Question 2: Start-Up
Now you will use another Tracker experiment to examine the displacement, velocity, and acceleration of a car in the x (horizontal) direction. To
begin your investigation, open the Tracker Car Start-up experiment.
Part A
Play () the video. At the end, rewind (H) and step forward (►) one frame ata time to observe the step-by-step changes in position. Based on
your observations, describe the car's displacement (distance and direction from the starting place) over time.

The matching of each method of transferring electric charge with the correct description should be explained below.

The friction means the transfer of electric charge via rubbing. The conduction means the transfer of electric charge via the direct contact.

Also, the induction means the transfer of the electric charge without the direct contact.

In this way it should be matched.

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Answer:

this is the answer that is correct

Answer:

75kg

Explanation:

[tex]F=mg[/tex]

[tex]m=\frac{F}{g}[/tex]

[tex]m=\frac{735}{9.8}[/tex]

[tex]m=75kg[/tex]

Therefore, the answer is the third option 75 kg

Answer:

λ = 698 nm (Approx)

Explanation:

Given:

Width d = 0.002 cm = 0.002 x 10⁻² nm

θ = 2°

Computation:

d sinθ = mλ

(0.002 x 10⁻²) sin2° = mλ

λ = 698 nm (Approx)

Answer:

I think C

Explanation:

if im wrong then what is it im taking the exam too

Answer: C. She can increase the number of wire loops.

Explanation:

[tex] \small\bf \: let \: the \: distance \: of \: the \: man \: from \: the \: cliff \: be \: x[/tex]

[tex] \small\bf \: thus \: time \: taken \: by \: sound \: to \: hit \: the \: cilff \: and \: return = \frac{2x}{v} = 1[/tex]

[tex] \bf \to \: x = \frac{320}{2} m = 160m[/tex]

[tex] \small \bf \: thus \: the \: distance \: between \: the \: cliffs \: = 160m \times 2 = 320m[/tex]

Answer: the required mass is 1.7628 × 10²⁵ μ

Explanation:

Given that;

energy released in the fission of one ²³⁵U₉₂ is 206.6 MeV

power p = 28.7 Mega Watts = 28.7 × 10⁶ W = 28.7 × 10⁶/ 1.6× 10⁻¹³ = 17.9375 × 10¹⁹ MeV/s

now fission need per second will be;

⇒ power / energy released i  fission

= 17.9375 × 10¹⁹ MeV /  206.6 MeV = 8.68 × 10¹⁷ per second

now fission need per day will be;

⇒ ( 8.68 × 10¹⁷ × 360 × 24 ) = 7.5 × 10²² per day

hence mass of ²³⁵U₉₂ that is consumed in one day in this reactor will be;

⇒ (235 × 7.5 × 10²²)μ

= 1.7628 × 10²⁵ μ

Therefore the required mass is 1.7628 × 10²⁵ μ

Answer:

basketball

Explanation: cause it is

Answer:

It honestly depends on which one you're better at.

Explanation: I mean that's what I think.

Answer: the length of the pendulum should be 1.4% longer

Explanation:

Given that;

when its noon, the clock reads 11:50 am,

i.e we have 10 minutes delay ⇒ 10min × 60 = 600secs

we know that in simple pendulum

T = 2π√(l/g)  

d means delay and c means correct;

24hrs = 86400 secs

Now

Td/Tc = (86400-600) / 86400  = 0.993 = [2π√(ld/gc)] / [2π√(ld/gc)] = √(ld/lc)

so ld/lc = 0.98616  

lc = 1.014 ld

lc/l.d = -1 + 1.014 = 0.014 × 100 = 1.4% longer

therefore the length of the pendulum should be 1.4% longer

Answer: the required height is 2.5 m

Explanation:

given the given data in question

Lime to come in rest;

using S = ((U1 + U2)/2) t

⇒ 0.5 = ((√(2gh) + 0)/2) t

t = (0.5 × 2) / √(2gh)

now fat = impulse = change in momentum dp

F = 5mg = dp/dt = [(m√(2gh))/(0.5 × 2)] × √(2gh)

5mg = 2mgh/(0.5 × 2)

5 = h / 0.5

h = 5 × 0.5

h = 2.5 m

Therefore the required height is 2.5 m

Answer:

The correct answer is "24 V".

