 Kg

4. The weight of an equal volume of water is 659912.4 Kg

5. The volume of water that would have a mass equal to the mass of air in the room is 8.588×10⁻² m³

We shall determine the volume of the air in the room by finding the volume of the room.

Volume of air = volume of room

Volume of air = 4.3 × 5.4 × 2.9

Volume of air = 67.338 m³

Density = mass / volume

Cross multiply

Mass = Density × volume

Mass of air = 1.2754 × 67.338

Mass of air = 85.88 Kg

W = mg

W = 85.88 × 9.8

Weight of air = 841.624 N

Density = mass / volume

Cross multiply

Mass = Density × volume

Mass of water = 1000 × 67.338

Mass of water = 67338 Kg

W = mg

W = 67338 × 9.8

Weight of air = 659912.4 Kg

Volume = mass / density

Volume of water = 85.88 / 1000

Volume of water = 8.588×10⁻² m³

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The final speed of the electron is 4.64 * 10^5 m/s.

Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.

E= F/q

F = Eq

F =  330 N/C * 1.6 * 10^-19 C

F = 5.28 * 10^-17 N

F = ma

a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg

a = 5.8 * 10^13 m/s^2

Using

v = u + at

u = 0 m/s because the electron was initially at rest

v = at

v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s

v = 4.64 * 10^5 m/s

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Answer:  It simply means that a freely falling object would increase its velocity by 9.8 m/s per second

Answer:

Hello!

"The acceleration due to gravity on the surface of the earth is 9.8 m/s ²." What does it means that

That every second an object is in free fall, gravity will cause the velocity of the object to increase 9.8 m/s. So, after one second, the object is traveling at 9.8 m/s.

A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest then

(a) The speed of the 0.350- kg puck after the collision - 2.40 m/s

(b) The direction of the velocity of the 0.350- kg puck after the collision is towards West.

c) The speed of the 0.950- kg puck after the collision is 2.82 m/s

d) The direction of the velocity of the 0.950- kg puck after the collision is towards East.

Mass of ice puck, m₁ = 0.350 kg

Mass of another puck, m₂ = 0.950 kg

Velocity of ice puck, v₁ = 5.22 m/s

Velocity of another puck, v₂ = 0 m/s

[tex]v^{'}_1= ?[/tex]

[tex]v^{'}_2= ?[/tex]

[tex]m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{2 }[/tex]

[tex]v_{1} - v_{2} = - ( v^{'}_1 - v^{'}_2 )\\v_{1} - v_{2} = - v^{'}_1 + v^{'}_2\\v^{'}_2 = v^{'}_1 + v_{1}\\\\\\m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{1 } + m_{2} v_{1 }[/tex]

[tex](0.35)(5.22) + 0 = 0.350 v^{'} _{1} + 0.950 v^{'}_{1 }+ (0.950)(5.22)\\1.827 = 0.350v^{'} _{1} + 0.950v^{'}_{1 } + 4.959\\1.827 - 4.959 = 1.3 v^{'} _{1}\\- 3.132 = 1.3 v^{'} _{1}\\v^{'} _{1} = -2.40 m/s\\\\\\[/tex]

Therefore, the speed of the 0.350- kg puck after the collision is - 2.40 m/s.

[tex]v^{'}_2 = v^{'} + v_1[/tex]

    [tex]= -2.40+5.22[/tex]

    [tex]= 2.82 m/s[/tex]

Therefore, the speed of the 0.950- kg puck after the collision is 2.82 m/s.

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Answer:

Sulfur (Has six valence electrons). It has maximum valency due to belonging to VI groups of the Periodic Table.

Explanation:

The electrons found in an element's outermost atomic shell are known as valence electrons.

Sulfur, which has an atomic number of 16, has an electrical configuration of 2, 8, 6, meaning it has six electrons in its outermost shell. As a result, its valence electrons will also be six.

However, in its natural condition, sulfur exists as the S8 molecule, which has the classic chair structure where each sulfur atom is covalently connected to two other sulfur atoms. In that sense, there will be 8 valence electrons.

Consequently, the answer will be 6 if you're asking about the "sulphur atom," but 8 if you're talking about sulfur in general.

