Please help me!!! I'm so lost!!

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:

gfdvcfffddfgffffdrddgfddddghtscgvfrggxfhxdg

Answer:

-.57

Explanation:

Slope from the points (0,6) and (7,2)

2-6/7-0=-.57

Answer:

C

Explanation:

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

The point A with the higher mass of 10kg will have the greatest moment of inertia.

This can be defined as the resistance of any physical object to a change in its velocity.

Masses which have the highest values and are further away form the axis of rotation have the greatest moment of inertia which was why point A was chosen as the most appropriate choice.

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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]

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

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

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

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).

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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.

(a) The work done in strecthing the spring is 122.5 N.

(b) The work done in compressing the spring is -122.5 N.

The work done in stretching a spring is positive since the applied force acts in the same direction as the displacement.

The work done in strecthing the spring is calculated as follows;

W = ¹/₂kx²

W = ¹/₂ x (5 x 10⁴) x (0.07)²

W = 122.5 N

The work done in compressing the spring is negative, since the applied force acts in opposite direction to the displacement of the spring.

The work done in compressing the spring is calculated as follows;

W = -¹/₂kx²

W = -¹/₂ x (5 x 10⁴) x (0.07)²

W = -122.5 N

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Answer:

1 m/s²

Explanation:

Let velocity at A be v,

Use d = v*t + at²2/2

From A to B,

6 = v(2) + a(2²)/2 = 2v + 2a, so

v + a = 3   (1)

From A to C,

6 + 10 = v(4) + a(4²)/2

16 = 4v + 8a  

or, v + 2a = 4   (2)

From (1) (2), get

a = 1 m/s²

Answer:

v=10 m/s

Explanation:

Given that,

Speed of a drone is 8 m/s due East with respects to the wind

Speed of a drone 6 m/s due North with respect to the ground.

We need to find the speed of the drone with respect to the ground. Let v is the speed. It is given by the resultant of two speeds.

[tex]v=\sqrt{8^2+6^2} \\\\v=10\ m/s[/tex]

So, the speed of the drone with respect to the ground is 10 m/s.

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]

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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:

Explanation:

For example 1, there is no friction, so the acceleration in example 1 is:[tex]\sum F=F=100=ma \rightarrow a=\frac{100}{50}=2 ms^{-2}[/tex]

For example 2, there is a friction, so the acceleration in example 2 is:

[tex]\sum F=F-f=100-40 = 60 = ma\rightarrow a_{2}=\frac{60}{50}=1.2 ms^{-2}[/tex]

Here we can say that "according to Newton's 2nd Law, the box in example 2 has less acceleration than in the example 1". So, the claim is denied.  

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:

60m/s

Explanation:

v=u+at

v=36+(2×12)

v=36+24

v=60m/s

Explanation:

The time needed for the block to travel from -A/2 to A/2 is mathematically given as

t=0.5sec

Generally, when the start point is the mean intial position and it goes from  A/2  down through the right goes through and fro back to the mean position, and this is after it must have gone through -A/2. Hence it has finished one cycle (0.1 sec.)

In conclusion, for an -A/2 to A/2 oscillation, then it will finish half-cycle given time period to be

t=1/2

t=0.5sec

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Answer:

9 :)

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:

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]

Answer:

Is it only me who can't understand your question?

Pls give the full question ...

Hope you have a great day!

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:

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:

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

Answer:

Tarala

Explanation:

It's Nepali

Answer:

Fluid meaning in Nepali is Tarala (तरल )