What does Weber's Law about 'just noticeable differences' predict about how much someone has to change the brightness of a light before we can notice the difference? a. It depends on how bright the light was in the first place - the brighter it was, the less change is needed before we realize it. b. It depends on how long we have been looking at the light - the longer we have been looking, the more change is needed. c. It is always the same amount - 7 lux. d. It depends on how bright the light was in the first place - the brighter it was, the more change is needed before we realize it.

the missing word is clockwise moment. I hope this helps good luck

Answer:

In Celsius it is 426.667

In Kevin it is 699.817

Explanation:

Answer:

Explanation: someone help me please you want free points right

Answer:

the last one

Explanation:

ii took the quiz and it was right... i think

Answer:

Low mass stars are: hydrogen burning in the core while on the Main Sequence. As the hydrogen fuel runs out, extreme pressure raises the temperature to 100 million degrees, where helium burning becomes possible.

Answer:

The tension in the web is 0.017738 N

Explanation:

Net Force

The net force exerted on an object is the sum of the vectors of each individual force applied to an object.

If the net force equals 0, then the object is at rest or moving at a constant speed.

The spider described in the question is hanging at rest. It means the sum of the forces it's receiving is 0.

A hanging object has only two forces: The tension of the supporting string (in our case, the web) and its weight. If the object is in equilibrium, the tension is numerically equal to the weight:

T=W=m.g

The mass of the spider is m=1.81 gr = 0.00181 Kg, thus the tension is:

[tex]T = 0.00181\ Kg\cdot 9.8\ m/s^2[/tex]

[tex]T=0.017738\ N[/tex]

The tension in the web is 0.017738 N

Answer:

36 N/m

Explanation: I got the answer right on the test ight noo cap

Answer:

30.63 m

Explanation:

From the question given above, the following data were obtained:

Total time (T) spent by the ball in air = 5 s

Maximum height (h) =.?

Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:

Total time (T) spent by the ball in air = 5 s

Time (t) taken to reach the maximum height =.?

T = 2t

5 = 2t

Divide both side by 2

t = 5/2

t = 2.5 s

Thus, the time (t) taken to reach the maximum height is 2.5 s

Finally, we shall determine the maximum height reached by the ball as follow:

Time (t) taken to reach the maximum height = 2.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

h = ½gt²

h = ½ × 9.8 × 2.5²

h = 4.9 × 6.25

h = 30.625 ≈ 30.63 m

Therefore, the maximum height reached by the cannon ball is 30.63 m

Answer: poor construction of houses

Explanation:

Majority of the people that died in Iran were as a result of poor building methods coupled with the fact that there was lack of proper regulation.

California experienced a similar earthquake but due to safer construction methods, about three people died.

Due to population boom in Iran and house shortage, this resulted in builders building cheap houses which were not strong enough.

Answer:

ok can u m a e the question make more sense like break ot down cs i wanna give u a answer but i dont really understand the question

Answer:

478.75 J

Explanation:

W=force* displacement

constant speed= (a=0) net F=0

Horizontal component of tension

Tcosx

125Ncos40= 95.76 N

W= (95.76 N)(5 m)

=478.75 J

The work done in moving the crate across the given distance is 478.75 J.

The given parameters;

The work done in moving the crate is the product of the horizontal component of the tension and the distance through which the crate is moved.

The work-done in moving the crate is calculated as;

W = Tcos(Ф) x d

W = 125cos(40) x 5

W = 478.75 J.

Thus, the work done in moving the crate across the given distance is 478.75 J.

Learn more here: https://brainly.com/question/19498865

Answer:

A) v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]

B)√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

C) µ_s = tan θ

D) µ_s = 0.4663

Explanation:

A) The forces acting on the car will be;

Force due to friction; F_f

Force due to Gravity; F_g

Normal Force; F_n

Now, let us take the vertical direction to be j^ and the direction approaching the centre to be i^ downwards and parallel to the road surface by k^.