Thank you ,

Eddie

Answer:

The distance is

=

7

m

Explanation:

Apply the equation of motion

s

(

t

)

=

u

t

+

1

2

a

t

2

The initial velocity is

u

=

0

m

s

−

1

The acceleration is

a

=

2

m

s

−

2

Therefore, when

t

=

3

s

, we get

s

(

3

)

=

0

+

1

2

⋅

2

⋅

3

2

=

9

m

and when

t

=

4

s

s

(

4

)

=

0

+

1

2

⋅

2

⋅

4

2

=

16

m

Therefore,

The distance travelled in the fourth second is

d

=

s

(

4

)

−

s

(

3

)

=

16

−

9

=

7

m

D. The minimum braking torque that must be applied is -11.2 Nm.

The moment of inertia of the uniform cylinder is calculated as follows;

I = ¹/₂MR²

where;

I = (0.5)(8)(0.1²)

I = 0.04 kgm²

τ = -Iα

where;

α = ω/t

α = (80,000 x 2π/rev x 1 min/60s) / (30 s)

α =  (80,000 x 2π)/(60 x 30)

α = 279.25 rad/s²

τ = - ( 0.04 kgm²) x (279.25)

τ =  -11.2 Nm

Thus, the minimum braking torque that must be applied is -11.2 Nm.

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When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

To find the answer, we have to know about the rules followed by drawing ray-diagram.

Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

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4. a poor insulator

If rest other things are kept constant or unchanged then a good conductor can be termed as a poor insulator.

Answer:

4 conductors and insulators are opposites of each other.

Explanation:

Answer:

100rad because it angular velocity

From the first circuit (see attached), it is correct to state that I₄ = 40mA. See the explanation below.

A path that is designed to transport electric current is that is referred to as a circuit.

I = 200mA

= 20mA + 140mA + I₄

I₄ = 200 - 160

I₄ = 40mA

From the above, we can rightly posit:

R₄ = 3/(40x 10⁻³)

= 75 Ω

Hence

V₆ =  V₄ - V₅

= 3v - 2v

= 1v

R₆ therefore, =

V₆/I₆

= 1/ (40 x 10 ⁻³)

= 14.28 Ω

I₃ = I₄ + I₆

= 40 + 140

= 180mA

R₃ = 9/(180 x 10⁻³)

= 50Ω

R₂ = 12/20 x (10⁻³)

= 60 Ω

Then,

Vbattery =

200 x 10⁻³ x90

= 18 volts

The above indices is used to complete the image. See Circuit Image II.

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The ball orbit the Earth, when launched from the theoretical cannon of Newton, is option B. it is magnetically attracted.

Newton's cannonball was a hypothetical situation. Isaac Newton once proposed that gravity, which he believed to be a universal force, was the primary factor behind the planetary motion. In this experiment, Newton imagines projecting a stone or a cannonball onto the summit of a very tall mountain. The body should move away from Earth in the direction it was projected if there were no effects from gravity or air resistance.

Depending on the projectile's initial velocity and the gravitational force acting on it, the bullet will travel in a different direction. Low speeds result in a simple fallback to Earth. The Earth's surface causes the cannonball to deviate from its elliptical route.

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Answer:

Approximately [tex]2.03 \times 10^{7}\; {\rm m}[/tex].

Explanation:

Assume that the radius of this orbit is [tex]r[/tex].

Let [tex]m[/tex] denote the mass of this satellite and let [tex]M[/tex] denote the mass of the Earth. At a distance of [tex]r[/tex] from the center of the earth, the magnitude of the gravitational attraction on this satellite would be [tex]G\, m\, M / (r^{2})[/tex].

The question implies that the gravitational pull from the earth is the only significant force on this satellite. Hence, the net force on this satellite would be also [tex]G\, m\, M / (r^{2})[/tex].

The acceleration of this satellite would thus be [tex]a = (\text{net force}) / (\text{mass}) = G\, M / (r^{2})[/tex].

Let [tex]\omega[/tex] denote the angular velocity of this satellite. Since this satellite in in a circular motion, the acceleration on this satellite would need to satisfy [tex]a = \omega^{2} \, r[/tex].

In other words:

[tex]\begin{aligned} \frac{G\, M}{r^{2}} = a = \omega^{2} \, r \end{aligned}[/tex].

[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3}\end{aligned}[/tex].

The question asks for a rotation of [tex]3\times (2\, \pi) = 6\, \pi\; {\text{rad}}[/tex] within a day, which is [tex]24 \times 3600\; {\rm s}[/tex]. The angular velocity of this satellite should be:

[tex]\begin{aligned}\omega = \frac{6\, \pi}{24 \times 3600\; {\rm s}} \end{aligned}[/tex].

Substitute this value into the expression for [tex]r[/tex] and evaluate:

[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3} \\ &= \left(\frac{(6.67 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}) \times (5.97 \times 10^{24}\; {\rm kg})}{((6\, \pi) / (24 \times 3600\; {\rm s}))^{2}}\right)^{1/3} \\ &\approx 2.03 \times 10^{7}\; {\rm m}\end{aligned}[/tex].

(Note that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].)

Answer: By dividing airflow by duct cross section.

Explanation:

In short, air velocity in the ducts is calculated by dividing airflow by duct cross-section. Airflow is expressed as a simple number. Example: Air conditioner has a max. airflow of 600 CFM.

