2 red and 2 blue overlapping balls in the center are surrounded by a green, fuzzy, circular cloud with a white line running through it. 2 green balls sit on the white line, and a line leading a bracket around the balls is labeled A. A line leading to a bracket overlapping the white line is labeled B. Which is a characteristic of the part of the atom marked "B”? It is very dense. It has a large mass. It is negatively charged. It is where neutrons move.

The tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (21 x 9.8)/(2 x sin66)

T = 112.64 N

T + F = W

F = W - T

F = (9.8 x 21) - 112.64

F = 93.16 N

Thus, the tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.

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The direction of torque τ this method is mathematically given as

D=X  

Option A is correct.

Generally, the equation for torque is  mathematically given as

τ = r X F

Hence to decipher the torque direction with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r

In conclusion, the direction of this method is

D=X   Option A.

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+ X axis.

The equation for torque is:
[tex]\tau = r \times F[/tex]

Torque is the CROSS-PRODUCT of 'r' and 'F'.

We can use the right-hand rule to determine the direction of torque.

Looking at the diagram, with an open hand, point your fingers towards the top of the picture (Along the +Z axis in the direction of the 'r' vector.)

Then, curl your fingers in the direction of the force vector, (along the -Y axis).

Your thumb will stick out in the direction of the torque, or the +X axis.

You can also just eliminate any options that involve axes that have already been used for 'r' and 'F'. In this case, there are components in the Z and Y axis, so this leaves the 'X' axis to be considered.

The portion that shows a decreasing velocity is  D to E.

The position time graph is used to describe the motion of a body. We know that the slope of a velocity time graph is the velocity of the graph.

If we look at the graph as shown in the image attached to the answer, we can see that the portion that shows a decreasing velocity is  D to E.

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Answer:

D - E

Explanation:

hope this help

Since the diameter of the star suddenly shrinks to 0.350 times its present size, the new period of the star is 4.2 days

To find the new period, we need to know Kepler's third law of planetary motion

Kepler's third law of planetary motion states that the square of the period of orbit of a planet, T equals the cube of its radius, R.

So, T² ∝ R³

and T₂²/T₁² = R₂³/R₁³ where

Given that the diameter suddenly shrinks to 0.350 times its present size and its radius is proportional to its diameter. So, its radius shrinks 0.350 times its initial size. Also, since there is a uniform mass distribution before and after, the mass remains the same.

So, R₂ = 0.350R₁

Making T₂ subject of the formula, we have

T₂ = [√(R₂/R₁)³]T₁

Substituting the values of the variables into the equation, we have

T₂ = [√(R₂/R₁)³]T₁

T₂ = [√(0.350R₁/R₁)³]20.30 days

T₂ = [√(0.350)³]20.30 days

T₂ = [√0.042875]20.30 days

T₂ = [0.2071]20.30 days

T₂ = 4.2 days

So, the new period of the star is 4.2 days

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(a) The angle the plane makes with respect to the horizontal is 27⁰.

(b) The magnitude of the normal force acting on the block is 135.3 N.

(c) The component of the force of gravity along the plane, fgx is 68.96 N.

(d) The expression for the net force on the block is mgsinθ - μmgcosθ = 0.

sin θ = h/L

where;

sin θ = 3.05/6.8

sin θ = 0.4485

θ = arc tan(0.4485)

θ = 26.7⁰

θ ≈ 27⁰

Fₙ = mgcosθ

Fₙ = (15.5)(9.8)cos(27)

Fₙ = 135.3 N

Fx = mgsinθ

Fx = (15.5)(9.8) sin(27)

Fx = 68.96 N

∑F = 0

Fx - Ff = 0

mgsinθ - μmgcosθ = 0

Thus, we can conclude the following;

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Light behaves as it relates to its nature (waves or particles). light travels as and interacts with matter as wave particles.

Electromagnetic radiation that is within the region of the electromagnetic spectrum that the human eye can see is referred to as light or visible light.

The main characteristics of light include intensity, polarization, frequency or wavelength spectrum, and propagation direction. One of the fundamental constants of nature is its speed in a vacuum, which is 299 792 458 meters per second (m/s).

