Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?

3.00 m is the magnitude of the resultant vector

137.1° is the directional angle of the resultant vector.

1) To find the magnitude of the resultant

Discover each vector's individual components first. 

We must consider the signs of the components because the angles in the figure are measured in various ways. 

In this case, both the y component of vector C and the x component of vector A are negative. 

The vectors' components are as follows, using a little trigonometry:  

Magnitude of A = 5 m

[tex]A_{x}[/tex] = - (5.00m) cos20° = -5 ×0.408 = -4.698 m

[tex]A_{y}[/tex] = + (5.00m)sin20° = +5 × 0.342 = +1.710 m

Magnitude of B = 5m

[tex]B_{x}[/tex] = +(5.00m)cos60° = 5 × 0.5 = +2.5m

[tex]B_{y}[/tex] = +(5.00m)sin60° = 5 × √3/2 = +4.33 m

Cx = 0

Cy = -4.00m

The sum of all three vectors, which we refer to as R, produces components.

Rₓ = Aₓ + Bₓ + Cₓ

   = -4.698 + 2.5 + 0

   = -2.198 m

[tex]R_{y}[/tex] = [tex]A_{y} + B_{y} + C_{y}[/tex]

    = +1.71 +4.33 - 4.00

    = 2.040 m

R = [tex]\sqrt{R_{x^{2} }+R_{y^{2} }[/tex]

  = [tex]\sqrt{(-2.198)^{2 } + (2.040)^{2} }[/tex]

  = 3.00 m

2) To find the directional angle of resultant

tan θ = 2.040/-2.198 = -0.928

θ = -42.9°

Such a vector would be in the so-called "fourth quadrant," as it is well known. However, we discovered that R has a negative x component and a positive y component, indicating that such a vector must reside in the "second quadrant.

The calculator accidentally returned an angle that is 180 degrees off, thus we must add 180 degrees to the naïve angle in order to get the right angle.

Therefore, R's actual direction is determined by-

θ = -42.9° + 180° = 137.1°

Hence, the magnitude of resultant vector is 3.00 m and directional angle is 137.1°

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(a) The time the baseball spends in the air is 0.92 s.

(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.

h = vt - ¹/₂gt²

-2.1 = (4.05 x sin 34)t  - ¹/₂(9.8)(t²)

-2.1 = 2.26t - 4.9t²

4.9t² - 2.26t - 2.1 = 0

t = 0.92 s

R = Vx(t)

R = (4.05 x cos 34)(0.92)

R = 3.1 m

Thus, the time the baseball spends in the air is 0.92 s.

The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.

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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 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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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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Answer:

unfournatletly

Explanation:

i have no clue sorry to waste ur time ill rather not say a answer that will be incorrect.

Answer:

Pluto

Explanation:

Three moons can fit inside the volume of the sun.

The moon is a non luminous body found in the space. It could cause a solar eclipse when it comes between the sun and the earth.

Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;

8,000 miles/ 2,000 miles = 3

Hence, three moons can fit inside the volume of the sun.

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The speed of the space craft relative to the earth is given as: 0.024c. This is solved using the the equation for time dilation.

Time dilation is the "slowing down" of a clock as determined by an observer in relative motion with regard to that clock under the theory of special relativity.

The formula is given as :

Δt = [Δr]/ √ 1 - (v²/c²)

Thus,

v = c√1 - (Δr/Δt)²

= c √(1 - (3600/3601)²

v = 0.024c

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Sound will travel fastest in air at 15°C.

The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;

Air at 100°C 387 m/s

Air at 0°C 331 m/s

From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.

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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 velocity with which the particle Q is fired is 15m/s upwards.

The vector quantity velocity (v), represented by the equation v = s/t, quantifies displacement (or change in position, s) over the change in time (t). Speed (or rate, r) is a scalar number, denoted by the equation r = d/t, that quantifies the distance traveled (d) over the change in time (t).

Assume the upward direction is positive and the downward direction is negative.

Velocity P, = 40m/s

Distance traveled by P,

Using the first equation of motion for particle P,

v = u + at

⇒ 0 = 40 + (-10)t

⇒ t = 4s

This is the time it takes for P to rise

Now, the maximum height(s) reached by the particle P is,

Using the second equation of motion,

s = ut + 1/2at²

⇒ s = 40×4 + 1/2 × (-10) × 4²

⇒ s = 80m

a) A particle P falls when Q is shot after 4 seconds from the initial time:

V² = U² + 2aH₁

⇒ H₁ = 15² - 0/ 2(-10)

⇒ H₁ = 11.25m

Particles P and Q meet at a distance from the ground (H₂).