In thicker materials the particles can move more easily, therefore the resistance has to decrease.

Hence Option (b) is correct.

Resistivity, which is sometimes represented by the Greek letter rho, is quantitatively equal to a specimen's resistance R times its cross-sectional area A times its length.

ρ = R × A/ l

R = ρ × l/ A

It means the Resistance of the material is inversely proportional to the area of the material.

As the area of the material increases, the resistance decreases and vice versa.

Hence, In thicker particles, there is a greater area available and that's why resistance has to decrease so that particles can move more easily.

Hence, Option (b) is correct.

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Answer:

an instrument measuring atmospheric pressure, used especially in forecasting the weather and determining altitude.

Explanation:

hope it helps ya

The thing that the presenter can do to make the presentation to appear informal is to use contractions such as won't, it's, let's, and we're.

The term report refers to any formal presentation. It is usually made as part of an academic course or as part of professional development. It could be given while a person is hoping for promotion at a job that he or she is doing.

We have to bear in mind that the audience to whom the report is being presented is a familiar audience hence the presentation can be quite informal and devoid of the formality of presentation of reports as we know it to be.

Now, the focus is to make the presentation to look informal. The thing that the presenter can do to make the presentation to appear informal is to use contractions such as won't, it's, let's, and we're.

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Answer:

B

Explanation:

friction opposes motion

Friction typically Slows down objects

While this is almost true for a wide range of low speeds, as speed increases and air friction is reckoned with, it has been found that friction depends not only on speed, but also on speed squared and sometimes on higher powers of friction. speed.

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The formular for calculating half life is:

T1/2=0.693/Π

The frictional force between the crate and the ramp is determined as 35.2 N.

Apply the principle of conservation energy to calculate the change in the energy of the crate.

Change in energy of the crate = energy loss to friction

P.Ei - K.Ef = E

mgh - ¹/₂mv² = E

where;

(30)(9.8)(2) - (0.5)(30)(2²) = E

528 J = E

E = Fd

where;

F = E/d

F = (528)/(15)

F = 35.2 N

Thus, the frictional force between the crate and the ramp is determined as 35.2 N.

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Answer:

150 J

Explanation:

Moving 3 times higher will increase the   P E  x 3   = 150 J

Answer:

B Gravitational potential energy

Explanation:

The electric field strength at a point 1.00 cm to the left of the middle is  -2.0 x 10⁷ N/C.

The magnitude of the force is 94.4 N and direction of the force on it towards the right.

The electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;

E = kq/r²

Electric field due to first charge

E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²

E1 = 1.35 x 10⁸ N/C

Electric field due to second charge

E2 =  -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²

E2 = - 1.35 x 10⁸ N/C

Electric field due to third charge

E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²

E3 = -2.0 x 10⁷ N/C

E = E1 + E2 + E3

E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - -2.0 x 10⁷ N/C

E = -2.0 x 10⁷ N/C

F = Eq

F = - 2.0 x 10⁷ x -4.72 x 10⁻⁶

F = 94.4 N

Thus, the direction of the force will be towards the right.

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Answer:

Approximately [tex]3.86\; {\rm N}[/tex] (given that the magnitude of this charge is [tex]-7.35\; {\rm \mu C}[/tex].)

Explanation:

If a charge of magnitude [tex]q[/tex] is placed in an electric field of magnitude [tex]E[/tex], the magnitude of the electrostatic force on that charge would be [tex]F = E\, q[/tex].

The magnitude of this charge is [tex]q = 7.35\; {\rm \mu C}[/tex]. Apply the unit conversion [tex]1\; {\rm \mu C} = 10^{-6}\; {\rm C}[/tex]:

[tex]\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}[/tex].

An electric field of magnitude [tex]E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}[/tex] would exert on this charge a force with a magnitude of:

[tex]\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}[/tex].

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

Answer:

The low energy X-ray calibration service is intended for thin-window plane-parallel chambers required for the dosimetry of superficial X-ray beams. Low-energy X-ray therapy is also known as superficial radiation therapy. This therapy uses beams of low-energy X-rays (radiation waves) to destroy skin cancer cells while not harming healthy tissue around or under the upper layers of skin.

The value of parameter B (angular frequency) for this function is 2.7.

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

f(x) = Acos(Bx)

where;

Angular frequency is  the angular displacement of any element of the wave per unit time.

From the blue color; at y = -0.9, x = -7.5 cm

-0.9 = cos(-7.5B)

7.5B = arc cos (0.9)

7.5B = 2.7

B = 7.5/2.7

B = 2.7

Thus, the value of parameter B (angular frequency) for this function is 2.7.

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The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

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