Also, we will assume that initially, F_n is in the negative k^ direction and that it will have a maximum possible value of; F_f = µ_s × F_n

Thus, sum of forces about the vertical j^ direction gives;

ΣF_j^ = F_n•cos θ − mg + F_f•sin θ = 0

Since F_f = µ_s × F_n ;

F_n•cos θ − mg + (µ_s × F_n × sin θ) =0

F_n = mg/[cos θ + (µ_s•sin θ)]

Also, sum of forces about the centre i^ direction gives;

ΣF_i^ = F_n(sin θ + (µ_s•cos θ)) = mv²/r

Plugging in formula for F_n gives;

ΣF_i^ = [mg/[cos θ + (µ_s•sin θ)]] × (sin θ + (µ_s•cos θ)) = mv²/r

Making v the subject gives;

v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]

B) What we got in a above is the minimum speed the car can have while going round the turn.

The maximum speed will be gotten by making the frictional force(F_f) to point in the positive k^ direction. This means that F_f will be negative.

Now, if we change the sign in front of F_f in the equation in part a that led to the minimum velocity, we will have the maximum as;

v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

Thus the range is;

√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

C) For the minimum speed to be 0, it implies that F_f will be in the negative k^ direction. Thus, Sum of the forces in the k^ direction gives;

ΣF_k^ = mg(sin θ - µ_s•cos θ) = 0

Thus;

mg(sin θ - µ_s•cos θ) = 0

Making µ_s the subject gives;

µ_s = sin θ/cos θ

µ_s = tan θ

D) If θ = 25.0°;

Thus;

µ_s = tan 25

µ_s = 0.4663

Answer:

Broadcasting is the method, not sure about the stage it is done in

Explanation:

Answer:

Broadcasting is the first (blank) second (blank) is Cultivation.

Explanation:

I took the test & got this answer correct.

Answer:

a)  vₓ = 100 + 7.07 = 107.07 knot     (East) ,   v_{y} = 7.07 knot            (North)

b) v_wind_ North = 0 ,   v_wind_West = 61.12 knot      (West)

Explanation:

a) This is a vector velocity addition problem, they tell us that the plane goes East at 100 Knots and the wind goes North East at 10 Knots

the components ask the total speed of the plane.

For this we decompose the wind speed

            cos 45 = v₂ₓ / V₂

            sin 45 = [tex]v_{2y}[/tex] / v₂

            v₂ₓ = v₂ cos 45

           v_{2y} = v₂ sin 45

            v₂ₓ = 10 cos 45 = 7.07 m / s

            v_{2y} = 10 sin45 = 7.07 m / s

the speed of the plane is

            vₓ = v_plane + v₂ₓ

            v_{y} = v_{2y}

            vₓ = 100 + 7.07 = 107.07 knot     (East)

            v_{y} = 7.07 knot            (North)

speed is

            v = (107.07 i ^ + 7.07 j ^) knot

b) Estimated time for Target VOR t = 10 min

for a time of t = 10 min and 15 s, it is at a distance of d = 300 m from the VOR in an easterly direction

find the average speed of time on the ride

in the North direction there is no deviation so the average wind speed is zero

            v_wind_ North = 0

In the east direction

as the estimated time was 10 min and the real time for the distance is 10 min 15 s

the time difference is t = 15 s to travel d = 300 m

with these data we can calculate the speed of the plane

              v = x / t

              v = 300/15

              v = 20 m / s

let's reduce this speed to knot

             v_real = 20 m / s (1knot / 0.5144 m / s) = 38.88 knots

therefore the wind speed is

              v_tral = v_plane + v_wind_Este

              v_wind_Este = v_real - v_avión

               v_wind_Este = 38.88 - 100

                v_wind_Este = -61.12 knot

the negative sign indicates that the wind is going west

                v_wind_West = 61.12 knot (West)

3)

* The compass is not calibrated, to correct the magnetic deviation, therefore it gives an appreciable error in the direction

* The wind speed is taken when leaving, but there is no constant monitoring, so a change in direction or wind speed in the trajectory can significantly affect the results.

* The aircraft may noir throughout the trajectory at level, due to pitch errors

Answer:

Usbe

Explanation:

Answer:

it has to be b contact forces

Explanation: because the two objects are next to each other but they still have force

Answer:

c

Explanation:

Answer:

the material

Explanation:

weight is defined as the amount of force on the object because of gravity. ping pong balls and bouncy balls are made out of different materials that are different weights. most bouncy balls are also not hollow, unlike ping pong balls. these factors affect the weight of these objects.

