An nfl caliber wide receiver can reach a top speed of 10.0 m/s in only 2.4 s. what is their acceleration during this time?

Answer:

"the maximum extent of a vibration or oscillation, measured from the position of equilibrium."

hope this helps

Answer:

amplitude and physics the maximum displacement or distance moved by point on a vibrating body or wave measured from its equilibrium position it is equal to 1/2 of the length of the vibration path

The plane's horizontal and vertical velocity components are 33.65 m/s and 53.85 m/s respectively.

What is velocity ?

velocity is defined as rate of change of displacement d of the object with respect to rate of change in time t. In mathematics It is written as :

v = d/t

it should be noted that, velocity is nothing but speed in particular direction. That is velocity is vector quality having both magnitude and direction where as speed is just the magnitude of velocity.

It is given that :

A plane moves with a velocity v of 63.5 m/s at 32 degrees South of East with the horizontal.

Therefore, the planes horizontal velocity components will be :

v₁ = vsinθ

v₁ = 63.5 x sin 32⁰

v₁ = 33.65 m/s

the planes vertical velocity components will be :

v₂ = vsinθ

v₂ = 63.5 x cos 32⁰

v₂ = 53.85 m/s

Therefore, the plane's horizontal and vertical velocity components are 33.65 m/s and 53.85 m/s respectively.

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The distance between Dina and Owen with respect to Ed is 1.44 × 10¹² m.

A frame of reference is a set of reference points—geometric points whose positions are known mathematically and physically—that define the origin, orientation, and scale of an abstract coordinate system.

If a body does not continuously adjust its position in relation to its environment throughout the course of time, it is said to be at rest.

In the frame S', Dina and Owen are at rest.

The speed of the ball with respect to Owen, u =0.800c

The speed of the frame S' with respect to frame S,  v = 0.600c

Distance between Dina and Owen, L(p) = 1.8 × 10¹² m

Speed of light, c = 3 × 10⁸ m/s

Therefore, the distance between Dina and Owen with respect to Ed is:

L = L(p) √[1-(v²/c²)]

L = (1.8 × 10¹² m)√[1-(0.6c)²/c²)]

L = (1.8 × 10¹² m)(0.8)

L = 1.44 × 10¹² m

The distance is 1.44 × 10¹² m.

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Answer:

The size of an object is directly proportional to the gravity

Explanation:

The size of an object has significant impact on the gravity exerted by such a body.

The more massive a body is, the larger the gravity it exerts.

The reason for this is because of the newton's law of universal gravitation.

Air pressure inside the balloon is 0.143 N/m²

PRESSURE : force applied perpendicular to the surface of object per unit area.

GIVEN -

Radius =  5.0m

Force = 45N

Air pressure inside the balloon is given by

       [tex]P = \frac{F}{A}[/tex]

Where

P = Pressure

F = Force

A = Area

Surface area of the sphere  = 4πr²

r is the radius of sphere

A = 4 ×3.14×(5)²

A = 314 m²

P = F/A

P =  45/314

P =  0.143 N/m²

Air pressure inside the balloon is 0.143 N/m²

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Answer:it’s a

Explanation:

Static electricity refers to the buildup or imbalance of electric charges on the surface of objects. The correct answer is B. An insulator prevents electric charge from moving through it.

An insulator prevents electric charge from moving through it. This statement describes an example of static electricity because insulators, such as rubber or plastic, do not allow electric charges to flow freely. When an insulator is rubbed or comes into contact with another object, it can cause the transfer of electrons between the two surfaces, leading to a buildup of static charge.

The charges remain stationary on the surfaces of the objects, resulting in an electrical imbalance without the flow of charges through the insulator.

Therefore, the correct answer is B. An insulator prevents electric charge from moving through it.

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Answer:

Speed is 0 m/s.

Distance is 200 meters.

Explanation:

In this scenario, we can treat speed as the absolute value of velocity and distance as absolute value of displacement since we are dealing with one-dimensional motion.

There’ll only exist one component of vector, by applying magnitude formula, it’ll return value as shown below:

Suppose we have a one-dimensional vector, [tex]\displaystyle{\vec v = 5\hat i}[/tex], this represents a vector with magnitude of [tex]\displaystyle{|\vec v | = \sqrt{5^2} = 5}[/tex].

Theorem I - Magnitude of One-Dimensional Vector

Suppose we have a vector [tex]\displaystyle{\vec v = a\hat i}[/tex] then the magnitude is [tex]\displaystyle{|\vec v| = \sqrt {a^2} = |a|}[/tex].

