Why does the ball orbit the Earth when launched from the theoretical cannon of Newton?
A. it gets stuck in the atmosphere
B. it is magnetically attracted
C. it is attached by a rope to the Earth
D. it falls at the same rate the Earth curves

Answer:Using the Nielsen form, determine the equation of motion for a mass m connected to a spring of constant k. Exercise 3.2 Using the

Explanation:

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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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:

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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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:

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.

Answer:

See below

Explanation:

29 mile/hr * 5280 f/mile / 3600  s/hr = 42.53  ft/ sec

   42.53 ft / sec    /   12 feet =  3.54  cycles / sec = 3.54 Hz

      for frequency I suppose....where is the referred to figure?

(a) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.

(b) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.

The speed of the shell at the bottom of the bowl is calculated by applying the principle of conservation of energy.

K.E(rot) + K.E(trans) = P.E

where;

¹/₂mv² + ¹/₂Iω² = mgh

where;

¹/₂mv² + ¹/₂(²/₃Mr²)(v/r)² = mgh

¹/₂v² + ¹/₂(²/₃r²)(v²/r²) = gh

¹/₂v² + ¹/₂(²/₃)(v²) = gh

¹/₂v² + ¹/₃v² = gh

⁵/₆v² = gh

v² = 6gh/5

v = √(6gh/5)

Let the vertical height from the edge of bowl to the bottom , h = R = 80 cm

v = √(6 x 9.8 x 0.8 /5)

v = 3.1 m/s

v = √(6gh/5)

v = √(6 x 9.8 x 0.1 /5)

v = 1.1 m/s

Thus, the speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.

The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.

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Answer:

B Gravitational potential energy

Explanation:

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:

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 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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Answer:

100rad because it angular velocity

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 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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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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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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A bottle of water weighs 25 pounds. If the bottle weighs 5 pounds, then it contains  0.599132136584 gal .

A bottle weighs 25 pounds ,

we know that 1 Gallon = 8.34 Lb

so, 25 lb = 2.995660682922 gal

and if the bottle weighs 5 pounds then ,

5 lb = 0.599132136584 gal

hence, 2.4 gal less.

Define pound.

a. a unit of weight equivalent to l6 ounces avoirdupois (453.59237 grams), the fundamental unit of weight in the FPS system;

b. a unit of weight equal to 12 ounces troy or 12 ounces apothecaries' (373.2418 grams).

It is an imperial unit of mass or weight measurement.

Define gallons.

In both imperial and US customary units, the gallon is a unit of volume.

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Answer:

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

Explanation:

hope it helps ya

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

The energy becomes 0.50 times in 6.72 s.

Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.

Et=0.50E

1

k(4₂)² = (0.5) - kA²

(4₂)² = (0.5) A²

At = 0.71A

So, the amplitude of the oscillator becomes 0.71 times its initial ar

0.71A = = A(0.96)¹2

log(0.71)

log(0.96)

8.4

n=

So, the time taken for n oscillation is obtained as,

t = n (0.800 s)

= (8.4) (0.800)

= 6.72 s

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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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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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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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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:

150 J

Explanation:

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

The formular for calculating half life is:

T1/2=0.693/Π

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:

lowering the ambient temperature causes chemical reaction to proceed more slowly,so a battery used in a low temperature produces less current than that at high temperature . As cold batteries run down they quickly reach the point where they cannot deliver enough current to keep up the demand.. hence discharge more quickly in cold weather..

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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