Answer:

a) v = 7.28 m/s

b) t = 0.74 s

Explanation:

a) The initial speed of the ball can be calculated using the following equation:

[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]

Where:

[tex]V_{f}[/tex] is the final speed = 0

[tex]V_{0}[/tex] is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 2.70 m

[tex] V_{0} = \sqrt{2gh} = \sqrt{2*9.81 m/s^{2}*2.70 m} = 7.28 m/s [/tex]

Hence, the initial speed of the ball is 7.28 m/s.

b) To find the time that takes the balls to reach the ceiling we can use the next equation:

[tex] V_{f} = V_{0} - gt [/tex]

[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{7.28 m/s}{9.81 m/s^{2}} = 0.74 s [/tex]

Therefore, the time it takes for the ball to reach the ceiling is 0.74 s.

I hope it helps you!

Answer:

Impact Craters.

Explanation:

An impact crater can be defined as a circular depression that is caused by impact on any planet or asteroids or any other celestial body's surface. When smaller body in galaxy impacts these larger bodies, they form a circular depression on it's surface.

This is a major feature found in solid object such as the Moon, Mercury, etc.

Therefore, the feature that a planet does not require is an impact crater. As other features are important to define a planet. Thus correct option is C.

Answer:

At 40 deg  Vy = V sin 40

at 70 deg Vy = V sin 70

So the ball launched at 70 deg has the greatest vertical velocity and will remain in the air the longest:

Since t = Vy / g    time for to reach zero vertical velocity and also the time for the ball to reach velocity Vy on the downward path

Mathis kicked a ball on a level surface at 30∘ and 60∘ with the same total speed as shown below.

Which launch angle results in the greater maximum height for the ball?

Answer: CORRECT (SELECTED)

60

Answer:

a)    F = -1.82 10⁻¹⁵ N,  b) K = 9.1 10⁻¹⁶ J

Explanation:

a) To calculate the force between the nucleus and the electrons, let's use the Coulomb equation

           F = k q Q / r²

as the nucleus occupies a very small volume compared to electrons, we can suppose it as punctual

let's calculate

          F = 9 10⁹ (-1.6 10⁻¹⁹) (79 1.6 10⁻¹⁹) / (10⁻¹⁰)²

          F = -1.82 10⁻¹⁵ N

b) they ask us for kinetic energy

let's use Newton's second law

         F = m a

acceleration is centripetal

         a = v² / r

we substitute

         F = m v² / r

         v = √ (F r / m)

         v = √ (1.82 10⁻¹⁵ 10⁻¹⁰ / 9.1 10⁻³¹)

         v = √ (0.2 10⁻¹⁶)

         v = 0.447 10⁸ m / s

kinetic energy is

          K = ½ m v²

          K = ½ 9.1 10⁻³¹ (0.447 10⁸)²

          K = 0.91 10⁻¹⁵ J

          K = 9.1 10⁻¹⁶ J

Answer:

b

Explanation:

when a ball falls from the sky, it falls down fast due to no friction.

Answer: positively charged

Explanation: basically when a neutral atom gets some electrons it becomes negatively charged and when a neutral atom loses elections it's positively charged Ik science is weird lol.

I hope this was the answer you were looking for.

Answer:

(assume moving in the positive direction, vi = + 20 m/s)

Explanation:

Answer:

(a). The magnitude of the net force is [tex](2.1\times10^{-18}\ N)k[/tex]

(b). The magnitude of the net force is [tex](4.23\times10^{-19}\ N)k[/tex]

(c). The magnitude of the net force is [tex](8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]

Explanation:

Given that,

Magnetic field [tex]B=-3.54\times10^{-3}i\ T[/tex]

Velocity = 2230j m/s

We know that,

The net force acting on the proton is equal to the sum of electric and magnetic force.