Speed and distance both are scalar quantity which means they only have magnitude but lack the direction. In one-dimensional motion, speed is defined to be the absolute value of velocity and direction is defined to be the absolute value of displacement, the reason is shown above regarding one-dimensional vector.

An example would be, consider a car moves at velocity of -4 m/s, this means that a car just moves to negative direction or opposite direction of positive at 4 m/s. The negative and positive sign refer to the direction that an object moves, to know how fast they are moving, we consider its magnitude which is speed.

This can also be said same to displacement and distance, but be aware that this can only be applied to straight horizontal or vertical line movement or else the motion will become two-dimensional.

From the problem, there’s no specific distance or displacement’s graph so we will assume that the car is moving in straight horizontal line. The car is moving at speed of 40 m/s and decelerates at 4 m/s².

This means that our acceleration is -4 m/s² since it’s decelerating and is moving at 40 m/s from start. Therefore, we can apply a general physic formula which is:

[tex]\displaystyle{v = u+at}[/tex]

Where v is final velocity, u is initial velocity, a is acceleration and t is time. As said, we can treat speed = positive velocity in this scenario so substitute u = 40 m/s and a = -4 m/s² in the formula.

[tex]\displaystyle{v=40-4t}[/tex]

Now substitute t = 10 in the formula:

[tex]\displaystyle{v = 40-4(10)}\\\\\displaystyle{v=40-40}\\\\\displaystyle{v=0 \ \sf{m/s}}[/tex]

Therefore, the speed of car at 10 seconds is 0 m/s, it’s not moving at this moment.

Next, we find distance which we can use the following formula:

[tex]\displaystyle{s = ut + \dfrac{1}{2}at^2}[/tex]

Substitute u = 40, t = 10 and a = -4 in:

[tex]\displaystyle{s=40(10)+\dfrac{1}{2}(-4)(10)^2}\\\\\displaystyle{s = 400-2(100)}\\\\\displaystyle{s=400-200}\\\\\displaystyle{s=200 \ \sf{m}}[/tex]

Therefore, in 10 seconds, it’ll be able to travel 200 meters.

Answer:

the last option is correct

Explanation:

Newton's third law of motion , the ground will exact the equal opposite force the rubber ball

Answer: Regarding the image posted, a unicellular paramecium relies on mitosis for reproduction.

I hope this helps!

Explanation:

The weight is 40 * 6.2 = 248N.

Answer:

248 N

Explanation:

Weight= Gravity*mass

W= 6.2*40

W= 248N

Thank you

According to sound power level, 80 decibel sound has greater intensity.

We need to know about sound power levels to solve this problem. The sound power level can be defined by a decibel unit. The decibel unit is proportional to the logarithm intensity ratio. It can be written as

dB = 10 log(I1 / I0)

where dB is sound power level, I1 is sound intensity and I0 is reference sound intensity (10¯¹² W/m²)

From the question we know that

Ia = 60 dB

Ib = 80 dB

Because the decibel unit is proportional to the logarithm intensity ratio, the sound with greater decibel will have greater intensity.

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Answer:

Acceleration= 2 m/s²

Time= 10 s

Final Speed= 24  m/s

Final Speed= 85  m/s

Starting Speed= 0 m/s

Explanation:

By using, v = u + at

Acceleration= 2 m/s²

Time= 10 s

Final Speed= 24  m/s

Final Speed= 85  m/s

Starting Speed= 0 m/s

Answer:

c: to show forces acting on an object

Answer:

a. 1/1000 sec

Explanation:

Shutter speed is the length of time that the film you’re photographing is being exposed to the scene in film photography. However, in digital photography, shutter speed is the length of time that the image sensor sees the scene the photographer is trying to capture.

For shutter speeds, the greater the denominator the higher the speed and the lower the denominator, the lower the speed.

Thus, the fastest one is option A.

The fastest shutter speed is represented as : ( D )  1/2000 sec

To determine the fastest shutter speed we have to the consider the numerator and denominator values of the given expression representing the relationship between the shutter speed and length of time. The greater the denominator the faster the shutter speed.

Therefore from the options listed the fastest shutter speed would be 1/2000 sec

Hence we can conclude that the fastest shutter speed is represented as 1/2000 sec.

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The object fall just during the 2nd second is 14.7m.

