form a hypothesis to explain the shape of this graph

Answer:

4

Explanation:

We can say the motion of the object is stopped when its velocity is 0. That means we just have to check how many times the line of the graph intersects the x-axis. Hence, we can say that the object of the motion is stopped 4 times.

Answer:

The dependent variable is the variable that is studied while the independent variable is the variable that is being manipulated

Answer: In this experiment, the

✔ fan speed

was intentionally manipulated. This was the independent variable.

The dependent variable measured was the

✔ acceleration

.

Explanation:

Answer:

I think it depends on the purpose of using Gamma rays. It can be both good and bad

Answer: 7.364 x 10^-4 N

Explanation:

[tex]$$Given:$$\begin{aligned}m_{1} &=7.5 \times 10^{5} \mathrm{~kg} \\m_{2} &=9.2 \times 10^{7} \mathrm{~kg} \\r &=2.5 \times 10^{3} \mathrm{~m}\end{aligned}$$Here, $m_{1}, m_{2}$, and $r$ are the first mass, the second mass, and the distance[/tex]

[tex]$$The gravitational force exerted between the masses is determined as follows.$$F=G \frac{m_{1} m_{2}}{r^{2}}$$Here, $G$ is the gravitational constant.Substitute the known values.$$\begin{aligned}F &=\left(6.67 \times 0^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2}\right) \frac{\left(7.5 \times 10^{5} \mathrm{~kg}\right)\left(2.5 \times 10^{7} \mathrm{~kg}\right)}{\left(2.5 \times 10^{3} \mathrm{~m}\right)^{2}} \\&\bold{=7.364 \times 10^{-4} \bold{\mathrm{~N}}}\end{aligned}$$[/tex]

The phenomenon known as the Emission Spectra.

Emission spectrum is the phenomenon provides evidence that the hydrogen atom has discrete energy levels.

Emission spectrum is the spectrum of the different energy levels. The occurrence of emission spectrum gives proof that the hydrogen atom has distinct energy levels.

Emission spectrum is the phenomenon provides evidence that the hydrogen atom has discrete energy levels

Hence, the phenomenon is known as emission spectrum.

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The angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].

The velocity of a particle when moving in the circular path.

Let speed of the bug with respect to ground is u.

Speed of bag with respect to ring will be

v = u - (- Rω) =

Then, u = v- Rω...............(1)

Angular momentum of ring and bug will remain conserved.

Initial  momentum: L ring + Lbug =0

Final  momentum:  -2m₁ R²ω + m₂uR =0...............(2)

Using equation (1) and (2), the angular velocity expression will be

ω =[m₂v / {(2m₁ +m₂)R}] in positive z direction

Thus, the angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].

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Hello!

We can use the following equation for magnetic force on a moving particle:

[tex]F_B = qv \times B[/tex]

[tex]F_B[/tex] = Magnetic Force (? N)

q = Charge of particle (1.28 × 10⁻⁵C)
v =velocity of particle (5.63× 10⁷m/s)

B = Strength of magnetic field (8.91 × 10⁻⁴T)

This is a cross-product, so the magnetic force depends on the SINE of the angle between the particle's velocity vector and the magnetic field vector.

Since the charge is moving PERPENDICULAR to the field, the angle between the velocity and magnetic field is 90°. The sine of 90° = 1, so we can simplify the equation to:

[tex]F_B = qvB(1) = qvB\\[/tex]

Plug in the values and solve.

[tex]F_B = (1.28*10^{-5})(5.63*10^7)(8.91*10^{-4}) = \boxed{\text{A. } 0.6421 N}[/tex]

Answer:

Different colour lights (RBG) uses additive properties to bring light of a certain color to our eyes.

Explanation:

:))

Answer: The answers are, 1) B. 2) A.  3) C.   4) C.

here is the proof maybe u will understand it more than the numbers and letters lm.ao

Explanation:

The red ball with arrows pointing away is a negatively charged particle, the blue ball is a positively charged molecule, and the grey ball is an uncharged molecule. Electric fields don't move uncharged molecules.

Analyse each scenario to identify representations:

A negatively charged particle is the red ball in a box with all arrows pointing away from it. The red ball generates an electric field, and negatively charged particles radiate electric field lines. The arrows indicate how a positive test charge would move in the electric field.

A positively charged molecule is the blue ball in a box with all arrows pointing to it. An electric field converges on the blue ball, signifying its destination. The electric field lines point towards positively charged particles, indicating the direction a positive test charge would move.

A boxed grey ball depicts an uncharged molecule. The ball has no arrows. Uncharged molecules do not emit or experience electric fields. The grey ball is electrically neutral because it lacks electric pitch lines.

An electric field will not move the grey ball in a box (uncharged molecule). Electric fields do not interact with uncharged molecules. Thus, an electric field will keep the grey ball still.

Science and engineering require understanding these representations and their electric field behaviour. Scientists and engineers can create and manipulate electrical systems, electronics, and other technologies that use electric charge interactions by recognising charged particles, uncharged molecules, and electric field effects.

