find the horizontal component of the electric field in n/c, taking east as positive.
Eh = - 209.625

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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Answer:

80GB= 80000000000 bytes

256GB= 274877906944 bytes

Explanation:80GB= 80000000000 bytes

256GB= 274877906944 bytes

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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The image distance is 33.3 cm while the image height is 14.2 cm.

A converging lens will always have a positive focal length hence, we have to find the object distance as follows;

1/f = 1/v + 1/u

1/12 = 1/v + 1/20

1/v = 1/12 - 1/20

1/v = 0.08 - 0.05

v =33.3 cm

Now;

Magnification =  33.3 cm/20.0 cm =1.67

M = Image height/Object height

1.67 =  Image height/8.50 cm

Image height = 1.67  * 8.50 cm

Image height = 14.2 cm

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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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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:

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! :)

(a) The surface charge density in the sphere of radius 7.4 is 0.0322 C/m³.

(b) The surface density on the 2nd sphere is  3.48  x 10⁻⁴ C/m³.

F = kq₁q₂/r²

Fr² = kq₁q₂

q₁q₂ = Fr²/k

where;

q₁q₂ = (0.62 x 0.38²) / (9 x 10⁹)

q₁q₂ = 9.95 x 10⁻¹² C

q₂ = 9.95 x 10⁻¹² C/q₁

From the question;

q₁ + q₂ = 55 x 10⁻⁶

q₁ + 9.95 x 10⁻¹² /q₁ = 55 x 10⁻⁶

q₁² + 9.95 x 10⁻¹²  =  55 x 10⁻⁶q₁

q₁² - 55 x 10⁻⁶q₁  + 9.95 x 10⁻¹²   = 0

solve the quadratic equation using formula method;

q₁ = 5.48 x 10⁻⁵ C

q₂ = 55 x 10⁻⁶ C - 5.48 x 10⁻⁵ =  1.82 x 10⁻⁷ C

V1 = ⁴/₃πr³

V1 = (⁴/₃ π)(0.074)³ =  1.7 x 10⁻³ m³

Surface charge density = (5.48 x 10⁻⁵ C) / (1.7 x 10⁻³ m³) = 0.0322 C/m³

V2 = (⁴/₃ π)(0.05)³ = 5.236 x 10⁻⁴ m³

Surface charge density = ( 1.82 x 10⁻⁷ C) / (5.236 x 10⁻⁴ m³) = 3.48  x 10⁻⁴ C/m³

Thus, the surface charge density in the sphere of radius 7.4 is 0.0322 C/m³.

The surface density on the 2nd sphere is  3.48  x 10⁻⁴ C/m³.

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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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From the calculations, the power developed is 337.5 W.

First we must obtain the acceleration from;

v = u + at

u = 0 m/s because the motion started rom rest

v = at

a = v/t

a = 12 m/s/ 1.6 s

a= 7.5 m/s^2

The force is obtained from;

F = ma = 7.5 kg * 7.5 m/s^2 = 56.25 N

Now the distance covered is obtained from';

v^2 = u^2 + 2as

v^2 = 2as

s = v^2/2a

s = (12)^2/2 *  7.5

s = 9.6 m

Now;

Work = Fs = 56.25 N *  9.6 m = 540 J

Power expended =  540 J/ 1.6 s = 337.5 W

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Conduction is a process through which energy is transferred from the branding iron to the handle.

When two objects with different surface temperatures come into close touch with one another, conduction happens. Up until they reach thermal equilibrium, or the point at which they are at the same temperature, the heat energy moves from the hotter to the cooler object.

The metal tip of the branding iron, which makes contact with the branded surface, becomes extremely hot when it is placed in the fire because of the high fire temperature.

Hence, conduction is a process through which energy is transferred from the branding iron to the handle.

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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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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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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:

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.

Infinity is the zero point for electric potential energy because Electric field strength is zero at infinity.

This conclusion is not "derived" from anything else; the electrostatic potential is zero at infinity because that is how we define it.

Define electric field and electric field strength.

The physical field that contains electrically charged particles and pulls or attracts all other charged particles in the vicinity is known as an electric field. Additionally, it describes the physical environment of a system of charged particles.

The intensity of an electric field at a specific location is expressed quantitatively as "electric field strength." The volt per meter (v/m or v m-1) is the common unit.

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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.

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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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.

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:

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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(a) The magnitude of the minimum magnetic field is 1.467 T.

(b) The direction of the minimum magnetic field will be eastwards.

F = qvB

mg = qvB

B = mg/qv

where;

B = (1.98 x 10⁻⁴ x 9.8) / (3.5 x 10⁻⁸ x 3.78 x 10⁴)

B = 1.467 T

The magnetic field will be directed towards the negative charge. Thus, the direction of the minimum magnetic field will be eastwards.

Thus, the magnitude of the minimum magnetic field is 1.467 T and the direction is eastwards.

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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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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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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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(a) The coil's self-inductance is 7.26 mH.

(b) The self-induced emf in the coil is 7.26 V

(c) The direction of the induced emf is from b to a.

L = N²μA/I

L = (600² x 4π x 10⁻⁷ x 6.9 x 10⁻⁴)/(0.043)

L = 7.26 x 10⁻³ H

L = 7.26 mH

emf = N(ΔBA)/t

where;

B = μNI/L

B1 = (4π x 10⁻⁷ x 600 x 5)/0.043

B1 = 0.0876 T

B2 =  (4π x 10⁻⁷ x 600 x 2)/0.043

B2 = 0.035 T

emf = NΔBA/t

emf = (600)(0.0876 - 0.035)(6.9 x 10⁻⁴) / (3 x 10⁻³)

emf = 7.26 V

The direction of the induced emf is always opposite to the direction of the applied current.

Thus, the direction of the induced emf is from b to a.

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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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Answer:

b

Explanation:

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