Explanation:

The given values are:

Current,

I = 0.50 A

Resistance,

R = 12 W

As we know,

⇒ [tex]I = 0.5\times (\frac{E}{2R})[/tex]

On substituting the given values, we get

⇒ [tex]0.5= (\frac{E}{4\times 12} )[/tex]

⇒ [tex]0.5= (\frac{E}{48} )[/tex]

⇒   [tex]E=24 \ V[/tex]  

Answer:

Test bullets fired from the suspected gun

Explanation:

This is the correct answer!! I took the test!!

Answer:

it comprises of the DNA/RNA bipolymer molecules

Answer:

there is a negative charge on electron.

Explanation:

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass

v is the velocity

From the question we have

[tex]k = \frac{1}{2} \times 1500 \times {8}^{2} \\ = 750 \times 64[/tex]

We have the final answer as

Hope this helps you

Answer:

hurricanes,Typhoons and cyclones

The question is incomplete. The complete question is :

First Law of Thermodynamics - Sign Convention The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is AU = Q-W. Here AU is the change in internal energy U of the system. Q is the net heat transferred into the system that is, Q is the sum of all heat transfer into (positive) and out of (negative) the system. W is the net work done by the system—that is, W is the sum of all work done by (positive) and on (negative) the system. We use the following sign conventions: if Q is positive, then there is a net heat transfer into the system; if W is positive, then there is net work done by the system. So positive Q adds energy to the system and positive W takes energy from the system. Thus AU = Q-W. Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. The first law of thermodynamics AU = 9 - W Ur-U, Heat Work System AU--W system w Qin: talu Qout:

The first law of thermodynamics AU = Q - W U-U Heat Work System AUQ-W Qin: ta Qout: - Wout: + WK w Win: - - volume expands t volume decreases o All answers can be positive or negative. (a) Suppose there is heat transfer of 42 ) into a system, while the system does 6 ) of work. Later, there is heat transfer of 22 J out of the system while 6 ) of work is done on the system. What is the net heat transfer? 20 Correct (100.0%) Submit What is the total work? Enter a number Submit (5 attempts remaining) What is the net change in internal energy of the system? Enter a number

What is the net change in internal energy of the system? Enter a number Submit (5 attempts remaining) (b) What is the change in internal energy of a system when a total of 140 J of heat transfer occurs out of (from) the system and 165 ) of work is done on the system? Enter a number Submit (5 attempts remaining) (c) An athlete doing push-ups performs 645 kJ of work and loses 440 kJ of heat. What is the change in the internal energy (in kJ) of the athlete? Enter a number Submit (5 attempts remaining) kJ (d) An athlete doing push-ups performs 690 kJ of work and loses 450 kJ of heat. Then he takes in 830 kJ of energy from eating food, What is the total change in the internal energy (in kJ) of the athlete? Enter a number kJ.

Solution :

a). Given :

[tex]$Q_1 = 42 \ J$[/tex] , [tex]$Q_2 = -22 \ J , \ W_1 = 6 \ J, \ W_2 = -6 \ J $[/tex]

Net heat transfer

[tex]$Q= Q_1+Q_2$[/tex]

   = 42 + (-22)

   = 20 J

Total work

[tex]$W= W_1+W_2$[/tex]

   = 6 + (-6)

   = 0 J    

∵ ΔU = Q - W

       = 20 - 0

        = 20 J

This is the net change in the internal energy of the system.

b). ΔU = Q + W

           = (-140) + (-165)

           = -305 J

c). ΔU = Q + W

           = (-440) + (645)

           = 205 J

d). ΔU = Q + W

           = (-450) + (690)

           = 240 J

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

[tex]F[/tex] - External force exerted on the sled, measured in newtons.

[tex]f[/tex] - Friction force, measured in newtons.

[tex]N[/tex] - Normal force from the ground on the mass, measured in newtons.

[tex]W[/tex] - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

[tex]W = m\cdot g[/tex] (1)

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

If we know that [tex]m = 50\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the weight of the sled is:

[tex]W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]W = 490.35\,N[/tex]

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

[tex]F_{min,s} = \mu_{s}\cdot W[/tex] (2)

Where:

[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.

[tex]W[/tex] - Weight of the sled, measured in newtons.