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Answer:

the one with v = 25 m/s

Explanation:

Momentum = m * v

 if they both have the same mass (15000 kg) , then the one with the higher v has more momentum...I think A= 25 m/s

a)The speed of the ball at the time of impact will be 8.1 m/sec.

b)The time taken by the ball impacts the surface of trampoline will be 1.1 second.

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. Its SI unit is m/sec.

Given data;

g( gravitational acceleration) = 9.81 m/s2

u (initial velocity of ball) = 2.5 m/s

h₁(initial height of the ball) = 3.0 m

v (speed of the ball at the time of impact ) ?

The time taken by ball for upward journey = u / g

t₁ = u / g

t₁  = (2.5 m/s) / (9.81 m/s2) =

t₁  =0.2548 s

From second equation of motion,upward vertical distance covered by ball is;

[tex]\rm h_2= \frac{1}{2} gt^2\\\\ h_2=\frac{1}{2} \times 9.81 \times (0.2548)^2\\\\ h_2=0.3184[/tex]

The total distance in the downward direction travelled;

[tex]\rm h_3=h_1+h_2 \\\\ h_3=3.0+0.3184 \\\\ h_3=3.3184 \ m[/tex]

The time taken for the downward motion is;

[tex]\rm t_2 = (\frac{2 \times h_3 }{g} )^\frac{1}{2}\\\\ t_2 = 0.825 \ m/sec[/tex]

The speed of the ball at the time of impact is found as;

[tex]\rm V=gt_2 \\\\ V= 9.81 \times 0.8225 \\\\ V= 8.07 \ m/sec[/tex]

The total time of the flight of ball is;

[tex]\rm t_3 = t_1+t_2 \\\\ t_3=0.2548+0.8225 \\\\ t_3=1.0774 \ sec[/tex]

Hence,the speed of the ball at the time of impact will be 8.1 m/sec and time taken by the ball impacts the surface of trampoline will be 1.1 second.

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Answer:

120 v

Explanation:

The two resistors have an equivalent of    20 * 30 /(20+30) = 12 ohms

  10 amps of current in the circuit

v = ir

  = 10 * 12 = 120 volts

Here is another way:

The two resistors are in prallel so the voltae across both is the same

  use the one on the right      v = ir     =   4 x 30 = 120 v

(a) The speed  of puck A before the collision is determined as 0.847 m/s.

(b) The change in the total kinetic energy of the system that occurs during the collision is 0.009 J.

The speed of puck A before the collision is calculated from the principle of conservation of linear momentum as follows;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.251u₁ + 0.371(0) = 0.251( -0.12) + 0.371(0.654)

0.251u₁ = 0.2125

u₁ = 0.2125/0.251

u₁ = 0.847 m/s to the right

Initial kinetic energy = ¹/₂ x 0.251 x (0.847)² + ¹/₂ x 0.371 x (0) = 0.09 J

Final kinetic energy = ¹/₂ x 0.251 x (-0.12)² + ¹/₂ x 0.371 x (0.654)² = 0.081 J

ΔK.E = 0.081 - 0.09 = 0.009 J ≈ 0 J

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Answer:

"light year" - about 6 million million miles

The nearest star is alpha Centauri - about 24 million million miles away

Answer:

Total resistance in parallel is Rp

So (1/Rp) = (1/12)+(1/16)+(1/20)

Rp = ?

impedance is 25

25 = Rp + Rs

put the value of Rp you'll get the answer.

Hello!

A)
We can solve for the reactances of each element using the following equations.

Capacitive reactance:
[tex]X_C = \frac{1}{\omega C}[/tex]

[tex]X_C[/tex] = Capacitive reactance (Ω)
ω = Angular frequency (2000 rad/sec)

C = Capacitance (2 μF)

Plug in the given values and solve.

[tex]X_C = \frac{1}{(2000)(0.000002)} = \boxed{250 \Omega}[/tex]

Inductive reactance:
[tex]X_L = \omega L[/tex]

[tex]X_L[/tex] = Inductive reactance (Ω)
L = Inductance (70 mH)

Solve:
[tex]X_L = 2000 * 0.07 = \boxed{140 \Omega}[/tex]

We can solve for impedance using the following equation:
[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]

Z = Impedance (Ω)

R = Resistance (200Ω)

Find the impedance using the values above and the given resistance:
[tex]Z = \sqrt{200^2 + (140 - 250)^2} = \boxed{228.254 \Omega}[/tex]

B)
The maximum current amplitude is found using Ohm's law:

[tex]I_{Max} = \frac{V_{Max}}{Z}[/tex]

[tex]V_{Max}[/tex] = 80V

Z = 228.254 Ω

Solve.

[tex]I_{Max} = \frac{80}{228.254} = \boxed{0.3505 A}[/tex]

To find the max voltages across each element, we can also use Ohm's Law.

For the resistor:
[tex]V_{R, max} = i_{Max} R\\\\V_{R, max} = (0.3505)(200) = \boxed{70.097 V}[/tex]

For the capacitor:

[tex]V_{C, max} = i_{Max}X_C\\\\V_{C, max} = (0.3505)(250) = \boxed{87.622 V}[/tex]

For the inductor:
[tex]V_{L, max} = i_{Max}X_L\\\\V_{L, max} = (0.3505)(140) = \boxed{49.068 V}[/tex]

C)
The maximum voltage across each element is greater than 80 V because of the transfer of energy between the capacitor and inductor that occur at different times (energy is stored and released) and being in addition to the voltage of the power source.

