You are driving along the New York State Thruway in a line of cars all travelling at a constant speed of 108.5 km/hr. The car in front of you applies its brakes for maximum acceleration. You then apply your brakes to achieve the same maximum acceleration after only a 1 s delay due to reaction time. What distance behind the car in front of you must you be to avoid a collision?

Answer:

The answer is a

Explanation:

Answer:

50280 meters

Explanation:

33.52 meters/seconds is 2011.2 meters/minutes (multiply by 60)

2011.2 meters/m * 25 minutes = answer

Answer:

Immune

Explanation:

because I remember taking that test and I I specifically remember that question and that's the answer

Answer:

The change in momentum is:  [tex]13.5\,\frac{kg\,m}{s}[/tex]

Explanation:

Let's define that vectors pointing up are positive, and vectors pointing down negative.

Then we express the initial momentum of the ball as the product of its mass times the velocity, and include the negative sign since this momentum is pointing down (velocity vector is pointing down):

[tex]P_i=-0.5\,(15) \frac{kg\,m}{s} =-7.5\, \frac{kg\,m}{s}[/tex]

the final momentum is positive (pointing up) and given by the product:

[tex]P_f=0.5\,(12) \frac{kg\,m}{s} =6\, \frac{kg\,m}{s}[/tex]

Therefore, since the change in momentum is defined as the difference between the final momentum minus the initial one, we get:

[tex]P_f-P_i=6-(-7.5)\,\frac{kg\,m}{s} =13.5\,\frac{kg\,m}{s}[/tex]

Answer:

C. Its velocity is perpendicular to its acceleration

Explanation:

Because acceleration is always perpendicular to the velocity when the velocity will change direction without change it's magnitude

Answer:

False

Explanation:

If a point charge has electric field lines point into it,then charge must be negative because electric lines point into negative charges and point out of positive charges

Speed=distance/time.

7.74mins=464.4secs.

72.1=distance/464.4.

distance=464.4×7.21.

=3348.324m.

Answer:

a) The police will take 18.056 seconds to catch the speedy car, b) The police will travel 326.019 meters before catching the speedy car, c) The speed of the police car when catches up with the speeder is 36.112 meters per second.

Explanation:

Let suppose that speeder moves in a uniform motion, whereas police car has an uniformly accelerated motion.

a) How long does it take for the police to catch the speeding car:

Kinematic equation of each vehicle's position are described:

Speeder

[tex]s_{A} = s_{A,o}+v_{A}\cdot t[/tex]

Police Car

[tex]s_{B} = s_{B,o}+v_{B,o}\cdot t + \frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]

If [tex]s_{A} = s_{B}[/tex], [tex]s_{A,o} = s_{B,o}[/tex], [tex]v_{A} = 18.056\,\frac{m}{s}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the resulting expression is done:

[tex]v_{A} \cdot t = v_{B,o}\cdot t +\frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]

[tex]\frac{1}{2}\cdot a_{B}\cdot t^{2}+(v_{B,o}-v_{A})\cdot t = 0[/tex]

[tex]t \cdot \left(\frac{1}{2}\cdot a_{B}\cdot t +v_{B,o}-v_{A} \right)= 0[/tex]

[tex]t = 0\,s\,\wedge\, t = \frac{2\cdot (v_{A}-v_{B,o})}{a_{B}}[/tex]

[tex]t = \frac{2\cdot \left(18.056\,\frac{m}{s}-0\,\frac{m}{s} \right)}{2\,\frac{m}{s^{2}} }[/tex]

[tex]t = 18.056\,s[/tex]

The police will take 18.056 seconds to catch the speedy car.

b) How far did the police car travel before police caught up with the speeder?

The distance travelled by the police is: ([tex]s_{B,o} = 0\,m[/tex],  [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and  [tex]t = 18.056\,s[/tex])

[tex]s_{B} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (18.056\,s)+\frac{1}{2}\cdot \left(2\,\frac{m}{s^{2}} \right) \cdot (18.056\,s)^{2}[/tex]

[tex]s_{B} = 326.019\,m[/tex]

The police will travel 326.019 meters before catching the speedy car.

c) What is the speed of the police car when catches up with the speeder?

The speed of the police car is represented by the following formula:

[tex]v_{B} = v_{B,o} + a_{B}\cdot t[/tex]

Where [tex]v_{B}[/tex] is the speed of the police car, measured in meters per second.

Given that [tex]v_{B,o} = 0\,\frac{m}{s}[/tex], [tex]t = 18.056\,s[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the final speed of the police car when catches up with the speeder is:

[tex]v_{B} = 0\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right)\cdot (18.056\,s)[/tex]

[tex]v_{B} = 36.112\,\frac{m}{s}[/tex]

The speed of the police car when catches up with the speeder is 36.112 meters per second.