Height, H₂ = s - H

⇒ H₂ = 80 - 11.25

         = 68.75m

Particles P and Q meet at a distance of 68.75m from the ground.

v = u + at₁

⇒ 15 = 0 + (-10) t₁

⇒ t₁ = 1.5s

It is equal to P's fall time and Q's rise time.

b) For particle Q

H₂ = u₂t₁ + 1/2at₁

⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5

⇒ u₂ = 15 m/s

Therefore,

The velocity with which the particle Q is fired is 15m/s upwards.

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The magnitude of the force on the left-hand pole is mathematically given as

f'=0.167N

Q=45 degrees

Generally, the equation for Force is  mathematically given as

F=mg/sinФ

Therefore

F(17.1*10^{-3})*9.8/sin 45

F=0.237N

Considering horizontal axis or plane

f'-fcos=0

Therefore

f'=0.237*cos45

f'=0.167N

In conclusion, the slope

tanФ=30/30

tanФ=1

Ф=45

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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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Solutions for the three problems are is mathematically given as

(a) First cosmic speed (arbitral velocily)

[tex]v=\sqrt{\frac{G M}{r}}\\v=\sqrt{\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24}}{5870 \times 10^{3}}}[/tex]

v=7338.9349

(b) Second cosmic speed (escape velo.)

$$

\begin{gathered}

[tex]V=\sqrt{\frac{2 G M}{r}}\\\\V=\sqrt{2} \sqrt{\frac{G M}{r}}\\\\=V\sqrt{2} \times 7338.9349 \\[/tex]

[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex]

(c) In conclusion, in a circular orbit, the gravitational force is gets balanced by centripetal force

[tex]&m_{q \omega \omega^{2}}=\frac{G M M}{r^{2}} \\&r^{3}=\frac{G M}{\omega^{2}}=\frac{G M}{4 \pi^{2}} T^{2} \\&r^{3}=\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24} \times(16.6 \times 3600)^{7}}{4 \pi^{2}} \\[/tex]

[tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]

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a)New position vector in vector form= r = 0.233404 î + 2.94056j m

b) Lies in second quadrant at 161.391°

c)Velocity =4.8675 m/s

d)It is moving in a direction making 161.391° with positive x-direction.

e)Acceleration will be centripetal acceleration (8.031 m/s²).

Given:

Mass of the object m = 3.95 kg

ω=1.65 rad/s

Radius of the circle = 2.95 m

a)

new position vector in vector form

=R cos1.65 î  + R sin 1.65 j

= 2.95 cos1.65 î  +2.95 sin1.65 j

= 2.95 x 0.07912 î + 2.95 x 0.9968 j

r = 0.233404 î + 2.94056j

b)

Angular Displacement = θ₀ = 9.10 rad

9.10 radian = 180/π× 9/10 degree

= 521.391°

=521.391°- 360°

=161.391°

This will lie in second quadrant.

Angle made with positive x-axis

=161.391°

c)

Velocity

v = ω R

= 1.65 x 2.95

=4.8675 m/s

d)

It is moving in a direction making 161.391° with positive x-direction.

e)

Acceleration will be centripetal acceleration.

= v²/R

=(4.8675)² / 2.95

=23.6925562 / 2.95

=8.031 m/s²

f) Position, Velocity and Acceleration graph:

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a) New position vector is r = 0.233404 î + 2.94056j m

b) Lies in second quadrant at 161.391°

c) Velocity =4.8675 m/s

d) It is moving in a direction making 161.391° with positive x-direction.

e) Acceleration will be centripetal acceleration (8.031 m/s²).

Given:

Mass of the object m = 3.95 kg

ω=1.65 rad/s

The radius of the circle = 2.95 m

a) new position vector in vector form

 r =R cos1.65 î  + R sin 1.65 j

 r = 2.95 cos1.65 î  +2.95 sin1.65 j

 r = 2.95 x 0.07912 î + 2.95 x 0.9968 j

 r = 0.233404 î + 2.94056j

b) Angular Displacement = θ₀ = 9.10 rad

9.10 radian = 180/π× 9/10 degree

 = 521.391°

 = 521.391°- 360°

 =161.391°

This will lie in the second quadrant.