The height of the mansion that the assassin shot the target from is determined as 1,164.9 ft.

The time of motion of the bullet is calculated as follows;

-h = vsinθ(t)  - ¹/₂gt²

-6.1 = (vt) sin63  - ¹/₂(32)t²

-6.1 = 0.89vt - 16t² ---(1)

X = vcosθ t

294 = (vt)cos63

294 = 0.454vt

vt = 294/0.454

vt = 647.577 ---(2)

solve (1) and (2) together;

-6.1 = 0.89(647.577) - 16t²

16t² = 576.344 + 6.1

16t² =  582.44

t² = 582.44/16

t² = 36.4

t = √36.4

t = 6.03 s

v = 647.577/t

v = 647.577/6.03

v = 107.4 ft/s

h = vt + ¹/₂gt²

h = (107.4 x sin63 x 6.03) + ¹/₂(32)(6.03)²

h = 1,158.8 ft

H = 1,158.8 ft + 6.1 ft = 1,164.9 ft

Thus, the height of the mansion that the assassin shot the target from is determined as 1,164.9 ft.

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Answer:

okay

Explanation:

power is rate of doing work

power is workdone over time taken

but you don't have information concerning the force

Power = work/time

Work = force x distance

Force = mass x acceleration

So Power = (mass x acceleration x distance) / (time)

acceleration = 8m/s / 2sec  =  4 m/s²

distance = (avg speed) x (time) = 8 m

time = 2 sec

we don't know the runner's mass

Power = (mass) x (4 m/s²) x (8 m) / (2 sec)

Power = (16 x mass in kg) watts

=============================================

Slightly easier way to do it:

Power = (runner's kinetic energy at the end) / (time)

Power = (1/2 m v²) / (time)

Power = (1/2 mass) (64 m²/s²) / (2 sec)

Power = (16 x runner's mass in kg) watts

Answer: 1.55 x 10 -19J on top

Explanation:

Find energy

[tex]\\ \rm\Rrightarrow E=-13.6\left(\dfrac{1}{6^2}-\dfrac{1}{4^2}\right)[/tex]

[tex]\\ \rm\Rrightarrow E=-13.6\left(\dfrac{1}{36}-\dfrac{1}{16}\right)[/tex]

[tex]\\ \rm\Rrightarrow E=-13.6(0.0278-0.0625)[/tex]

[tex]\\ \rm\Rrightarrow E=-13.6(-0.347)[/tex]

[tex]\\ \rm\Rrightarrow E=0.47eV[/tex]

Now

[tex]\\ \rm\Rrightarrow E=h\nu[/tex]

[tex]\\ \rm\Rrightarrow \nu=\dfrac{E}{h}[/tex]

[tex]\\ \rm\Rrightarrow \nu=\dfrac{0.47\times 10^{-19}}{6.626\times 10^{-34}}[/tex]

[tex]\\ \rm\Rrightarrow \nu=0.071\times 10^{15}[/tex]

[tex]\\ \rm\Rrightarrow \nu=7.1\times 10^{13}Hz[/tex]

Answer:

S = V t + 1/2 g t^2    where the ball has a speed of V and falls for .2 sec

6 = V t + 4.9 * .04 = V * .2 + .2

V = (6 - .2) / .2 = 29 m/s      speed entering area

T = V / g = 29 / 9.8

T = 2.96 sec      time to reach speed V

H = 1/2 g t^2   time to fall a distance H

H = 4.9 * 2.96^2 = 42.9 m      height from which ball was dropped

Check using first equation:

S = 29 * .2 + 4.9 + .2^2 = 6 m

Jenny and Tina play jumping rope,Ben and Ted plays soccer and football and Bovet and Cris perform aerobic exercises​ are the following situations is reaction time exhibited. Option A,B and D are correct.

The reaction time of an organism is a measurement of how quickly it responds to a stimulus.

The time interval between the presentation of the stimulus and the manifestation of a suitable voluntary response in the subject is defined as RT.

In the following situations is reaction time exhibited is;

A. Jenny and Tina play jumping rope.

B. Ben and Ted play soccer and football .

D. Bovet and Cris perform aerobic exercises​.

Hence,option A,B and D are correct.

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Answer:

i think its 400 miles driven

Explanation:

attempted to do it...

Answer:

12 cm/s

Explanation:

Quite simply, you are looking for  cm/s

  so   60 cm / 5s   = 12 cm/s

Answer:

12 cm/second

Explanation:

To calculate the speed of an object, simply obtain the distance traveled by the ball and divide it by the total time it took to travel 60 centimeters.

⇒ Distance/Time = Speed

In this case, the ball traveled 60 centimeters in 5 seconds.

Therefore, the speed of the ball is;

60 centimeters (Distance)/5 seconds (Time) = 12 cm/second

Answer:

5 m/s

Explanation:

Rate of movement is   distance / time

6 m / 1.2 s = 5 m/s