How can you drop two eggs the fewest amount of times, without them breaking?

The solution for the acceleration of gravity is given as

This is further explained below.

Generally,

Mass of earth [tex]$M=5.97 \times 10^{24} \mathrm{~kg}$[/tex]

Radius of earth [tex]$R=6371 \mathrm{~km}$[/tex]

Gravitational const. [tex]$G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$[/tex]

height [tex]$h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$[/tex]

[tex]R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}[/tex]

In conclusion, acceleration due to gravity at this point will be

[tex]g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex]

for [tex]$h_{2}=7900 \mathrm{~km}$[/tex]

[tex]R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]

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The final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.

In an elastic collision, both momentum and energy are conserved. But only momentum is conserved in inelastic collision.

Given that two hockey pucks are moving towards each other. The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path.

The given parameters are;

The mathematical representation of the above question will be in two components.

Horizontal component

M1U1 - M2U2 = M1V1cosФ - M2V2cosФ

Substitute all the parameters

0.13 x 1.11 - 0.16 x 1.21 = 0.13 x V1 cosФ - 0.16 x 1.16cos42

0.1443 - 0.1936 = 0.13V1cosФ - 0.1379

0.13V1cosФ = 0.0886

V1cosФ = 0.0886/0.13

V1cosФ = 0.6815 ........ (1)

Vertical component

0 = M1V1sinФ - M2V2sinФ

M1V1sinФ = M2V2sinФ

Substitute all the parameters

0.13 x V1 sinФ = 0.16 x 1.16sin42

V1 sinФ = 0.1242/0.13

V1 sinФ = 0.9553 ......... (2)

Divide equation 2 by 1

V1 sinФ / V1 cosФ = 0.9553/  0.6815

Tan Ф = 1.40

Ф = [tex]Tan^{-1}[/tex](1.4)

Ф = 54.5°

Substitute Ф into equation 2

V1 sin54.5 = 0.9553

V1 = 0.9553 / 0.8141

V1 = 1.17 m/s

Therefore, the final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.

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The position of the object is = -68cm

The magnification of the mirror= 0.3

The image distance = 20.5cm

The focal length= R/2 = 31.5/2= 15.75

The object distance= ?

Using the lens formula,1/f = 1/v-1/u

1/u = 1/v- 1/f

1/u = 1/20.5 - 1/15.75

1/u = 0.0489- 0.0635

1/u = -0.0146

u = -68cm

The magnification of the mirror is image size/object size

= 20.5cm/-68cm

= 0.3

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Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

[tex]\Sigma \tau = 0[/tex]

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
[tex]\tau = r \times F[/tex]

Doing the summation using their respective lever arms:

[tex]0 = L Tsin\theta - dF_g[/tex]

[tex]dF_g = LTsin\theta[/tex]

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

[tex]tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o[/tex]

Now, let's solve for 'T'.

[tex]T = \frac{dMg}{Lsin\theta}[/tex]

Plugging in the values:
[tex]T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}[/tex]

5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .

The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.

Wavelength of green light = 520 nm

f = c / λ

where, f = Frequency

            c = Speed of light = 3 × [tex]10^8[/tex] m/s

            λ = Wavelength of light

∴ f = c / λ

  f = [tex]\frac{3*10^8}{520 * 10^-^9}[/tex]

    = 5.77 ×[tex]10^1^4[/tex] Hz

Therefore,  5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .

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The next time the string will have the same appearance that it did at t=0s is 2.29 s.

v = fλ

f = v/λ

where;

half of the upward pulse is a quarter of wavelength = ¹/₄ x 4 m = 1 m

f = 3.5/1

f = 3.5 Hz

t1 = 4/3.5 = 1.143 s

The next time the string will have the same appearance that it did at t=0s.

d = 4 m x 2 = 8 m

t2 = 8/3.5

t2 = 2.29 s

Thus, the next time the string will have the same appearance that it did at t=0s is 2.29 s.

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The speed of a satellite in a circular orbit around the Earth is 4,188 m/s.

The speed of the satellite is calculated as follows;

v = √GM/r

where;

v = √[(6.67 x 10⁻¹¹ x 5.98 x 10²⁴) / (3.57 x 6.37 x 10⁶)]

v = 4,188 m/s

Thus, the speed of a satellite in a circular orbit around the Earth is 4,188 m/s.