As we know the equation of motion

equation off motion are equation that describe the behavior of physical system in terms of its motion as a function of time.

using 2nd equation of motion

S = ut +gt²/2

S= distance covered

u = initial velocity

g = acceleration due to gravity

t = time taken (t= 1 sec)

a = g = 9.8 m/s²

u = 0 m/s

At time t=1 sec

S₁ = 0 + 9.8 (1)²/2  = 4.9m

S₁  = 4.9m

S1 is the distance covered in ist second.

At t = time (t= 2 sec)

S₂ = 0 + 9.8 (2)²/2  = 19.6m

S₂ = 19.6m

S2 is the distance covered after 2nd second.

The object fall just during the 2nd second is

S = S₂ - S₁

S = 19.6 - 4.9  = 14.7m

The object fall just during the 2nd second is 14.7m.

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The final velocity of the rock before hitting the ground is 34.42m/s

Velocity of an object can be said as, the rate of change of displacement with respect to time of an object . Its unit is m/s and it is a vector quantity.

V = Δx/Δt

In here , V is velocity ,  Δx is change in displacement and Δt is change in time of an object.

We are given that,

The rock thrown downward with initial velocity = (-u) = -20m/s

The height of the building before and after hitting the ground = -Δy = -40 m

Therefore , to get the value of the final velocity (v) of the rock before hitting the ground , then the equation of motion cab be written as,

v² = u² + 2a(Δy )

Here , (a) is the acceleration due to gravity and taken negative (-9.81m/s²) due downward direction of the building then putting all the values in above equation of motion ,

v² = (-20m/s)² + 2(-9.81m/s²) (-40m)

v² = 1184.42m/s

v = 34.42m/s

Hence, the final velocity of the rock before hitting the ground would be 34.42m/s .

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Answer:

The speed of the electron is 0.98c, given that the rest energy of an electron is 0.511MeV and the rest energy of a proton is 938MeV, with both particles having kinetic energies of 2.00MeV.

Explanation:

A particle's total energy and the rest energy follow the laws of physics and the theory of relativity. The energy of any particle is equal to mass times the speed of the light-squared relatively.

Where;  E=γ.〖mc〗^2

γ is the ratio of total energy to rest energy of the particle.

m_0 is known as the rest mass of the particle.

Rest energy can be written as: E_0=〖m_0 c〗^2

γ=K/E_0 +1         (1)

γ=1/(√1-v^2/c^2)     (2)

Putting the values of the electron energy into equation (1):

γ=2.00MeV/0.511MeV+1

γ= 3.91 + 1

γ=4.91

From equation (2),

γ=1/(√1-v^2/c^2)

Squaring both sides of the equation,

γ^2=1/(1-v^2/c^2)

1/γ^2 =1-v^2/c^2  

v^2/c^2 =1-1/γ^2  

v/c=√1-1/γ^2    

Making v subject of the formula,

v=√1-1/γ^2 ×c    (3)

Inserting the values into Equation 3,

v=√1-1/〖4.91〗^2 ×c

v=√1-1/24.11×c

v=√0.959×c

v=0.98×c

v=0.98c

Therefore, the speed of the electron is 0.98c.

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When the parallel-plate capacitors are disconnected from a battery and plates are pulled a small distance farther apart then Q remains the same, C decreases, Change in potential increases, Energy stored increases and E stays the same.

Solution:

Let's take C as the capacitance of parallel plates and Q as the Charge stored and 'd' is the distance between a plate of capacitors.

In the case of disconnection between plate capacitors, the capacitance of a parallel-plate capacitor becomes equal.

Then, charge Q remains the same i.e.

Q = εA / d = Q / Ed = Q/V

Here Q refers to the charge stored in a capacitor, E is the Electric field, V is the potential difference, and 'd' is the separation between plates.

There is an inverse relationship between potential difference and distance against the capacitance between a plate of the capacitor.

Similarly, the potential difference will increase when capacitance decreases.

E = V/d

While the values of V and d show a gradual decrease, E remains constant, therefore energy stored in the capacitor will be E = 1/2CV^2

The decrease in C causes an increase in V, as a result of which energy stored in the capacitor increases.