Answer:

Explanation:

When a projectile is launched at some angle to the horizontal with some speed vi , and air resistance is negligible , it is definitely a freely falling body .

It is so because it is free to accelerate towards the earth with acceleration of g . Air has no resistance , hence no force is acting on it except the gravitational force . Hence it is a freely falling body .

b )

The acceleration in the vertical direction is due to force exerted by the earth that is gravitational force on it  . Hence its  acceleration is equal to g in vertically downward direction .

c )

It has zero acceleration in horizontal direction . It is so because no force is acting on it in horizontal direction . So no acceleration will be present in horizontal direction . It will move in horizontal direction with constant speed of vi cos θ where θ is the angle vi make with the horizontal .

Answer:

0.00004666N

Explanation:

We know that

intensity (I) = P/ A

Where

P= power

A= Area

So lets say that the power absorbed

Will be = Intensity x Area

Which Is = 1.4 x 10^3 x(10)

So

14000 Watt = 14 kWatt

However we know that radiation pressure is equal to

time-averaged intensity all over the speed of light in free space

So

P = (1.4 x 1000)/c

But

F= P x A

So

((1.4 x 1000)/(3 x1 0^8)) x 10

Which is

=0.000046666N

Explanation:

Answer:

The electric field inside the conductor is zero.

Explanation:

In electrostatic equilibrium charges are assumed to be in rest. Therefore no current.

I = 0

From the ohm's law

ΔV= IR = 0

⇒ΔV =0

Therefore, potential V is same through the conductor.

Also, E = -ΔV d = 0

Hence, The electric field inside the conductor is zero.

Answer: his distance from the pole is 90.16 ft

Explanation:

based on the diagram of the question i will upload along this answer;

45 - d is equal to part of the pole below the horizontal line

d/x = tan(16) ; x = d/tan(16)............equ1

(45-d) / x = tan(12); x = (45-d) / tan(12)-----------equ2

∴ d/tan(16) = (45-d) / tan(12)

d.tan(12) = (45-d).tan(16)  

d.tan12 = (45×tan16) - d.tan16

0.2125d = 12.9034 - 0.2867d

0.2125d + 0.2867d = 12.9034

0.4992d = 12.9034

d = 12.9034/0.4992

d = 25.85 ft

now substitute value of d into any of our previous equation, lets take equation 1

x = 25.85 / tan(16)

x = 25.85 / 0.2867

x = 90.16 ft

Therefore his distance from the pole is 90.16 ft

Answer:

We are given x= bt +ct²

So

A. bxt= m

Because m/s*s= m

So b= m/s and c= m/s²

B.

x= bt-ct²

So at x=0 t=0

x=0 t= 2

We have

bt = ct² so t = b/c at x= 0

So b-2ct= 0

B. To find velocity we use

dx / dt = b - 2 Ct

C. At rest wen V= 0

We have t= b/2c

D. To find acceleration we use

dv / dt = - 2C

When We are given x= bt +ct²

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Answer:

it is B 1:50

Explanation:

just did it on apex

Answer:

The  value  is  [tex]t = 10 \  s[/tex]

Explanation:

From the question we are told that

  The  height is  [tex]h = 500 \ m[/tex]

  The acceleration is  [tex]g =  10 \  m/s[/tex]

Generally from the second equation of motion we have that

    [tex]h  =  ut  +  \frac{1}{2} a t^2[/tex]

Here  the initial velocity u =  0  m/s  given that the stone started from rest

       So

            [tex]500 =  0 (t) +0.5 *  (10) t^2[/tex]

=>         [tex]t = 10 \  s[/tex]

Distance = (speed) x (time)

Distance = (10 meter/second) x (2 minutes)

Distance = (10 meter/second) x (2 minutes) x (60 second/minute)

Distance = (10 x 2 x 60) (meter-minute-second / second-minute)

Distance = 1,200 meters

Answer:

lamp oil.

Explanation: the ping pong ball doesnt sink through the lamp oil.

◈ Electric current :

Electric current is defined as the rate of charge flow. (per unit time)

I denotes current

Q denotes charge

t denotes time

Given : I = 2A and t = 2s

To Find : Charge flow in wire (Q)

[tex]:\implies\sf\:I=\dfrac{Q}{t}[/tex]

[tex]:\implies\sf\:2=\dfrac{Q}{2}[/tex]

[tex]:\implies\:\boxed{\bf{\red{Q=4\:C}}}[/tex]

A flow of electrical charge carriers, generally electrons or electron-deficient atoms, is referred to as current.  The charge that will be in the wire is 4 C.