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The tangential speed of the wheel is determined as 4.786 m/s.

The tangential speed of the wheel is calculated as follows;

v = ωr

where;

v = (2.17 x 2π rad)/s x 0.351 m

v = 4.786 m/s

Thus, the tangential speed of the wheel is determined as 4.786 m/s.

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The buoyant force acting on the cylinder is, [tex]Fb = \rho Ahg[/tex]. Here A is the cross-sectional area of the cylinder, h is the height of the cylinder, ρ is the density of the cylinder, and g is the acceleration due to gravity.

This buoyant force is also equal to the volume of the fluid displaced. [tex]Fb = \sigma h(A-x)g[/tex]. Here σ is the density of the fluid.

Equate the above two equations and solve for x.

[tex]\rho Ahg = \sigma A(h-x)g\\\rho h = \sigma h - \sigma x\\x = \frac{(\sigma - \rho)h}{\sigma}[/tex]

So, the distance x depends on the density of the fluid, density of the cylinder and the height of the cylinder.

1. The density of the cylinder is same and distance x is independent of the diameter of the cylinder. Therefor, there will be no change in the distance x. Hence, the correct answer is No change.

2. Now the height is changing keeping the density same. As the distance x is directly proportional to the height, the distance x will increase.

3. The density of the added liquid is greater than of the water and it does not mix with the water. So, the liquid will settle down and there will be no change in the distance x.

4. The density of the added liquid is less than that of the water and it does not mix with the water. So, the liquid will not settle down and the distance x will change. The change in distance x can be determined as follow:

[tex]\rho Agh = \sigma' Axg + \sigma A(h-x)g\\\rho h=\sigma' x + \sigma h - \sigma x\\x=(\frac{\sigma - \rho}{\sigma - \sigma'})h[/tex]

Here, σ' is the density of the added liquid.

From the above relation it is clear, that on adding the liquid of the density less than that of water, the denominator term become small ando so the value of x will increase.  

5. On removing some of the water inside the glass, the height of the water column will decrease, but the value of x does not depend on the height of the water column. So, there will be no change in the distance x.

6. The density of the new cylinder is smaller than that of the earlier one. So, the numerator term will increase. Therefore, the value of x will increase.

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The highest point of a roller coaster is almost always the first hill. In the majority of roller coasters, the hills get smaller as the train travels down the track.

To find the answer, we have to know more about the mechanical energy of a system.

                  Mechanical energy = U = mgh

Where m represents the car mass, g represents gravity, and h represents height

Finally, by applying the principle of energy conservation, we may determine that, the initial hill must be the highest.

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1. The speed with which the ball hits the ground is 17.1 m/s

2. The magnitude of the average force of air resistance exerted on it is 0.77 N

v² = u² + 2gh

v² = 2gh

Take the square root of both side

v = √(2 × 9.8 × 15)

v = 17.1 m/s

We'll begin by calculating the time to reach the ground. This is illustrated below:

h = ½gt²

15 = ½ × 9.8 × t²

15 = 4.9 × t²

Divide both side by 4.9

t² = 15 / 4.9

Take the square root of both side

t = √(15 / 4.9)

t = 1.75 s

Now we can determine the force. This can be obtained as illustrated below:

F = m(v –u) / t

F = 0.149(9 – 0) / 1.75

F = 0.77 N

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The escape velocity from the surface of the planet X is 2,249.2 m/s.

[tex]v = \sqrt{\frac{2GM}{r} } \\\\v^2 = \frac{2GM}{r}\\\\v^2r = 2GM\\\\G = \frac{v^2r}{2M}[/tex]

where;

[tex]\frac{v_x^2 \ r_x}{2M_x} = \frac{v_y^2 \ r_y}{2M_y} \\\\\frac{v_x^2 \ r_x}{M_x} = \frac{v_y^2 \ r_y}{M_y} \\\\v_x^2 = \frac{v_y^2 \ r_yM_x}{M_yr_x}\\\\v_x^2 = \frac{(1695)^2 (r_y)(1.59M_y)}{M_y(0.903r_y)} \\\\v_x^2 = 5,058,814.78\\\\v_x = \sqrt{5,058,814.78} \ \ = 2,249.2 \ m/s[/tex]

Thus, the escape velocity from the surface of the planet X is 2,249.2 m/s.