If we know that [tex]\mu_{s} = 0.3[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to start the sled moving is:

[tex]F_{min,s} = 0.3\cdot (490.35\,N)[/tex]

[tex]F_{min,s} = 147.105\,N[/tex]

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

[tex]F_{min,k} = \mu_{k}\cdot W[/tex] (3)

Where [tex]\mu_{k}[/tex] is the kinetic coefficient of friction, dimensionless.

If we know that [tex]\mu_{k} = 0.1[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to keep the sled moving at a constant velocity is:

[tex]F_{min,k} = 0.1\cdot (490.35\,N)[/tex]

[tex]F_{min,k} = 49.035\,N[/tex]

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Answer:

[tex]I_{rms}=2.12\ A[/tex]

Explanation:

Given that,

The instantaneous current in the circuit is giveen by :

[tex]I=3\sin\theta\ A[/tex]

We need to find the rms value of the current.

The general equation of current is given by :

[tex]I=I_o\sin\theta[/tex]

It means, [tex]I_o=3\ A[/tex]

We know that,

[tex]I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A[/tex]

So, the rms value of current is 2.12 A.

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

Answer:

The value of the inductance is 1.364 mH.

Explanation:

Given;

amplitude current, I₀ = 200 mA = 0.2 A

amplitude voltage, V₀ = 2.4 V

frequency of the wave, f = 1400 Hz

The inductive reactance is calculated;

[tex]X_l = \frac{V_o}{I_o} \\\\X_l = \frac{2.4}{0.2} \\\\X_l =12 \ ohms[/tex]

The inductive reactance is calculated as;

[tex]X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2 \pi f}[/tex]

where;

L is the inductance

[tex]L = \frac{12}{2 \pi \times \ 1400} \\\\L = 1.364 \times \ 10^{-3} \ H\\\\L = 1.364 \ mH[/tex]

Therefore, the value of the inductance is 1.364 mH.

Given values are:

As we know the formula,

→ [tex]X_l = \frac{V_0}{I_0}[/tex]

By substituting the values, we get

       [tex]= \frac{2.4}{0.2}[/tex]

       [tex]= 12 \ ohms[/tex]

Now,

The inductive reactance will be:

→ [tex]X_l = \omega L[/tex]

or,

→ [tex]X_l = 2 \pi f L[/tex]

Hence,

The Inductance will be:

→ [tex]L = \frac{X_i}{2 \pi f}[/tex]

By putting the values, we get

      [tex]= \frac{12}{2 \pi\times 1400}[/tex]

      [tex]= 1.364\times 10^{-3} \ H[/tex]

      [tex]= 1.36 \ mH[/tex]

Thus the above answer is right.

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Answer:

0.068kg

Explanation:

In compressed state, a spring 3.00 meters long hanging from the ceiling has a potential energy of 2.0 joules. If the entire potential energy is used to release a ball of mass

Potential energy is expressed using the formula

PE = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height =  3.0m

PE = 2.0Joules

Required

Mass of the ball

Substitute the given values into the formula as shown;

2.0 = m(9.8)(3)

2 = 29.4m

m = 2/29.4

m = 0.068kg

Hence the mass of the ball is 0.068kg

This question is incomplete, the complete question is;

An l-c cirucit with a 70 mh inductor and a 0.54 μF capacitor oscillates. The maximum charge on the capacitor is 11.5 μC. What are the oscillation frequency and the maximum current in this circuit ;

Options

a) 1.07 kHz, 63.4 mA

b) 4.38 kHz, 101.3 mA

c) 6.74 kHz, 55.7 mA

d) 2.31 kHz, 93.5 mA

e) 0.82 kHz, 59.1 mA

Answer:

the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

Explanation:

Given that;

inductor L = 70 mH = 70 × 10⁻³ H

Capacitor C = 0.54 μf = 0.54 × 10⁻⁶ f

Qmax on capacitor = 11.5 μf = 11.5 × 10⁻⁶ c

oscillation frequency in L-C circuit;

f = 1/2π√(LC)

we substitute our values;

f = 1/2π√(70 × 10⁻³ × 0.54 × 10⁻⁶  )

f = 0.0818 × 10⁴ Hz

f = 0.082 × 10³ Hz ≈ 0.82 kHz

Maximum circuit in L-C circuit is given by

I_max = Qmax/√(LC)

we substitute

I_max = 11.5 × 10⁻⁶ / √(70 × 10⁻³ × 0.54 × 10⁻⁶  )

= 59.1 × 10³ A ≈  59.1 mA

Therefore the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.

so Option e) 0.82 kHz, 59.1 mA is the correct answer

Answer:

Since binary is only 1 and 0, you can use a flashlight to display something similar to Morse code (see explanation below)

Explanation:

In binary, 1 means "on" and 0 means "off". A way you can use visible light is through turning on and off a flashlight. If the flashlight is turned on, it would represent a 1. If the flashlight is turned off, it would represent a 0. To make the message easier and more accurately understood for the receiver make sure to flash the lights in a consistent pattern (ex. each flash lasts no longer than half a second, one second between each digit, etc.)