A flow of electrical charge carriers, generally electrons or electron-deficient atoms, is referred to as current. The capital letter I is a typical sign for current. The ampere, abbreviated as A, is the standard unit.

As it is given to us that the current in the wire is 2A, while the time for which the current flows through the wire is 2 seconds.

[tex]\rm I=\dfrac Qt[/tex]

Substitute the value we will get,

[tex]\rm 2\ A = \dfrac{Q}{2\ sec}\\\\Q = 4\ C[/tex]

Thus, the charge that will be in the wire is 4 C.

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Answer:

less precise than

Explanation:

Answer:

less precise than

Explanation:

its right on edge2020

Answer:

b. disorder in the universe is increasing with the passage of time

Explanation:

The second law of thermodynamics states that the total entropy of an isolated system  increases over time . The entropy of a system measures the randomness or disorder  in the system . So the we can state,  in terms of disorder , the second  law of thermodynamics as follows .

The disorder in the universe is increasing with the passage of time .

The second law of thermodynamics leads us to conclude that disorder in the universe is increasing with the passage of time. So, here, option b is the correct option.

Entropy is a measure of the disorder or randomness in a system. The second law of thermodynamics states that in any spontaneous process, the total entropy of an isolated system (including both the system and its surroundings) tends to increase or, at best, remain constant. This means that over time, natural processes tend to lead to an increase in the overall disorder or randomness of the system. It's important to note that while the increase in entropy is a statistical tendency on a macroscopic scale, there can still be localized decreases in entropy within a system. 

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#SPJ6

You haven't described whether they're connected in series or in parallel.  Actually, there are eight (8) different ways they can be arranged, and each way has a different equivalent resistance.

==> All 3 resistors in series.  Equivalent resistance = (2+3+4) = 9.000 ohms

==> All 3 resistors in parallel.

Equivalent resistance = 1 / (1/2 + 1/3 + 1/4) = (12/13) ohm or 0.923 ohm

==> The 2 ohm resistor in series with (the 3 and the 4 in parallel)

Equivalent resistance = 2 + 1/(1/3 + 1/4) =  3.714 ohms

==> The 3 ohm resistor in series with (the 2 and the 4 in parallel)

Equivalent resistance = 3 + 1/(1/2 + 1/4) =  4.333 ohms  

==> The 4 ohm resistor in series with (the 2 and the 3 in parallel)

Equivalent resistance = 4 + 1/(1/2 + 1/3) =  5.200 ohms

==> The 2 ohm resistor in parallel with (the 3 and the 4 in series)

Equivalent resistance = 1.556 ohms

==> The 3 ohm resistor in parallel with (the 2 and the 4 in series)

Equivalent resistance = 2.000 ohms

==> The 4 ohm resistor in parallel with (the 2 and the 3 in series)

Equivalent resistance = 2.222 ohms

Answer:

q = -2.19 x 10⁻⁵ C

Explanation:

Given;

mass of the particle, m = 1.5 g = 0.0015 kg

magnitude of electric field, E = 670 N/C

Electric field is given by;

[tex]E = \frac{F}{q}[/tex]

where;

q is the magnitude of the

f is the force of the charge

f = mg

[tex]E = \frac{F}{q}\\\\E = \frac{mg}{q}\\\\q = \frac{mg}{E}\\\\q = \frac{0.0015*9.8}{670} \\\\q = 2.19*10^{-5} \ C[/tex]

Since the electric field is acting downward, the force on the charge must be acting upward. Therefore, the charge must be negative

q = -2.19 x 10⁻⁵ C

Answer:

455.9 Hz, Flat

Explanation:

The beat frequency is basically the difference between the frequency of the B4 note and the frequency of the A4 tuning fork. This means the B4 note is 15.9 Hz off of 440Hz, and 67.4 Hz off of 525.3 Hz. As B4 is between the two notes, it would make sense to find its frequency by adding 15.9 to 440 and subtracting 67.4 from 523.3, both if which give us a frequency 455.9 Hz for the B4 key. This is because the note doesn't change for the different turning forks so both differences should result in the same frequency. Because the note should be 493.9 frequency but instead has a frequency of 455.9 Hz, it is flat because the frequency is lower than it is supposed to be.

Hope this helped!

This question deals with the phenomenon of "Beat Frequency".

The frequency being played by the B4 key is "455.9 Hz", which is "lower (flat) than it should be".