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If you speed through a construction zone while workers are present, your fines could be as much as one thousand dollars.

This is referred to as the amount which is instructed by a court or authority to be paid as a result of it being a penalty for various types of offences. each crime has its specific fine which helps to serve as deterrent for unlawful behavior in the community.

it is always best not to speed when within a construction zone which has workers present in the area. This is ideal as it helps to prevent incidences of accidents or death.

it is therefore the reason why a fine of 1000 dollars is to be paid so that people can think of such high amount before performing certain types of activities when driving and makes it the most appropriate choice.

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The tension in the cable is equal to 323.5 N.

The tension, T in the cable is determined by taking moments about the pivot  marked X.

The angles of the boom and the cable with the horizontal are first calculated.

Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°

The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80

Taking moments about the pivot:

175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1

Tension = 241.68/0.747

Tension = 323.5 N

In conclusion, the tension in the cable helps to suspend the crate.

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The resultant force is 8N  

Given that mass is 2kg , v= 40m/s, u =20m/s and we need to calculate resultant force
F=ma

m is given
so for a
v-u/t=a { first equation of motion }

40-20/4= 4
so a=4

F = ma =2*4 = 8N
The difference between the forces that are acting on an object as part of a system is known as the resultant force.
v = u + at is the first equation of motion. Here, v denotes the end speed, u the starting speed, an acceleration, and t the passage of time. The first equation of motion is provided by the velocity-time relation, which may be used to calculate acceleration.

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The altitude or height of the pole vaulter as she crosses the bar is 4.04 m.

The height of the pole vaulter is determined from the change in kinetic energy which is equal to the potential energy at that height.

h = (v - u)²/2g

h = (10 - 1.1)²/2 * 9.8

h = 4.04 m.

In conclusion, the height is determined from the potential energy at that height.

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The orbiting velocity of the satellite is 4.2km/s.

To find the answer, we need to know about the orbital velocity of a satellite.

=4.2km/s

Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.

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(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

(b) The net work done on the crate while it is on the rough surface is 23.7 J.

(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

F(net) = F - μFf

F(net) = 280 - 0.351(92 x 9.8)

F(net) = -36.46 N

W = F(net) x L

W = -36.46 x 0.65

W = - 23.7 J

a = F(net)/m

a = -36.46/92

a = - 0.396 m/s²

v² = u² + 2as

v² = 0.845² + 2(-0.396)(0.65)

v² = 0.199

v = √0.199

v = 0.45 m/s

Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

The net work done on the crate while it is on the rough surface is 23.7 J.

The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

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Answer:

the branch of science that is concerned with nature and properties of matter and energy.

Explanation:

a study of the basis of what does what in science.

The kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.

To find the answer, we need to know about the concept of collision and kinetic energy.

                 [tex]TE=KE+PE\\T=PE=m_1gh=(0.4*9.8*20)=78.4 J[/tex]

                   [tex]TE=KE=78.4J[/tex]

                          [tex]U'=m_1gh'=mg\frac{h}{2} \\U'=39.2J[/tex]

                      [tex]TE=78.4 J\\KE'+PE'=78.4J\\KE'=78-U'=78.4-39.2=39.2J[/tex]

                            [tex]KE=\frac{1}{2} mv^2=78.4J\\v=\sqrt{\frac{2KE}{m} } =4m/s[/tex]

               [tex]\frac{1}{2}m_1u_1^2+ \frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]

                 [tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

                            [tex]m_1=m_2=m\\u_1=4m/s\\u_2=0\\v_1=?\\v_2=?[/tex]

                       [tex]\frac{1}{2}m*4^2=\frac{1}{2}m(v_1^2+v_2^2)\\(v_1^2+v_2^2)=16[/tex]  from resolving KE equation.

                        [tex]4m=m(v_2+v_1)\\4=v_2+v_1\\v_1=4-v_2[/tex] From resolving momentum conservation.

                            [tex]v_2=4m/s\\v_1=0[/tex]

Thus, we can conclude that, the kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.

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The initial speed of the rock as it left the astronaut's hand is 168 m/s.