Light can be categorized according to two theories: the first categorizes light as particles, and the second categorizes light as waves.

The main difference between the wave and particle natures of light is that the former asserts that light can act as an electromagnetic wave while the latter asserts that light is made up of tiny particles known as photons.

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Answer: 1.waves 2.particles

Explanation: just did it!<3

For a ball that slows down as it rolls up a hill the value of the change in its kinetic energy, ΔEk is: Negative

Kinetic energy can be defined as the ability assumed by an object in motion.

Kinetic energy is also said to be a form of mechanical energy that is possessed by an object to be in motion.

Normally kinetic energy is believed to be positive by when it begins to slow down a negative effect is experienced

In conclusion, For a ball that starts rolling down a ramp the value of the change in its kinetic energy, ΔEk is: Positive

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(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

Apply Newton's second law of motion;

∑F = ma

where;

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

           →                 →              Ф                         ←

         1st child      friction       wagon                2nd child

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

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Answer:

47.24 in

Explanation:

120 / 2.54 = 47.24

Answer

its c

Explanation:

When resistors are connected in parallel combination the voltage drop across each resistor is same. However , the total current gets branched into different amount .

Question :

— What must be the same for two resistors that are connected in parallel ?

AnswEr :

Option B seems to be correct answer .

Answer:

• black hole     -   one of the most dense objects in the universe

• red giant       -    a very large star

• white dwarf   -   an old, very dense hot star that is cooling

• nebula           -   a mass of gas and dust

Answer:

[tex]T_0=80695.17162...[/tex]

Explanation:

Given equation:

[tex]\ln \left(\dfrac{T_0-100}{T_0}\right)=-0.00124[/tex]

To solve the given equation:

[tex]\textsf{Apply log rules}: \quad e^{\ln (x)}=x[/tex]

[tex]\implies \dfrac{T_0-100}{T_0}=e^{-0.00124}[/tex]

Multiply both sides by T₀:

[tex]\implies T_0-100=T_0e^{-0.00124}[/tex]

Add 100 to both sides:

[tex]\implies T_0=T_0e^{-0.00124}+100[/tex]

Subtract [tex]T_0e^{-0.00124}[/tex] from both sides:

[tex]\implies T_0-T_0e^{-0.00124}=100[/tex]

Factor out the common term T₀:

[tex]\implies T_0(1-e^{-0.00124})=100[/tex]

Divide both sides by [tex](1-e^{-0.00124})[/tex]

[tex]\implies T_0=\dfrac{100}{1-e^{-0.00124}}[/tex]

Carry out the calculation:

[tex]\implies T_0=\dfrac{100}{1-0.99876...}[/tex]

[tex]\implies T_0=\dfrac{100}{0.001239231...}[/tex]

[tex]\implies T_0=80695.17162...[/tex]

Answer:

1.37 seconds before.... splat !

Explanation:

y = y0 + v0 t + 1/2 a t^2      y0 = 30       y = 0

                    v0 = 0 ( from rest)

                            a = - 32.2 ft/s^2

0 = 30 - 1/2 * 32.2 * t^2

t=1.366 seconds                    Note that   mass   is irrelevant.

Answer:

the watt is the unit of power or radiant flax

The energy lost to the air due to friction is 37.7 J.

The energy lost to the air due to friction is determined from the change in kinetic energy of the ball.

ΔK.E = K.Ef - K.Ei

ΔK.E = ¹/₂mv² - ¹/₂mu²

ΔK.E =  ¹/₂m(v² - u²)

ΔK.E = ¹/₂(0.45)(19.89² - 15.1²)

ΔK.E = 37.7 J

Thus, the energy lost to the air due to friction is 37.7 J.

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The new period will be 2.486 days.

The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.

Given data;

Mass of a star,m= 1.210×10³¹ kg

The time period for one rotation of the star, T = 20.30 days

D' = 0.350 D

R' = 0.350 R

From the law of conservation of angular momentum;

[tex]\rm I \omega = I' \omega' \\\\ \frac{2}{5} MR^2 \times \frac{2 \pi }{T}=\frac{2}{5} MR'^2 \\\\ R^2 \times \frac{1}{T}= R'^2 \times \frac{1}{T} \\\\ T' = \frac{R'^2T}{R^2} \\\\ T' = \frac{(0.350 R)^2 \times 26.1 }{R^2} \\\\T' = 0.1225 \times 20.30 \\\\ T'= 2.486 \ days[/tex]

Hence, the new period will be 2.486 days.