When two waves with a slightly different frequency interfere with each other, they produce a pattern known as beat frequency. The beat frequency is the difference between the frequencies of the interfering wave. Since the frequency of B4 key produces beat frequency with both the C5 tuning fork (which has a frequency of 523.3 Hz) and the A4 (which has a frequency of 440 Hz). Hence, the frequency of the B4 key must be between the range of these frequencies, that is, 440 Hz to 523.3 Hz.

ANALYZING THE BEAT FREQUENCY WITH A4 TUNING FORK:

Beat Frequency with A4 tuning fork = Frequency of B4 - Frequency of A4

Frequency of B4 = Frequency of A4 + Beat Frequency with A4 tuning fork

Frequency of B4 = 440 Hz + 15.9 Hz

Frequency of B4 = 455.9 Hz

ANALYZING THE BEAT FREQUENCY WITH C5 TUNING FORK:

Beat Frequency with C5 tuning fork = Frequency of C5 - Frequency of C5

Frequency of B4 = Frequency of C5 - Beat Frequency with C5 tuning fork

Frequency of B4 = 440 Hz - 67.4 Hz

Frequency of B4 = 455.9 Hz

Both the answers validate each other.

Since the expected frequency of the B4 key is 493.9 Hz, which is greater than the actual frequency produced by the B4 key. Hence, we can say that "the frequency of the key lower than it should be ("flat")"

The attached picture shows the general formula of the beat frequency.

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Answer:

  [tex]\sqrt{\dfrac{2g\lambda^3}{m}}[/tex]

Explanation:

We can write the expression here, but the point of the problem seems to be to see if you can manipulate the controls on the answer box to reproduce that expression.

  [tex]\boxed{\sqrt{\dfrac{2g\lambda^3}{m}}}[/tex]

The momentum of the truck is "56250 kgm/s".

According to the question,

Mass,

Velocity,

As we know,

→ [tex]p = mv[/tex]

By putting the values, we get

→    [tex]= 2250\times 25[/tex]

→    [tex]= 56250 \ kgm/s[/tex]

Thus the answer is right.

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Complete Question

A boy throws a ball on a spring scales which oscillates about the equilibrium position with a period of T = 0.5 sec. The amplitude of the vibration is A = 2 cm. Assume the ball does not bounce from the scales’s surface afterwards. Assume the vibration of the scale is expressed mathematically as x(t) = Acos(t + ). Find:

a) frequency

b) the maximum acceleration

c) the maximum velocity

Answer:

a

[tex]f =  2  \ Hz[/tex]

b

[tex]a_{max} =  3.2 \ m/s^2[/tex]

c

[tex]v_{max} =  0.25 \  m/s[/tex]

Explanation:

From the question we are told that

   The period is  [tex]T  =  0.5 \ sec[/tex]

    The  amplitude is  [tex]A =  2 \ cm  =  0.02 \ m[/tex]

    The  vibration of the scale is  [tex]Acos(wt +  \phi )[/tex]

Generally the frequency is mathematically represented as

   [tex]f =  \frac{1}{T}[/tex]

=>   [tex]f =  \frac{1}{0.5}[/tex]

=>    [tex]f =  2  \ Hz[/tex]

The  maximum acceleration is mathematically represented as

    [tex]a_{max} =  A *(2 \pi f)^2[/tex]

=>    [tex]0.02 *  (2 * 3.142  *  2)^2[/tex]

=>  [tex]a_{max} =  3.2 \ m/s^2[/tex]

The  maximum velocity is mathematically represented as  

     [tex]v_{max} =  A *  (2 \pi f)[/tex]

=>    [tex]v_{max} =  0.02 *  (2 * 3.142 *  2)[/tex]

=>    [tex]v_{max} =  0.25 \  m/s[/tex]

The magnitude of Dot product of two vectors is 3.

     [tex]A= 4j - 2i - 3k[/tex]  and [tex]C= -4i -2j - 3k[/tex]

        [tex]A\cdot C=(-4*4)+(-2*-2)+(-3*-3)\\\\A\cdot C=-16+4+9\\\\A\cdot C=-16+13=-3[/tex]

Thus,  The magnitude of Dot product of two vectors is 3.

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Answer:

A. Using

I= P/ A

So = P/ 2π R²

So=777E3 watt/ 2*πx (²5.5E3m)

4.3*10^ -3W/m²

So pressure= 2I/c

=2* 4.3*10^ -3W/m²/ 3*10^ 8

2.7*10^-11p.a

B.

Using Emax= √2I/Eoc

So

√ 2*4.3*10^-3/8.8E-12*3*10^8

= 5.48V/m

So

Bmax= Emax/c

5.58/3E8

= 1.83*10-9T

D.average density is given by

= Eo(Emax/√2)²

= 8.85E-12(5.48/√2)²

1.3*10^10W/m²