The speed of the satellite is 60.2 m/s.

g = GM/R²

where;

g = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(6.32 x 10⁴)²

g = 0.132 m/s²

v² = u² - 2gh

0 = u² - 2gh

u² = 2gh

u = √2gh

u = √(2 x 9.8 x 1440)

u = 168 m/s

v = √GM/r

v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(1.45 x 10⁵)]

v = 60.2  m/s

Thus, the initial speed of the rock as it left the astronaut's hand is 168 m/s.

The speed of the satellite is 60.2 m/s.

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The answer is A. Decreasing the distance between the particles by a factor of 3.

The Universal Law of Gravitation is :

F = Gm₁m₂ / r² (where 'r' is the distance between them)

Since force is inversely proportional to the square of the distance between them, distance has to be decreased by a factor of 3 to increase the force to 9 times the original force.

Answer:

Explanation:

2.1 x 10^2 - 20J

From the calculation, the normal force is 6161.2 N.

The normal force is given by the expression;

N - mg = ma

Then;

N = mg + ma

m = 84.4 kg

g = 9.8 m/s^2

a = 63.2 m/s2

Now we have;

N = m(g + a)

N = 84.4 (9.8 + 63.2)

N = 6161.2 N

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Force necessary to support the object on piston 2 is 24× 10⁴ N.

To find the answer, we need to know about the force and pressure on piston 1 and piston 2.

= 991 Kg × 9.8 = 9414.5N

= 9414.5N × π(9.46² cm² /4)

= 9414.5N × π(9.46²× 10^(-4)m²/4)

= 66.2 N/m²

= 66.2/ π(1.87² cm² /4)

= 66.2/ π(1.87²×10^(-4)m² /4)

= 24× 10⁴ N

Thus, we can conclude that force necessary to support the object on piston 2 is 24× 10⁴ N.

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Answer: The mass of the object is 25kg.

The given question deals with Newton's second law of motion and its applications.

Explanation: Given force, F=500N

                                 acceleration, a=20 m/[tex]s^{2}[/tex]

  From Newton's 2nd law of motion , we have

                             F=ma where m=mass of the object

                         ⇒500=m×20

                         ⇒m=500/20=25

                 ∴ Mass of the object is 25 kg .

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The image is present at 20cm from the crown glass spherical surface.

To find the answer, we need to know about the lens formula.

=> (1/V)+5=10

=> 1/V= 5

=> V=0.2m = 20cm

Thus, we can conclude that the image is present at 20cm.

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a) Her feet are in the air for 0.73+0.41 = 1.14 seconds

b) Her highest height above the board is 0.82 m

c) Her velocity when her feet hit the water is 7.16 m/s

t = Time taken

u = Initial velocity = 4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

a) Her feet are in the air for 0.73+0.41 = 1.14 seconds

s = ut + [tex]\frac{1}{2}[/tex]at²

2.62 = 0t + [tex]\frac{1}{2}[/tex] ₓ 9.8  ₓ [tex]t^{2}[/tex]

t = 0.73 s

b) Her highest height above the board is 0.82 m

The total height she would fall is 0.82+1.8 = 2.62 m

v = u + at

0 = 4 ₋ 9.8 ₓ t

t = 0.41 s

s = ut +[tex]\frac{1}{2}[/tex] at²

s = 4 ₓ 0.41 ₊ [tex]\frac{1}{2}[/tex] ₓ ₋9.8 ₓ 0.41 [tex]t^{2}[/tex]

c) Her velocity when her feet hit the water is 7.16 m/s

[tex]v = u + at \\v = 0 + 9.8[/tex] ₓ [tex]0.73[/tex]

v = 7.16 m/s

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A grandfather paradox is a situation where individual travels to the past and then introduces a change which affects or contradicts the present.

A paradox is a situation or statement which involves two contradictions.

A grandfather paradox is a situation which is defined by the ability of an individual to travel to a time in the past usually before the birth of their grandfather and then introduces a change which affects or contradicts the present. For example, killing the grandfather to prevent their birth.

In conclusion, a grandfather paradox is is an event which contradicts the present as a result of a change done to the past.

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The change in momentum in time interval, given the data will be F × Δt

Momentum is defined as the product of mass and velocity. It is expressed as

Momentum = mass × velocity

This is defined as the change in momentum of an object.

Impulse = change in momentum

But

Impulse = force × time

Therefore

Force × time = change in momentum

Force × time = change in momentum

F × Δt = change in momentum

Change in momentum = F × Δt

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