the meaninn of dynamics
​

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

Answer:

No you can't cuz,if you put water instead of clock oil in Millikan oil drop your experiment will fail and it won't turn out the way you wanted it to be

Answer:

150 J

Explanation:

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

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 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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The distance of separation between the lenses is determined as 12 cm.

The distance of the image formed by the first lens is calculated as follows;

1/d₁ = 1/f - 1/di₀

1/di₁ = 1/4 - 1/12

1/di₁ = 2/12

1/di₁ = 1/6

di₁ = 6 cm

Use magnification formula as shown below;

M = (di₁di₂)/(d₀₁d₀₂)

1 = (di₁di₂)/(d₀₁d₀₂)

di₂ = (d₀₁d₀₂)/(di₁)

di₂ = (12 x d₀₂)/(6)

di₂ = 2d₀₂

1/f = 1/ d₀₂ + 1/di₂

1/f =  1/ d₀₂ + 1/2d₀₂

1/f = 1.5/d₀₂

d₀₂ = 1.5f

d₀₂ = 1.5(4 cm) = 6 cm

D = di₁ + d₀₂

D = 6 cm + 6 cm

D = 12 cm

Thus, the distance of separation between the lenses is determined as 12 cm.

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The amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.

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.

From the graph, the amplitude of the red colored wave is 1 unit.

From the graph, the amplitude of the red colored wave is 2.1 unit.

Thus, the amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.

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(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:

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..

Explanation:

getting me up to date with recent mission development

The maximum rate at which energy can be added to the circuit element mathematically given as

[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]

Generally, the equation for P is  mathematically given as

[tex]P=\ln s \frac{\Delta T}{\Delta t}[/tex]

Therefore

[tex]Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}[/tex]

[tex]\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}[/tex]

[tex]\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}[/tex]

Max temp Change

[tex]MaxT=5.6^{\circ} \mathrm{C}[/tex]

[tex]\text { time }=3 \times 60[/tex]

t=180s

In conclusion, Max Energy Rate

[tex]MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}[/tex]

[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]

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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?

The formular for calculating half life is:

T1/2=0.693/Π

Answer:

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

Explanation:

hope it helps ya

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 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 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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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 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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Angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.

The principle of conservation of angular momentum states that the total angular momentum acting on an object is constant, provided there is no  external torque acting on the object.

Angular momentum of a system is conserved as long as there is no net external torque acting on the system.

Thus, angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.

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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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Answer:the

8/9 h

Explanation:

Height  =   1/2 a T^2     now change to T/3

 now height = 1/2 a (T/3)^2   = 1/9  1/2 a T^2     <===== it is 1/9 of the way down   or   8/9 h

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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Answer:

100rad because it angular velocity

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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The hypothesis will be:

H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.

H₁ =  The amount of salt added to ice will affect the rate at which the ice will melt.

The independent variable which is can be changed by increasing the rate of salt added to the equation.

The dependent variable which is ice will change or melt in response as it will decrease if the rate of the salt added increases.

Salt does not really lower the temperature of an ice cubes, it is known to just lowers their freezing point, that is lowers their melting point.

Note that if salt is around, ice cubes are known to be colder to be solid, and they tend to melt at a temperature that is said to be lower than the freezing point of pure water.

If  the ionic compound salt is known to be added, it tends to lowers the freezing point of the water, which implies that the ice on the ground is not able to freeze that layer of water at all.

Therefore, The hypothesis will be:

H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.

H₁ =  The amount of salt added to ice will affect the rate at which the ice will melt.

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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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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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Answer:

44.7 seconds

Explanation:

D = 500 m

   500 = .25  t^2

    500/.25 = t^2

          2000 = t^2

                t = 44.7 seconds