(based on 15-103 in the text) At an initial instant, a 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a speed (vB)1 = 4.8 ft/s. The attached cord is then pulled down through the hole with a constant speed vr = 2.2 ft/s. a. Determine the ball's speed at the instant r2 = 2 ft. Neglect friction and the size of the ball. Note that particle path is no longer of constant radius, and the particle has velocity components in both tangential and radial directions. b. How much work is done to pull down the cord from the initial instant to the instant when r2 = 2 ft? Neglect friction and the size o

Question :-

Answer :-

Explanation :-

As per the provided information in the given question, we have been given that the Velocity of the ball is 12 m/s . Time is given as 36 sec . And, we have been asked to calculate the Acceleration .

For calculating the Acceleration , we will use the Formula :-

[tex] \bigstar \: \: \boxed{ \sf{ \: Acceleration \: = \: \dfrac{v \: - \: u}{t} \: }} [/tex]

Where ,

Therefore , by Substituting the given values in the above Formula :-

[tex] \dag \: \: \: \sf { Acceleration \: = \: \dfrac{Final \: Velocity \: - \: Initial \: Velocity}{Time} } [/tex]

[tex] \longmapsto \: \: \sf { Acceleration \: = \: \dfrac{0 \: - \: 12}{36}} [/tex]

[tex]\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 12 \: }{36}}[/tex]

[tex]\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 1 \: }{3}}[/tex]

[tex] \longmapsto \: \bf {Acceleration \: = \: 0.33 \: m/s^{2}} [/tex]

Hence :-

[tex] \underline {\rule {212pt} {4pt}} [/tex]

Answer:

50 V; step-down transformer

Explanation:

10,000 : 5,000 = 100 : V

V = 50 V

From 100 V change to 50 V, so it is a step-down transformer

#a

Now

[tex]\\ \tt\Rrightarrow Q=\dfrac{I}{t}[/tex]

[tex]\\ \tt\Rrightarrow Q=\dfrac{76.2\times 10^{-3}}{624}[/tex]

[tex]\\ \tt\Rrightarrow Q=0.122\times 10^{-3}[/tex]

[tex]\\ \tt\Rrightarrow Q=122\times 10^{-6}C[/tex]

[tex]\\ \tt\Rrightarrow Q=122\mu C[/tex]

#b

Answer:A).3.16 s B).56.92

A).12.5m/s

B).7.81m

Explanation:

Answer: B

Explanation:  A force perpendicular to the momentum changes the direction of the momentum but not its magnitude.

Answer:

a. To change the magnitude of the momentum you need to apply a force with a component parallel to the momentum (either in the same direction as the momentum or the opposite direction).

d. To make an object turn to the left, something has to exert a force to the right on the object.

Explanation:

For motion along a circle or curve, the net force on the particle must be parallel with the change in momentum.

This change in momentum of the system is given by;

[tex]\delta P_{sys} = F_{net} \delta t[/tex]

Thus, option A is correct.

a. To change the magnitude of the momentum you need to apply a force with a component parallel to the momentum (either in the same direction as the momentum or the opposite direction).

Also, applying the principle of torque, to make an object turn to the left, something has to exert a force to the right on the object.

Option D is also correct.

Answer:

60m/s

Explanation:

v=u+at

v=36+(2×12)

v=36+24

v=60m/s

Answer:

350 ft/s²

Explanation:

First, convert mph to ft/s.

58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s

Given:

v₀ = 85.1 ft/s

v = 0 ft/s

t = 0.24 s

Find: a

v = at + v₀

a = (v − v₀) / t

a = (0 ft/s − 85.1 ft/s) / 0.24 s

a = -354 ft/s²

Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².

Answer:

The ball has potential energy just like the Longbow has when it is stretched. The other aspect of energy that the bow portrays is kinetic energy. Kinetic energy is the energy of motion. When the string is released, the potential energy is converted to kinetic energy in movement of the arrow

Explanation:

Answer:

It has potential energy not kinetic

Explanation:

Answer:

C

Explanation:

Gravity pulls objects towards earth's core without needing any contact with objects

Answer:

They create enormous amounts of sound energy, much like explosion. When an aircraft passes through air, it creates a series of pressure waves just like the waves created by the boat. As the speed of aircraft increases these waves are forced to compress.

Answer:

What happens when a plane passes the speed of sound?

Answer - You hear a loud BOOM then you cant hear it

How does it sound?

Answer - there is no sound

Is the sound super high since the wavelength is super low or maybe even non-existent?

Answer - When its about to break the sound barrier it makes a high pitched sound

How does it look and why does it look that way?