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Answer:

See below

Explanation:

Set up your pendulum

  measure its length  and time the period ( you could time 100 of them and divide the time result by 100 to get the period, T)

    then use  

             T = 2 pi  sqrt (L/g)        T = period    L = length   g = gravity

The sound waves with the frequency 20 Hz, 200 Hz, 2,000 Hz, and 20,000 Hz have a velocity of 344 m/s.

A sound wave is a sample resulting from the motion of strength traveling thru a medium (inclusive of air, water, or every other liquid or stable matter) because it propagates from the supply of the sound. Sound waves are created and convey strain waves, for example, a cellphone.

The frequency can be given as:

Frequency = Velocity/ Wavelength

Velocity = Frequency * Wavelength

The velocity of the following sound waves is given as:

Wavelength = 17.2 m

Velocity = 20 * 17.2 m/s

Velocity = 344 m/s

Wavelength = 1.72 m

Velocity = 200 * 1.72 m

Velocity = 344 m/s

Wavelength = 17.2 cm = 0.172 m

Velocity = 2,000 * 0.172 m/s

Velocity = 344 m/s

Wavelength = 1.72 cm = 0.0172 m

Velocity = 20,000 * 0.0172 m/s

Velocity = 344 m/s

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The speed will be 0.2 m/s and the velocity will be 0 m/s.

Speed = Total Distance / Total time

We have given total distance as ( 50 + 95 ) metres and total time as 12 minutes or we can say 720 seconds.

Speed = 145/ 720 m/s

Speed = 0.2 m/s

Velocity = Total Displacement / Total time

As the initial and final is the home, hence the net displacement is 0 in that case.

In this case also the total time we have given is 12 minutes or we can say 720 seconds.

Velocity = 0 / 720 m/s

Velocity = 0 m/s

So to conclude with we can say that the speed is 0.2 m/s and the velocity is 0 m/s.

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Answer:

Approximately [tex]31\; {\rm m\cdot s^{-1}}[/tex] assuming that there is no friction on the rollercoaster, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].

Explanation:

Make use of the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where:

If there is no friction on the rollercoaster, the acceleration of the rollercoaster would be equal in magnitude to the gravitational field strength, [tex]g[/tex]:

[tex]a = g = 9.81\; {\rm m \cdot s^{-2}}[/tex].

The initial velocity of this rollercoaster is [tex]0\; {\rm m\cdot s^{-1}}[/tex] (that is, [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) since the rollercoaster was initially stationary. The displacement of this rollercoaster would be [tex]x = 49\; {\rm m}[/tex] (same as the height of the hill.)

Rearrange the equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find an expression for the final velocity [tex]v[/tex] of this rollercoaster:

[tex]\begin{aligned}v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 9.81\; {\rm m\cdot s^{-2} \times 49\; {\rm m} + (0\; \rm m\cdot s^{-1}})^{2}}\\ &\approx 31\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Answer:

F = - F x       force acting on cube

F = -106 N/n * .14 m = -14.8 N         direction opposite to force

a = F / m = -14.8 N / .66 kg = -22.5 m/s^2        acceleration of cube

direction is towards equilibrium point (opposite F)

The speed of cars at A and B will be 25 m/sec.

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.