For example, let's say you're trying to send the message "11001"

  on     on    off     off     on

0       1       2       3       4       5      Numbers represent seconds

As you can see above the message starts at 0 seconds. Between 0 and 1 seconds the flashlight is turned on once. Between 1 and 2 seconds the flashlight is turned on again, Between 2 and 3 seconds as well as 3 and 4 seconds the flashlight is not turned on at all. And finally between 4 and 5 seconds the flashlight is turned on.

Answer:

ok fine I'll answer u on comment

Answer:

1832

Explanation:

From;

Δp Δx = h/4π

Δp = uncertainty in momentum

Δx = uncertainty in position

h= Plank's constant

But p =mv hence, Δp= Δmv

m= mass, v= velocity

mass of electron = 9.11 * 10^-31 Kg

Mass of proton = 1.67 * 10^-27 Kg

since m is a constant,

Δv = h/Δxm4π

For proton;

Δv = 6.6 * 10^-34/4 * 3.14 * 1.67 * 10^-27 * 1 * 10^-10

Δv = 315 ms-1

For electron;

Δv = 6.6 * 10^-34/4 * 3.14 * 9.11 * 10^-31 * 1 * 10^-10

Δv = 577000 ms-1

Ratio of uncertainty of electron to that of proton = 577000 ms-1/315 ms-1= 1832

Answer:

the second pendulum ( also called the royal pendulum) o.994m (39.1 in) long, in which each swing takes one second, became widely used in quality clocks.

Answer:

No they just orbit the sun

Explanation:

Answer:

The options are

a. Sally is a very creative kind of person who likes to build things.

b. Jerry only works because he receives a very large income.

c. Rikki is usually shy, but at work she appears to be quite outgoing.

d. Maury gives money to charities because he wants other people to think he is very generous.

The answer is c. Rikki is usually shy, but at work she appears to be quite outgoing.

Lewin's interactionist perspective explains that an individual’s behavior is usually dependent on his personal behavior/ trait and the environment. The best option is that Rikki is usually shy which is her personal behavior but at work she appears to be quite outgoing due to her environment.

Answer:

Object A

Explanation:

The object that would make you feel worse if you're hit by it is the object possessing the highest momentum. Thus, we need to find the momentum of the two objects.

Momentum of an object is the product of its mass and that of it's velocity. Momentum is given by the formula

P = M * V, where

P = momentum

M = mass of the object

V = velocity of the object

Now, solving for object A, we have

P(a) = 1.1 * 10.2

P(a) = 11.22 kgm/s

And then, solving for object B, we have

P(b) = 2 * 5

P(b) = 10 kgm/s

The object when the highest momentum is object A, and thus would make you feel worse when hit by it

Answer:

5.35 rad/s

Explanation:

From the question, we are toldthat an Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick.

Solving this question, the potential energy of the particles must equal to the Kinectic energy i.e

P.E=K.E

Mgh= m½Iω²-------------eqn(*)

Where M= mass of the particles

g= acceleration due to gravity= 9.81m/s^2

ω= angular speed =?

h= height of the particles in the stick on the metre stick= ( 50cm + 80cm)= (0.5m + 0.8m)= 130cm=1.3m

If we substitute the values into eqn(*) we have

m×9.81× (1.3m)= 1/2× m×[ (0.5m)² + [(0.8m)²]× ω²

m(12.74m²/s²)= 1/2× m× (0.25+0.64)× ω

m(12.74m²/s²)= 1/2× m× 0.89× ω²

We can cancel out "m"

12.74= 1/2×0.89 × ω²

12.74×2= 0.89ω²

25.48= 0.89ω²

ω²= 28.629

ω= √28.629

ω=5.35 rad/s

Hence, the angular speed of the meter stick as it swings through its lowest position is 5.35 rad/s