Answer - Its air you cant see it ;-;

How will it affect people on ground?

you will hear the sound and it might hurt but nothing else

Explanation:

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux  GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;  

transferred energy + generated energy = 0

(radiation + convection) +  generated energy = 0

[tex][\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0[/tex]

[tex][\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s[/tex]

(a) the electrical power generated for still summer day

[tex]T_s = T_{\infty} = 35 ^oC = 308 \ k[/tex]

[tex][0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k[/tex]

[tex]P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W[/tex]

(b)the electrical power generated for a breezy winter day

[tex]T_s = T_{\infty} = -15 ^oC = 258 \ k[/tex]

[tex][0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k[/tex]

[tex]P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W[/tex]

Answer:

The time of travel and the distance between neighboring dots is always the same. So, the speed between neighboring dots is constant. This speed will be the same as the average speed in part F. hope it helps!

Answer:

9 :)

Explanation:

Answer:

When one side of the tug-of-war rope has a larger net force than the other side then the basket in the middle will be pulled to whichever side has a larger net force.

Explanation:

When one side of the tug-of-war rope has a larger net force than the other side then the basket in the middle will be pulled to whichever side has a larger net force.

Answer:

[tex]a=-10.86\ m/s^2[/tex]

Explanation:

Given that,

Mass of a ball, m = 0.75 kg

It is dropped off a balcony Sandra standing on the ground below the balcony, catches the ball by exerting a 15.5N force upwards on the ball.

We need to find the acceleration of the ball while Aisha is catching it.

The net force acting on the ball is given by :

F = ma

In this case, when the ball is accelerating downward,

mg-F=ma

a is the acceleration of the ball and g is the acceleration due to gravity

So,

[tex]a=\dfrac{mg-F}{m}\\\\a=\dfrac{0.75\times 9.81-15.5}{0.75}\\\\a=-10.86\ m/s^2[/tex]

A negative sign shows that the ball is accelerating in a downward direction.

Explanation:

Answer:

The current is   [tex]I  =  1.1434*10^{-5}}\  A[/tex]

Explanation:

From the question we are told that

   The radius of the kite string is  [tex]R =  2.02 mm =  0.00202 \ m[/tex]

   The  distance it extended upward is   [tex]D =  0.823 km = 823 \  m[/tex]

   The thickness of the water layer is [tex]d = 0.506 mm  =  0.000506 \  m[/tex]

   The resistivity is  [tex]\rho =  159\ \Omega  \cdot m[/tex]

   The potential  difference is  [tex]V  =   186 MV =  186 *10^{6} \  V[/tex]

Generally the cross sectional area of the water layer is mathematically represented as

      [tex]A =  \pi r^2[/tex]

Here  r is mathematically represented as

      [tex]r =  [(R + d ) - R][/tex]

=>   [tex]r =  [(0.00202 +  0.000506 ) - 0.00202][/tex]

=>  [tex]r =  0.000506[/tex]

=>     [tex]A = 3.142 *  [0.000506]^2 [/tex]  

=>     [tex]A = 8.0447*10^{-7}\ m^2 [/tex]  

Generally the resistance of the water is mathematically represented as

    [tex]R =  \frac{\rho  * D }{A}[/tex]

=>   [tex]R =  \frac{159  *823 }{8.0447*10^{-7}}[/tex]

=>   [tex]R = 1.62662 * 10^{11} \  \Omega  [/tex]

Generally the current is mathematically represented as

      [tex]I  =  \frac{V}{R}[/tex]

=>    [tex]I  =  \frac{186 *10^{6} }{1.62662 * 10^{11}}[/tex]

=>    [tex]I  =  1.1434*10^{-5}}\  A[/tex]

The work W needed to stretch/compress a spring from rest by a distance x is

W = 1/2 kx²

where k is the spring constant.

This means the work needed to change the length of this spring by 10 cm = 0.01 m is

W = 1/2 (200 N/m) (0.01 m)² = 0.01 J

and by 15 cm = 0.015 m is

W' = 1/2 (200 N/m) (0.015 m)² = 0.0225 J

Then the total work performed on the spring by stretching from 10 cm to 15 cm is

∆W = W' - W = 0.0225 J - 0.01 J = 0.0125 J

The Final angular velocity of the sphere is : Half of its initial value ( W/2 )

Given that The kinetic energy of a rotating body ( K ) = 1/2 IW² --- ( 1 )

where : I = moment of inertia,  w = angular velocity

I = [tex]\frac{2}{5} MR^{2}[/tex]

Therefore equation ( 1 ) becomes

K = [tex]\frac{1}{5}MR^2W^2[/tex]   ----- ( 2 )

After doubling the radius

R' = 2R

K' = K

Mass = unchanged

therefore :

  [tex]\frac{1}{5}MR'^2W'^2 = \frac{1}{5}MR^2W^2[/tex]

= ( 2R )² W'² = R²W²

Resolving equation above

Hence : W' ( new angular velocity ) =  W / 2

In conclusion The Final angular velocity of the sphere is : Half of its initial value ( W / 2).

Learn more about Kinetic energy of a rotating body :https://brainly.com/question/25959744

The time required for the block to travel or complete it's half-cycle is 0.05s

Data;

This is the maximum displacement of a body from it's mean position.