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The electric field strength of the electron is 212143.75 N/C

To calculate the electric field strength,we use the formula below,

Formula

•E=ma/q............. equation 1

where,

E=electric field strength of the electron

m=mass of the electron

q=charge of an electron

first we need to calculate the acceleration of electron

•

[tex]v {}^{2} = u {}^{2} + 2as..........equation \: 2[/tex]

from the question,

given:

[tex]v = 3.9 \times 10 {}^{7} \frac{m}{s} [/tex]

[tex]u = 2.5 \times 10 {}^{7} \frac{m}{s} [/tex]

[tex]s = 1.6 cm = 1.2 \times 10 {}^{ - 2} [/tex]

substitute the values into equation 2

[tex](3.9 \times 10 {}^{7} ) {}^{2} = (2.5 \times 10 {}^{7} ) {}^{2} + 2(1.2 \times 10 {}^{ - 2} )a[/tex]

solve for a

[tex]1.521 \times 10 {}^{15} = 6.25 \times 10 {}^{14} + 0.024a[/tex]

[tex]a = 1.521 \times 10 {}^{15} - 6.25 \times 10 {}^{14} \div 0.024[/tex]

[tex]a = 8.96 \times 10 {}^{14} \div 0.024[/tex]

[tex]a = 3.73 \times 10 {}^{16} \frac{m}{s {}^{2} } [/tex]

Also given the following constant

[tex]m = 9.1 \times 10 {}^{ - 31} kg[/tex]

[tex]q = 1.6 \times 10 {}^{ - 19} c[/tex]

substitute these values in equation 1

[tex]e = \frac{(9.1 \times 10 {}^{ - 31})(3.73 \times 10 {}^{16} )}{1.6 \times 10 {}^{? - 19} } [/tex]

[tex] = 212143n \div c[/tex]

Hence,the electric field strength of the electron is 212143n/c

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The model house on remote renewable energy will generate (20mmm + 30sss) kilowatt-hours of energy in 10 years.

Given:

Remote Renewable House= 2 windmills + 3 turbines

1 windmill generate mmm kilowatts-hours in a year

Therefore,

2 windmills will generate,

2×mmm=2mmm kwh in a year

In 10 years, 2 windmills will generate

10×2mmm=20mmm

1 solar panel generate sss kilowatt-hour each in 1 year

3 solar panels will generate yearly

3×sss=3sss

In 10 years, 3 solar panels will generate

10×3sss=30sss

Therefore, the total energy generated in the renewable model house is in 10 years

(20mmm+30sss)  kwhr

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Based on the graph, the correct option is that Marty is right because the change in concentration increases with time; option C.

Decomposition reactions are reactions in which a large molecule splits or breaks down into two or more smaller molecules.

The general equation of a decomposition reaction is given below;

The decomposition of hydrogen peroxide to give water and oxygen occurs when it is exposed to air.

The given graph of the decomposition of hydrogen peroxide shows that the concentration of hydrogen peroxide decreases with time.

In conclusion, the decomposition of hydrogen peroxide in the presence of air gives oxygen and water.

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Answer:

increases your heart rate

Explanation:

Warming up increases your heart rate and therefore your blood flow which enables more oxygen to reach your muscles.

Answer:

Radiation

Explanation:

I hope that help's

The constant acceleration of the green car is -2m/s2

Given,at t= 0

d=220m(distance between the cars)

let v0 be the final velocity of the green car

And,a be the green car acceleration

Then,for the red car with V1= 20km/h=5.56m/s the passing point

X1=44.5m

t1=X1/V1 =8.01s

then,X1-d=Vot1+1/2at2.......... equation (1)

And,for the red car with V2=40km/h=11.11m/s,the passing point

X1=76.6m

t2=X2/V2= 6.894s

then,X2-d= V0 t2+1/2at2.......equation (2)

on solving equation (1) & (2),

We obtain the following results;

The initial velocity of the green car is = -2m/S2 (the negative sign means that it's along the -X direction)

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Disclaimer: The question was given incomplete on the portal.Here is the complete question.

Question:a red car and a green car, identical except for the colour, move toward each other in adjacent lanes and parallel to the x−axis. At time t=0, the red car is at x r =0, and the green car is at x g =220m. If the red car has a constant velocity of 20km/h, the cars pass each other at x=44.5m, and if it has a constant velocity of 40km/h, they pass each other at x=76.6m. What is the constant acceleration of the green car?

As a result, it is a noise without a pitch or wavelength. When the time comes, part of it is incidental and radiates away from you in all directions.

Following that, there is a temporal delay of 60 cm, or 0.6 m divided by 340 m/s, or 0.002 seconds. This is the amount of time it takes for the sound impulse to travel from one riser to the next. Any material will now reflect a significant portion of the sound that reaches it, and the reflected wave will sound substantially different from the shop. first sound. If there are 20 rows of seats and you hear a tone with 20 crests coming from the stands,

We are informed about this sound wave problem that a loud, sharp sound is produced in the middle of a field in an empty stadium. We hear an echo with a distinct pitch but no frequency or Waveland in the pulse.

