Starting from rest, a panther accelerates at 3 m/s/s for 10 seconds. What is the final velocity of the panther

Answer:

a= 18.4m

b= 26.2m

Explanation:

Please see the attached picture for full solution.

Answer:Newton’s law also states that larger bodies with heavier masses exert more gravitational pull, which is why those who walked on the much smaller moon experienced a sense of weightlessness, as it had a smaller gravitational pull. To help explain his theories of gravity and motion, Newton helped create a new, specialized form of mathematics.

Explanation:

Answer: newtons 2nd law of motion because in order to push the object you must put force to the mass and move it causing motion

Explanation:

Answer:

lamp oil.

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

Answer:

50280 meters

Explanation:

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

2011.2 meters/m * 25 minutes = answer

Answer:

Heat required in converting 1g of ice at −10∘C into steam at 100∘C is: latent heat of fusion= 80 cal/g80 cal/g. latent heat of vaporization=540 cal/g.

Answer:

The answer is a

Explanation:

The complete question is;

In electronic circuits it is not unusual to encounter currents in the microampere range. Assume a 35 μA current, due to the flow of electrons. What is the average number of electrons per second that flow past a fixed reference cross section that is perpendicular to the direction of flow?

Answer:

2.185 × 10^(14) electrons/seconds

Explanation:

We are given current as 35 μA = 35 × 10^(-6) Amperes

The value of the charge on one electron is; e = 1.602 × 10^(-19) coulombs

Now, from conversions;

One ampere = one coulomb/second =

Then, 35 × 10^(-6) Ampere's would give;

35 × 10^(-6) Coulombs/sec

Now,

1.602 × 10^(-19) coulombs = 1 electron

Thus;

35 × 10^(-6) Coulombs/sec gives;

(35 × 10^(-6))/(1.602 × 10^(-19)) = 2.185 × 10^(14) electrons /sec

The average number of electrons per second that flow past a fixed reference cross-section perpendicular to the direction of flow is 2.185 × 10⁻¹⁴ electrons /sec

The question is as given below:

In electronic circuits, it is not unusual to encounter currents in the microampere range. Assume a 35 μA current, due to the flow of electrons. What is the average number of electrons per second that flow past a fixed reference cross-section that is perpendicular to the direction of flow?

Given information:

Current I = 35 μA = 35 × 10⁻⁶ A

charge on electron e = 1.602 × 10⁻¹⁹C

We know that:

I = de/dt, rate of flow of charge.

1A = 1C/s

Thus,

35 × 10⁻⁶ A = 35 × 10⁻⁶ C/s

Now, one electron has a charge of  1.602 × 10⁻¹⁹C

So, the number of electrons for a current of 35 × 10⁻⁶ C/s will be:

n = (35 × 10⁻⁶ C/s) ÷ (1.602 × 10⁻¹⁹C)

n = 2.185 × 10⁻¹⁴ electrons /sec

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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:

it is B 1:50

Explanation:

just did it on apex

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

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:

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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Answer:

Immune

Explanation:

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

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:

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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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:

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

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]

You probably already took this but if anyone wanted to know the answer it's D. "A system where matter and energy cannot enter or leave the system".

(I also took the quiz for it and got it correct)

A closed physical system is a system where matter and energy cannot enter or leave the system. The correct option is D.

A physical system is any item or part of an object that can be analyzed using physics laws.

An atom and lake water are both examples of physical systems. Everything outside the system is referred to as the environment, and it is ignored in the analysis except for its effects on the system.

A physical system is a collection of parts or elements that, when combined, exhibit behavior that the constituents do not. Matter and energy make up physical systems.

A closed system is a natural physical system that does not allow the transfer of matter into or out of the system, though it does allow the transfer of energy in contexts such as physics, chemistry, or engineering.

Thus, the correct option is D.

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◈ 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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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:

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:

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]

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]

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:

  [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]

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: 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