[tex]F=F_{e}+F_{B}[/tex]

(a). If the electric field is in the positive z direction and has a magnitude of 5.25 V/m,

We need to calculate the magnitude of the net force acting on the proton

Using formula of net force

[tex]F_{net}=e(E+v\times B)[/tex]

Put the value into the formula

[tex]F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(j\times i))[/tex]

[tex]F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(-k))[/tex]

[tex]F_{net}=(2.1\times10^{-18}\ N)k[/tex]

(b). If the electric field is in the negative z direction and has a magnitude of 5.25 V/m,

We need to calculate the magnitude of the net force acting on the proton

Using formula of net force

[tex]F_{net}=e(E+v\times B)[/tex]

Put the value into the formula

[tex]F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(j\times i))[/tex]

[tex]F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(-k))[/tex]

[tex]F_{net}=(4.23\times10^{-19}\ N)k[/tex]

(c). If the electric field is in the positive x direction and has a magnitude of 5.25 V/m

We need to calculate the magnitude of the net force acting on the proton

Using formula of net force

[tex]F_{net}=e(E+v\times B)[/tex]

Put the value into the formula

[tex]F_{net}=1.6\times10^{-19}(5.25i+2230\times-3.54\times10^{-3}(j\times i))[/tex]

[tex]F_{net}=(8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]

Hence, (a). The magnitude of the net force is [tex](2.1\times10^{-18}\ N)k[/tex]

(b). The magnitude of the net force is [tex](4.23\times10^{-19}\ N)k[/tex]

(c). The magnitude of the net force is [tex](8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]

Answer:

114 m/s

Explanation:

see the image below

Answer:

b- how much work can be done in a given time

how much work can be done in a given time

Answer:

Since the ball was thrown horizontally, there was no vertical component in that force. and hence, the initial vertical velocity of the ball is 0 m/s and the initial horizontal velocity is r.

We are given:

initial velocity  (u) = 0 m/s     [vertical]

final velocity (v) = v m/s  [vertical]

time taken to reach the ground (t) = t seconds

acceleration (a) = 10 m/s/s   [vertical , due to gravity]

height from the ground (h) = 130 m

displacement (s) = 53 m [horizontal]

Solving for time taken:

From the third equation of motion:

s = ut + 1/2 at²

130 = (0)(t) + 1/2 * (10) * t²

130 = 5t²

t² = 26

t = √26 seconds  or   5.1 seconds

Final Horizontal velocity of the ball

Since the horizontal velocity of the ball will remain constant:

the ball covered 53 m in 5.1 seconds [horizontally]

horizontal velocity of the ball = horizontal distance covered / time taken

Velocity of the ball = 53 / 5.1

Velocity of the ball = 10.4 m/s

Answer:

51.51519 m/s

Explanation:

Given: [tex]a_{x} =0[/tex] [tex]a_{y} -g[/tex] [tex]v_{yo} =0[/tex] [tex]x_{o} =0[/tex] [tex]x=53[/tex][tex]y_{o} =130[/tex]

X-direction                           | Y-direction

[tex]x=x_{o} +v_{xo}t[/tex]                         | [tex]y=y_{o} +v_{yo}t+\frac{1}{2}a_{y}t^2[/tex]

[tex]53=0v_{xo}(5.15078)[/tex]                 | [tex]0=130+\frac{1}{2}(-9.8)t^2[/tex]

[tex]53=v_{xo} (5.15078)[/tex]                    | [tex]-130=-4.9t^2[/tex]

[tex]\frac{53}{5.15078} =v_{xo}[/tex]                             |  [tex]\sqrt{\frac{-130}{-4.9} }=\sqrt{t^2}[/tex]

[tex]10.2897=v_{xo}[/tex]                            | [tex]5.15078=t[/tex]

[tex]v=\sqrt{v_{y}^2+ v_{x}^2}[/tex]                            | [tex]v_{y}^2 =v_{yo}+2a_{y} d[/tex]

[tex]v=\sqrt{(50.27771)^2+(10.2897)^2}[/tex] | [tex]\sqrt{v_{y}^2} =\sqrt{2(-9.8)(0-130)}[/tex]

[tex]v=51.51519 m/s[/tex]                        | [tex]v_{y}=50.47771[/tex]