Assuming a system start from its mean position i.e mean initial position and go to A/2 towards right and then back to his mean position after going through -A/2 then it is said to have completed one cycle. If the block oscillate only between -A/2 to A/2 then it will complete half cycle then it's time period will be 0.05 sec.

Learn more on amplitude here;

https://brainly.com/question/21124447

Answer:

true.

Explanation:

The forces on a thrown ball after it leaves the thrower's hand are the force of gravity .

Answer:

GIVE ME BRAINLIEST!!!!!!!!!!

Why does Venus appear so bright to our eyes?

Venus is so bright because its thick clouds reflect most of the sunlight that reaches it (about 70%) back into space, and because it is the closest planet to Earth. Venus can often be seen within a few hours after sunset or before sunrise as the brightest object in the sky (other than the moon).

it is much larger than either Mercury or Mars

Mars, the fourth planet from the sun, is the second smallest planet in the solar system; only Mercury is smaller.

it gets closer to us than does any other planet

In other words, Mercury is closer to Earth, on average, than Venus is because it orbits the Sun more closely. Further, Mercury is the closest neighbor, on average, to each of the other seven planets in the solar system.

All of the items listed helps us understand why Venus is so bright.

Venus is so bright because its thick clouds reflect most of the sunlight that reaches it (about 70%) back into space, and because it is the closest planet to Earth. Venus can often be seen within a few hours after sunset or before sunrise as the brightest object in the sky (other than the moon).

its sulfuric acid cloud cover reflects almost 60% of the sunlight

Clouds. Venusian clouds are thick and are composed mainly (75–96%) of sulfuric acid droplets. These clouds obscure the surface of Venus from optical imaging, and reflect about 75% of the sunlight that falls on them. The geometric albedo, a common measure of reflectivity, is the highest of any planet in the Solar System

it lies closer to the Sun than we do, so sunlight is more intense there

When it is winter in the northern half of Earth, the southern hemisphere, tilted toward our Sun, has summer. During fall and spring, some locations on Earth experience similar, milder, condition.

Answer:

if an astronaut weighs 981 N on Earth and only 160 N on the Moon, then what is his mass on the Moon?

ANSWER 256n

The mass of the astronaut on the moon as compared to the earth will be  [tex]M_m=97.85\ lg[/tex]

The mass of any substance or body is defined as how much quantity of matter is present.

Now it is given in the question:

Weight of the astronaut on earth  [tex]W_E=981\ N[/tex]

Weight of the astronaut on earth  [tex]W_M=160\ N[/tex]

The mass of the astronaut on the moon will be calculated as:

Weight on the moon will be  given as:

[tex]W_M=M_M\times g_m[/tex]

Weight on the earth will be given as:

[tex]W_E=M_E\times g_e[/tex]

The ratio of the gravity of the earth to the moon is given as

[tex]\dfrac{g_e}{g_m} =\dfrac{9.81}{1.62} =6[/tex]

The mass of the earth will be calculated as

[tex]W_E=M_E\times ge[/tex]

[tex]M_E=\dfrac{981}{9.81} =100\ kg[/tex]

Now taking the ratio of the weight of the earth to the moon :

[tex]\dfrac{W_E}{W_M} =\dfrac{M_E\times g_e}{M_M\times g_m}[/tex]

[tex]M_M= \dfrac{W_M\times M_E}{W_E} \times \dfrac{g_e}{g_m}[/tex]

Now by putting the value in the formula we get;

[tex]M_M=\dfrac{160\times 100}{981} \times 6=97.85\ kg[/tex]

Thus the mass of the astronaut on the moon as compared to the earth will be  [tex]M_m=97.85\ lg[/tex]

To know more about the Mass of the moon follow

https://brainly.com/question/13499822

Answer:

D refraction

Explanation:

I had a question like this so and I got it right so I believe this is the correct answer

Answer:

the after image is showing precipitation. i dont know what the before image is though.

Explanation:

Answer:

V * ρB = WB        volume of ball * density of ball = weight of ball

V * ρW = Ww        volume of ball * density of water = buoyant force

ρB / ρW = WB / Ww = 80 / 20 = 4    water provides 20 N of buoyant force

ρB = 4 ρW = 4 gm/cm^3

ρW = 1000 kg / m^3 = 1 gm / cm^3

(1 gm/cm^3) = .001 kg / .000001 m^3 = 1000 kg/^3

ρW = 1 gm / cm^3

multiply by 9.8 or 980 to get weight densities

In this case

4 * 1000 kg/m^3 * 9.8 m/s^2 = 39200 N /m^3      weight density of ball

Answer:

Tarala

Explanation:

It's Nepali

Answer:

Fluid meaning in Nepali is Tarala (तरल )

Recall that

v² - u² = 2 a ∆x

where u and v denote initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled.

Then we get

(18 m/s)² - (26 m/s)² = 2 a (52 m)

a = ((18 m/s)² - (26 m/s)²) / (104 m)

a ≈ -3.4 m/s²