the distance between crest and the adjacent trough of water waves is 3m, they pass a given point at rate of 5m/s. what is the frequency and the speed of water waves?

Answer:

The Fundamental Quantity is a physical quantity that exists independently and cannot be expressed by any other physical quantity.

Explanation:

Examples of these could be : Length, electric current and mass

Hope this helps! :)

According to Newton's 3rd law, there will be equal and opposite force on the astronaut which is  -6048 N

The law state that in every action, there will be equal and opposite reaction.

Given that a rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. We are to calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit.

Let us first calculate the force involved in the acceleration of the rocket by using the formula

F = ma

Where mass m = 87.4 kg, acceleration a = 69.2 m/s2

Substitute the two parameters into the formula

F = 87.4 x 69.2

F = 6048.08 N

According to the Newton's 3rd law, there will be equal and opposite force on the astronaut.

Therefore, the normal force acting on the astronaut is -6048 N approximately

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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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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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Answer: A candle 4.75 cm tall is 40.0 cm to the left of a plane mirror. Then, the image formed by the mirror will be to the right of the mirror and the height of this image is same as that of the object.

Explanation: To find the answer, we need to know about the reflection at a plane mirror.

There are some points about the reflection at a plane surface.

Thus, we can conclude that, a candle 4.75 cm tall is 40.0 cm to the left of a plane mirror. Then, the image formed by the mirror will be to the right of the mirror and the height of this image is same as that of the object.

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In response, a plane mirror is 40.0 cm to the left of a candle that is 4.75 cm tall. The mirror's generated image will then be to the right of the mirror and have the same height as the object.

In order to determine the solution, we must understand how a plane mirror reflects light.

As a result, we may say that a flat mirror is 40.0 cm to the left of a candle that is 4.75 cm tall. The mirror's generated image will then be to the right of the mirror and have the same height as the object.

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Answer:

b

Explanation:

the answer is b because if you do the math it equals to b

The height, h, of the water level in the tank is  mathematically given as

h=1.3675

Generally, the equation for kinematics is  mathematically given as

S=v_o t+(1/2)at^2

Therefore

y=0+(1/2)gt^2

Where

t=(2y/g)^{1/2}

t=(2*1.72/9.8)^{1/2}

t=0.5925s

The horizontal exit velocity will be given as

[tex]Vx=x/t[/tex]

Therefore

[tex]Vx=\frac{0.3.13}{0.5925}[/tex]

Vx=0.5283

In conclusion, applying Bernoulli's Law the tank bottom and tank surface

P+(1/2)pv0^2+pgh=P+(1/2)pvx^2+pgh

(1/2)(p)(0m/s)^2+pgh=(1/2)pvx^2+pg(0m)

gh=(1/2)vx^2

h=(0.5283m/s)^2/(2)(9.8)

h=1.3675

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The distance between two consecutive nodes and the amplitude after 0.56s are m/2 and 1.75×10^(-4) m respectively.

= 2π/(4π÷m)

= m/2

Displacement after 0.56 s = 0.008×cos(50π×0.56s)

=1.75×10^(-4) m

Thus, we can conclude that the distance between consecutive nodes and displacement after 0.56 s are m/2 and 1.75×10^(-4) m respectively.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The particle displacement y of air molecules due to a sound wave is given by y=0.008coswtsinkz where k=4π÷m and w=50π rads/s.

Calculate:

I) the distance between 2 consecutive nodes

ii) the amplitude after 0.565s

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The altitude of the pole vaulter as she crosses the bar is 5 m.

v² = u² - 2gh

where;

h = (u² - v²)/2g

h = (10² - 1.5²)/(2 x 9.8)

h = 5 m

Thus, the altitude of the pole vaulter as she crosses the bar is 5 m.

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Answer:

VR = Velocity of effort / Velocity of Load

Explanation:

The ratio of the out-lever to the in-lever (length of the resistance arm to effort arm) is known as the velocity ratio (VR). It indicates that the distance covered by effort is four times that covered by the load.

The load-to-effort ratio of a machine, or alternatively, the output-to-input ratio of a machine, is known as its mechanical advantage. Another definition of velocity ratio is the ratio of the velocity of effort to the velocity of load.

When a machine has a force ratio of 4 and a velocity ratio of 4, this indicates that the weight moved is multiplied by 4 and the distance moved by the effort is multiplied by 4 at the same time.

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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:

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:

compound

Explanation:

a substance made from two or more different elements that have been chemically joined. Examples of compounds include water (H2O), which is made from the elements hydrogen and oxygen, and table salt (NaCl), which is made from the elements sodium and chloride.

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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The force required to pull the two hemispheres apart 53696.25N and  the minimum number of horses required is 37 .

( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 )

(1)The contact area between the hemispheres is (π x 0.430²) = 0.5805m²

Pressure difference = (940 - 15) = 925millibars.

(925 x 100) = 92,500N/m^2.

(92500 x 0.5805) = 53696.25N. is the force required to part the hemispheres.

(2)[tex]\frac{53696.25N}{1450N}[/tex] = 37 horses for each side .

37 + 37 = 74 horses will be required.

Force is something which can change the motion of an object, stop it or move it, change its shape or size with it. There are two types of forces, contact forces and non-contact forces. Here, it is a contact force at first, then when the horses come it becomes non-contact force.

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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]

From the calculations, the power expended is 43650 W.

Now we can find the acceleration from;

v = u + at

u = 0 m/s

v =  95 km/h or 26.4 m/s

t =  6.8 s

a = ?

Now

v = at

a = v/t

a = 26.4 m/s/ 6.8 s

a = 3.88 m/s^2

Force = ma = 850-kg * 3.88 m/s^2 = 3298 N

The distance covered is obtained from;

v^2 = u^2 + 2as

v^2 = 2as

s = v^2/2a

s = (26.4)^2/2 * 3.88

s = 696.96/7.76

s = 90 m

Now;

Work = Fs

Work =  3298 N * 90 m = 296820 J

Power =  296820 J/ 6.8 s

= 43650 W

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The the magnitude of the average acceleration during the impact is 9.8 m/s² and the time of motion is 1.11 s.

The lead weight will fall under the influence of gravity with magnitude of 9.8 m/s².

t = √2h/g

where;

t = √((2 x 6.004)/9.8)

t = 1.11 s

Thus, the the magnitude of the average acceleration during the impact is 9.8 m/s² and the time of motion is 1.11 s.

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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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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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I will go to school tomorrow .....is this present tense or past tense or future tense

Answer:

D

Explanation:

The given vector can be broken down into two components (as shown in the figure).

Which results ; vector (s) = 3x + 4y.

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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Answer:

80GB= 80000000000 bytes

256GB= 274877906944 bytes

Explanation:80GB= 80000000000 bytes

256GB= 274877906944 bytes

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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Hello!

a)

To begin, let me first clarify that no section of the wire along the axis of the point 'P' will contribute to the magnetic field (Aka, the straight part of the wire) because the radius vector and current vectors would be parallel.

Now, let's use Biot-Savart's Law:
[tex]dB = \frac{\mu_0}{4\pi }\frac{id\vec{l}\times \hat{r}}{r^2}[/tex]

B = Magnetic field strength (? T)
μ₀ = Permeability of free space (Tm/A)

i = Current in wire ('I' A)
dl = differential length element

r = distance from wire to point P ('R' m, remains constant!)

We can rewrite Biot-Savart as:
[tex]B = \frac{\mu_0}{4\pi } \int \frac{id\vec{l}\times \hat{r}}{r^2}[/tex]

First, let's mind the cross-product.

The angle between the radius vector (Along 'R') and the current vector is ALWAYS 90° since the two vectors are perpendicular along the arc. At a certain point, think about the current as being "tangential" to the differential length and thus perpendicular to the radius.

Therefore, we can disregard the cross-product since sin(90) = 1.

Let's plug in what we already know, replacing 'dl' with 'ds' since this is an arc:

[tex]B = \frac{\mu_0}{4\pi } \int \frac{ids}{R^2}[/tex]

We have a semi-circle, so we are integrating from 0 to π radians.

[tex]B = \frac{\mu_0}{4\pi } \int\limits^{\pi}_{0} {\frac{ids}{R^2}}[/tex]

Simplifying to make the integral easier, we can take constants out of the integral.

[tex]B = \frac{\mu_0i}{4\pi R^2 } \int\limits^{\pi R}_{0} {} \, ds[/tex]

Evaluating:
[tex]B = \frac{\mu_0i}{4\pi R^2} (\pi R- 0) = \boxed{\frac{\mu_0 i}{4R}}[/tex]

b)

Using the current Right-Hand-Rule at the top of the arc, point your thumb to the right. Curl your fingers as if you are gripping the wire over the top and all the way over to the other side of the wire (Where point 'P' would be). Your fingers should point INTO THE PAGE.

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:

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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(1) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.

(2) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.

(3) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.

v² = u² - 2gh

where;

when the lander's velocity = 0

0 = u² - 2gh

u² = 2gh

h = u²/2g

h = (2.8²)/(2 x 9.8)

h = 0.4 m

h = (u² - v²)/2g

h = (2.8² - 1.7²)/(2 x 9.8)

h = 0.25 m

h = (u² - v²)/2g

h = (2.8² - 1.7²)/(2 x 9.8)

h = 0.25 m

Thus, the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.

The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.

The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.

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