Given data;

Mass of roller coaster,m =750 kg

Height at location 1,h₁ = 40 m

Height at location 2,h₂ = 302 m

The velocity of car A,[tex]\rm v_A[/tex]

The velocity of the car B, [tex]\rm v_B[/tex]

The velocity of the car A and B is the same is found as;

[tex]\rm v_A = v_B = v = \sqrt{v_0^2 + 2g(h_1-h_2} )\\\\ v = \sqrt{15^2 + 2 \times 10 (40-30)} \\\\ v = \sqrt{225+400} \\\\ v = 25 \ m/sec[/tex]

Hence the speed of cars at A and B will be 25 m/sec

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Answer:

A2/A1 = 9

V2/V1 = 27

Explanation:

Area of Sphere 1; A1 = 4πr²

Volume of Sphere 1; V1 = (4/3)πr³

Area of Sphere 2; A2 = 4π(3r)² = 9*4πr²

Volume of Sphere 2; V2 = (4/3)π(3r)³ = 27*(4/3)πr³

Thus;

A2/A1 = 9*4πr²/4πr² = 9

V2/V1 = [27*(4/3)πr³]/[(4/3)πr³] = 27

To create a tragic past for your character, you would have to write a plot that shows a sad ending to a good story.

This refers to a type of story that has negative emotions such as sadness, death, etc.

Hence, we can see that for example, you could write about two lovers and how despite all the struggles they faced, eventually died in mysterious circumstances.

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(a)The initial velocity of the shell will be 594.27 m/sec

(b)The maximum height it reaches will be 9000 m.

c)101.249 m meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface of the ship.

d)The error could be significant compared to the size of a target. Option C is correct.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

m is the mass of the block = Kg.

u is the initial velocity of fall = m/sec

h is the distance of fall =  m

g is the acceleration of free fall = m/sec²

v is the hitting velocity of =?

a)

The range of the projectile is;

[tex]\rm R = \frac{u^2 sin 2 \theta }{g} \\\\ 36 \times 10^ 3 = \frac{U^2 sin 45^0}{9.81} \\\\ U= 594.27 \ m/sec[/tex]

b)

The maximum height of the projectile is;

[tex]\rm H = \frac{u^2 sin 2 \theta }{2g} \\\\ H = \frac{(594.27)^2\times (sin 45)^2}{2 \times 9.81 } \\\\ H = 9000 \ m[/tex]

c)

The distance between its surface and the ship, measured in a horizontal arc parallel to the surface, will be 36.0 kilometers. The distance from the lower surface is found as;

[tex]\rm( R_e + h)^2 = R_e^2+(36)^2 \\\\ (R_e)^2 = h^2+2R_e h= R_e^2 + 12196 \\\\ h^2 + 12800 h - 1296 = 0 \\\\ h = 101.249 \ m[/tex]

d)

An error is a mistaken or erroneous action. In some contexts, an error is interchangeable with a mistake.

The difference between the calculated value and the original value is known as the error. The inaccuracy may be large in comparison to the target's size. Option C is correct.

Hence the initial velocity of the shell, maximum height, and the distance from the lower surface will be 594.27 m/sec,9000 m, and 101.249 m and option c for question d are correct respectively.

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The  position of the car at the given time is 18 m.

The position of the car at the given time is calculated as follows;

x = vt

where;

x = 3 m/s x 6.0 s

x = 18 m

Thus, the  position of the car at the given time is 18 m.

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The speed and pressure at another end will be 9 m/sec and -31500 pascals.

The difference between absolute pressure and atmospheric pressure is known as gauge pressure. Relative pressure is another name for gauge pressure.

Given data in problem is;

For horizontal pipe, Z₁=Z₂

Pressure at end 1, P₁ = 4500Pa

Speed at end 1, V₁ = 3 m/sec

Speed at end 2, V₂ = ? m/sec

Area of cross-section at end 1,A₁ =A

Area of cross-section at end 1,A₂ = A/3

From the continuity equation;

A₁V₁=A₂V₂

A× 3  = (A/3)×V₂

V₂ = 9 m/sec

From Bernoulli's equation;

[tex]\rm \frac{P_1}{\rho g} + \frac{v_1^2}{2g} +Z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} +Z_2 \\\\ \frac{P_1-P_2}{\rho g } = \frac{v_2^2-v_1^2}{2g} \\\\ \frac{P_1-P_2}{\rho} = \frac{v_2^2-v_1^2}{2} \\\\ \frac{4500-P_2}{1000} = \frac{9^2-3^2}{2} \\\\ 4500 -P_2 = 36000 \\\\ P_2 = - \ 31,500 \ pa[/tex]

Hence the speed and pressure at another end will be 9 m/sec and -31500 pascals.

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