Angle made with the positive x-axis =161.391°

c) Velocity

   v = ω R

   v = 1.65 x 2.95

   v = 4.8675 m/s

d) It is moving in a direction making 161.391° with positive x-direction.

e) Acceleration will be centripetal acceleration.= v²/R

   =(4.8675)² / 2.95

   =23.6925562 / 2.95

   =8.031 m/s²

f) Position, Velocity, and Acceleration graph:

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Answer:

It will still hover until the spaceship "hits" or exerts a force on it.

Explanation:

Remember, if there is no net force, there is no acceleration or movement.

In this case, our ball is hovering in the spaceship, and in space, we can assume there is no [tex]F_g[/tex], and we can assume there is no [tex]F_N[/tex], nor no forces acting against it.

So, the ball would not move.

However, once the spaceship starts accelerating, the ball would still hover until the spaceship exerts a force on it.

This is because of the same thing as explained above, no forces acting on it, therefore, no acceleration.

Think about it this way.

Imagine you jumped up, then someone threw a ball at you. Now let's imagine you can't move until you hit the floor, meaning that in an ideal situation only  [tex]F_g[/tex] is acting on you. Now again, let's imagine time slows really down for you, but not the ball. Before the ball comes and hits you, you are "hovering" like a ball. But after the ball hits you, you move a little because the ball exerted a force on you.

If you did not understand what I meant above, just forget about it, and think about the fact that if there is a Net force (all the force values added up), then there is acceleration and movement.

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:

(a) The electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

(b) The total flux is 1.24 Vm.

(c) The total amount of charge that is inside the box is  1.1 x 10⁻¹¹ C.

The area of the left face is calculated as follows;

A1 = 0.03 m x 0.02 m = 0.0006 m²

Ф1 = EA1

Ф1 = (400 V/m)( 0.0006 m²) = 0.24 Vm

Let the dimension of the right face = 5 cm by 2 cm

A2 = 0.05 m x 0.02 m = 0.001 m²

Ф2 = EA2

Ф2 = (1000 V/m)( 0.001 m²) = 1 Vm

Ф = Ф1 + Ф2

Ф = 0.24 Vm + 1 Vm = 1.24 Vm

Ф = Q/ε

Q = εФ

Q = (8.85 x 10⁻¹²)(1.24)

Q = 1.1 x 10⁻¹¹ C

Thus, the electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

The total flux is 1.24 Vm and the total amount of charge that is inside the box is  1.1 x 10⁻¹¹ C.

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Answer:

D

Explanation:

momentum can be negative as it's a vector quantity. positive momentum indicates that the object is traveling in the positive direction as defined by the coordinate system. negative momentum indicates that the object is moving in the opposite direction (backwards). the momentum equation itself is p = mass x velocity.

and only if you consider velocity a a directed vector, can you have a negative momentum. this equation shows that the momentum is in the same direction as velocity.

so, the sign is only for the direction.

the force is the same (going with the absolute value). a negative momentum is not smaller than a positive momentum.

It is possible that momentum can be negative if an object is moving backward. The correct option is D.

A vector quantity called momentum depends on an object's mass and velocity. It is described as the result of the mass and the velocity of an object.

When an item is moving against a selected positive direction, its velocity will be negative since velocity is a vector variable that comprises both magnitude and direction. As a result, the negative velocity will yield a negative value for momentum when it is calculated.

Thus the correct option is D.

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Acceleration is the rate of change of velocity. Usually, acceleration means the speed is transforming, but not always. When an object moves in a circular path at a steady speed, it is still accelerating, because the focus of its velocity is changing.

a = 4,552 m / s²,   b)  a = 2,588 m / s²

Newton's second law is

F = ma

a = F / m

in this case the force remains constant

indicate us

* for a mass m₁

a₁ = F/m₁

a₁ = 12, m/ s²

* for a mass m₂

a₂= 3.3 m / s²

acceleration  m = m₂-m₁

substitute

[tex]$\begin{aligned}&a=\frac{F}{m_{2}-m_{1}} \\&1 / a=\frac{m_{2}}{F}-\frac{m_{1}}{F}\end{aligned}$[/tex]

let's calculate

1/a=1/3.3 - 1/12  = 0.21969

 a = 4,552 m / s²

a= 4,552m/s²

Speed ​​estimates the rate of movement of an object, that is, the distance traveled per unit of time. Acceleration calculates the rate of change of velocity, that is, the change in velocity between two different moments