Now that the sound has been accounted for, we need to estimate its frequency, wavelength, and duration in terms of orders of magnitude. We will now graph the air pressure versus time and model the Lord Sharp sound impulse as a single narrow peak.

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Answer:

Explanation:

The force required can be found by using the formula

[tex]f = \frac{w}{d} \\ [/tex]

w is the workdone

d is the distance

From the question we have

[tex]f = \frac{25}{0.5} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

1.78m/s

Explanation:

Given parameters:

Velocity of the arrow = 22.8m/s

Height of the wall  = 15.5m

Unknown:

Time taken for the arrow to reach the ground  = ?

Solution:

To solve this problem we employ the appropriate motion equation:

     H  = ut + [tex]\frac{1}{2}[/tex] g t²

 H  is the height

  u is the initial velocity = 0

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

  t is the unknown time

  15.5 = [tex]\frac{1}{2}[/tex] x 9.8 x t²

  15.5 = 4.9t²

     t² = 3.16

      t = 1.78m/s

The shuttlecock that is kicked upward from the roof edge of a tall building at the speed of 4m/s after 3s will have a velocity of: -25.4 m/s (the value is negative due to the object is going down)

The formula for the vertical launch upward and the procedure we will use is:

vf = v₀ - g * t

Where:

Information about the problem:

Applying the final velocity formula we get:

vf = v₀ - g * t

vf = 4 m/s - 9.8 m/s² * 3 s

vf = 4 m/s - 29.4 m/s

vf = -25.4 m/s

this negative value indicates that the object is going down

In physics vertical launch upwards is the motion described by an object that has been launched vertically upwards in which the height and the effect of the earth's gravitational force on the launched object are taken into account.

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Answer:

Explanation:

The solution along with the diagram can be found in the attachment. Feel free to raise doubts.

Centripetal force is defined as the force that allows the object to follow a curvature path.

The velocity will be 19.80 m/s.

Given:

Normal Force = 1.2 x weight

Normal Force = 1.2 Mg

Mass = Mass of car + occupants

Centripetal Acceleration = [tex]\text a_c[/tex]

Now, applying the equilibrium equation:

[tex]\begin{aligned} \sum \text F&= \text m \times \text a_c\\\\\text F_n - \text{mg} &= m \times \text a_c\\\\1.2 \text{Mg} - \text{Mg} &= m \times \text a_c\\\\\text a_c &=0.2 \;\text g\\\end[/tex]

Now, substituting the value in the formula:

[tex]\begin {aligned} \text a_c &=\dfrac{\text v^2}{\text r}\\\\0.2 \text g &= \dfrac {\text v ^2}{\text r}\\\\\text v &= \sqrt{0.2 \times 9.8 \times 200}\\\\\text v &= 19.80 \;\text {m/s}\end[/tex]

Thus, the velocity will be 19.80 m/s.

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The force on each side of the loop is [tex]2.26\times10^{-4}N[/tex] away from the centre .

It is given that solenoid has 30.0 turns/cM which means we have given number of turns per unit length [tex]n=\frac{N}{L} =30\times10^{2}m[/tex] and carries a clockwise current of 15.0A .

Putting above values in equation (1) , we get

[tex]B=4\pi \times10^{-7}\times 30\times10^{2} \times15.0\\B=0.0565T[/tex]

The magnetic force exerted due to their magnetic field on the sides of the loop will be away from the center of loop and is given by equation (2) .

It is given that current in the loop is  0.200 A and length of the each side of the loop is  2.00 cm [tex]=2\times10^{_2}m[/tex] and plane of the loop perpendicular to the magnetic field of the solenoid.

Putting above values in equation (2) , we get

[tex]F=0.0565\times0.2\times2\times10^{-2}\times sin90\\F=2.26\times10^{-4}N[/tex]

Direction of force is found by Fleming' Left Hand rule which states that when the thumb, index finger and middle finger of the left hand are placed perpendicular to each other,  the thumb points in the direction of the magnetic force, the index finger points in the direction of the magnetic field, and the middle finger points in the direction of the flow.

According to Fleming's left hand rule, the magnetic force exerted due to their magnetic field on the sides of the loop will be away from the center of loop and is given by equation (2) .

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