deriving projectile motion formulas

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 lowest carrier frequency for which each component of the modulated wave S(t) is uniquely determined by m(t) is

"-w,w -1kHz,1kHz" are the frequency components of the detector output in this case.

This is further explained below.

Generally, the equation for the modulated wave containing frequency components  is  mathematically given as

[tex]Fc,Fc+w,Fc-W = > 1.5k,2.5k,0.5kHz[/tex]

[tex]-Fc,-Fc+W,-Fc-wW = > -1.5k,-0.5k,-2.5kHz[/tex]

"-w,w -1kHz,1kHz" are the frequency components of the detector output in this case.

b)

In conclusion,, then the modulated wave contains frequency components as

[tex]Fc,Fc+W,Fc-W = > 0.75k,1.75k,-0.25kHz[/tex]

[tex]-Fc,-Fc+w,-Fc-W = > -0.75k,0.25k,-1.75kHz[/tex]

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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.

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 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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Answer: The speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth is 4188.11 m/s.

Explanation: To find the answer, we need to know about the equation of motion of a satellite around earth.

                 [tex]F_g=\frac{GMm}{r^2}[/tex]

                 [tex]F_c=\frac{mv^2}{r} \\[/tex]

                    [tex]\frac{GMm}{r^2} =\frac{mv^2}{r} \\thus,\\v=\sqrt{\frac{GM}{r} }[/tex]

                  [tex]r=3.57 R_E=3.57*6.37*10^3=22.74*10^3 km\\M=5.98*10^24kg\\G=6.67*10^{-11}Nm^2/kg^2[/tex]

                  [tex]v=\sqrt{\frac{6.67*10^{-11}*5.98*10^{24}}{22.74*10^6m} } =4188.11 m/s[/tex]

Thus, we can conclude that the speed of satellite will be 4188.11 m/s.

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In a circular orbit with a radius that is 3.57 times the mean radius of the Earth, a satellite moves at a speed of 132.43km/s.

In order to get the solution, we must understand the satellite's planetary motion equation.

                          [tex]F_g=\frac{GMm}{r^2}[/tex]

                         [tex]F_c=\frac{MV^2}{r}[/tex]

                          [tex]\frac{GMm}{r^2}=\frac{mV^2}{r}\\V=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10{-11}*5.98*10^{24}}{22.74*10^3} } \\V=132.43*10^3m/s[/tex]

As a result, we may estimate that the satellite will move at a speed of 132.43 km/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 type of lens involved is a convex lens and the object is positioned at the center of the curvature of the convex lens.

According to the question, a real, inverted, and same-sized image of the object is formed.

A concave lens is a diverging lens and always forms virtual images. But convex lenses are converging lenses and are the only type of lens that produce real and inverted images of the corresponding objects.

When an object is placed at the center of the curvature of a convex lens, its corresponding image is formed on the opposite side of the convex lens. The image formed is real and inverted.

The distance of the image from the lens is equal to the distance of the object from the lens.

For a convex lens, the distance of the center of curvature from the lens is double the focal length of the lens.

That's why the convex lens and center of curvature are the correct answer to this question.

Explanation:

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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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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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 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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The acceleration due to gravity is 3 m/s^2 and the work done is -4.3 J.

Now we must use the formula;

h = ut + 1/2gt^2

Since it was dropped from a height u = 0 m/s

h = height

u = initial velocity

g = acceleration due to gravity

t = time

h = 1/2gt^2

g = 2h/t^2

g = 2 * 2.85 /(1.38)^2

g = 5.7/1.9

g = 3 m/s^2

The work that must be done is against gravity hence;

W = -(mgh)

W = - (0.5 kg *  3 m/s^2 * 2.85 m)

W = - 4.3 J

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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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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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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]

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 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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(a) The magnetic field the line produced at ground level is 3.75 x 10⁻⁶ T.

(b) The lines magnetic field as a percent of the earth's magnetic field is 7.5%.

(c) No, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.

B = (μ x I) / (2πr)

B = (4π x 10⁻⁷ x  150) / (2π x 8)

B = 3.75 x 10⁻⁶ T

x = 3.75 x 10⁻⁶ /0.5G

x = (3.75 x 10⁻⁶ ) / (0.5 x 10⁻⁴)

x = 0.075

x = 0.075 x 100% = 7.5%

Thus, we can conclude that, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.

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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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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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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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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 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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Answer:

Explanation:

2.1 x 10^2 - 20J

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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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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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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323.5 N is the tension in the cable.

Given

Mass of crate(M) = 175.5 kg

Mass of boom(m) = 94.7 kg

The tension, T in the cable can be calculated by taking moments of force about the central point of marked X.

The Angle of the boom with the horizontal can be calculated by

tanθ = 5/10

θ = tan⁻¹(5/10) = 26.56°

Angle of the boom with horizontal is 26.56°

The angle of cable with horizontal can be calculated by

tan B = 4/10

B = tan⁻¹(4/10) = 21.80°

Angle of cable with horizontal is 21.80°

Taking moments of force about the point X

(Mcosθ + mcosθ) 0.5 = T(sin(θ +B)1

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

By calculating, we get

Tension(T) = 241.68/0.747

Tension(T) = 323.5 N

Hence, 323.5 N is the tension in the cable.

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