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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 application of total Internal Reflection occurs in a mirage. Option A

1) We know that a charge can be positive or negative. If the charge is negative, we call it an electron. If the charge is positive, we call it a proton. Now we know that the magnitude of charge on each electron is 1.6×10-19 C. Hence, the magnitude of charge on five electrons = 5 * 1.6×10-19 C = 8.0×10-19 C. Option D

2) The details of question 2 are not shown hence the question can not be answered.

3) Looking at the electronic configuration of the element;  1s22s22p63s23p5, it is clear to see that it has the ns2 np5 outermost configuration that is common to halogens thus it;

4) The atom is composed of the protons, neutrons and electrons. Given the nuclide identified as 230/92U we have  92 protons, 138 neutrons and 92 electrons. Option C

5) The application of total Internal Reflection occurs in a mirage. Option A

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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 magnitude of the force on the left-hand pole of the thin flexible gold chain of uniform linear density with a mass of 17.1 and, hangs between two 30.0 cm long vertical sticks, which are a distance of 30.0 cm apart will be, 0.167N.

To find the correct answer, we have to know more about the Basic forces that acts upon a body.

                                    [tex]tan\alpha =\frac{30}{30}\\ \alpha =45^0[/tex]

                                 [tex]y-direction\\T_2sin\alpha =mg\\\\T_2=\frac{mg}{sin\alpha } =\frac{17.1*10^{-3}*9.8}{sin45} \\\\T_2=0.236N[/tex]

                                 [tex]x-direction\\T_1-T_2cos\alpha =0\\T_1=T_2cos\alpha \\T_1=0.236*cos45=0.167N[/tex]

Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.

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The force acting on the left-hand pole of a thin, flexible gold chain of uniform linear density weighing 17.1 grams that is suspended between two vertical sticks that are 30.0 cm apart will be 0.167 N in magnitude.

We need to learn more about the fundamental forces that affect a body in order to arrive at the right answer.

                              [tex]\alpha =tan^{-1}(\frac{30}{30})=45^0[/tex]

                                [tex]T_2sin\alpha =mg\\T_2=\frac{mg}{sin\alpha } =0.236N[/tex]

                                     [tex]T_1=T_2cos\alpha \\T_1=0.167N[/tex]

Thus, we may infer that the force acting on the left-hand pole will have a value of 0.167N.

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Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Explanation: To find the answer, we need to know about the different equations of planetary motion.

                    [tex]m_g=7.88*10^{18}kg\\r_g=6.32*10^4 m.\\h_{max}=1.44*10^3m.\\[/tex]

                    [tex]v=\sqrt{2gh}[/tex]

                  [tex]g_g=\frac{GM_g}{r_g^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} =0.132m/s^2.[/tex]

                  [tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s.[/tex]

                [tex]v=\sqrt{ \frac{GM}{r}}=\sqrt{\frac{7.88*10^{18}*6.67*10^{-11}}{1.45*10^5 }} =3.624km/s.[/tex]

Thus, we can conclude that,

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

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1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.

2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.

Understanding the planetary motion equations is necessary in order to determine the solution.

                             [tex]v=\sqrt{2gh}[/tex]  

                    [tex]a=\frac{GM}{R^2} =0.132m/s^2[/tex]

                  [tex]v=\sqrt{2*9.8*0.132} =19.46m/s[/tex]

                        [tex]v=\sqrt{\frac{GM}{R} } =3.624km/s\\where, M=7.88*10^{18}kg[/tex]

Consequently, we can say that

1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.

2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.

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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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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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Answer:

d

Explanation:

Gravitational force is a type of force that pulls objects to the Earth's surface.

Answer:a)45cos40 b)45sin40 c) 2.95 d)102

Explanation:

a) [tex]v_{x} = 45 cos40[/tex]

b) [tex]v_{y}=45sin40[/tex]

c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -g. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: [tex]v_{f} = v_{i} + at[/tex].

[tex]0 m/s = 45sin40 + (-9.8m/s^2)t\\t = 2.95s[/tex]

d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.

[tex]x=vt\\x=45cos40(2.95) =102m[/tex]

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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