Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one?
a) 16F
b) 8F
c) 4F
d) F
e)F/4

Answer:

584.3 m at [tex]135.7^{\circ}[/tex] above the positive x-axis

Explanation:

We have to find the magnitude and direction of the resultant vector. In order to do that, we have to resolve each vector along the x- and y- direction first.

Resolving the vector L:

[tex]L_x = L cos \theta_L = 303 \cdot cos 205^{\circ} =-274.6m\\L_y = L sin \theta_L = 303 \cdot sin 205^{\circ} = -128.1 m[/tex]

Resolving the vector M:

[tex]M_x = M cos \theta_M = 555 \cdot cos 105^{\circ} =-143.6 m\\M_y = M sin \theta_M = 555 \cdot sin 105^{\circ} = 536.1 m[/tex]

Now we can find the components of the resultant vector by adding the components of each vector along each direction:

[tex]R_x = L_x + M_x = -274.6 +(-143.6)=-418.2 m\\R_y = L_y + M_y = -128.1 +536.1 = 408 m[/tex]

So the magnitude of the resultant vector is

[tex]R=\sqrt{R_x^2 + R_y^2}=\sqrt{(-418.2)^2+(408)^2}=584.3 m[/tex]

While the direction is given by:

[tex]\theta = tan^{-1} \frac{R_y}{R_x}=tan^{-1} \frac{408}{418.2}=44.3^{\circ}[/tex]

But since [tex]R_x[/tex] is negative and [tex]R_y[/tex] is positive, it means that this angle is measured as angle above the negative x-axis; so the direction of the vector is actually

[tex]\theta = 180^{\circ} - 44.3^{\circ} = 135.7^{\circ}[/tex]

Answer:

13.82 m/s

Explanation:

You can see it in the pic.

Answer:

vo = 13.74 m/s

Explanation:

The goal is to find the initial velocity vo. According to the exercise, the maximum height that the ball will reach and the initial height will be respectively, yo = 1.5 m and ymax = 7 m, tmax will be equal to the time it takes for the ball to reach the maximum height. Having the equation:

vx = vo * cos∝o

clearing vo:

vo = vx/cos∝o

vy = vo * sin∝o + gtmax

0 = sin∝o * (vx/cos∝o) + gtmax

0 = vx * tan∝o + gtmax

gtmax = -vx * tan∝o

clearing tmax:

tmax = - (vx * tanao/g)

ymax = i + (vo * sin∝o) * tmax + (gtmax^2/2)

Replacing:

ymax = i - (vx^2 * tan^2∝o/2 * g)

clearing the angle a or:

∝o = arctan (((2 * g * (i - ymax)) / vx^2)^1/2)

substituting values:

∝o = arctan (((2 * (- 9.8) * (1.5 - 7)) / (9^2))) = 49.08 °

We find the initial speed with the following formula:

vo = vx/cos∝o = 9/cos49.08 ° = 13.74 m/s

Answer: 3250 N in the opposite direction

Explanation:

According to Newton's third law of motion:

"Each action has an equal and opposite reaction"

To understand it better:  

When two bodies interact between them, appear equal forces and opposite senses in each of them.  For example, if a body A exerts a force on another body B, it performs on A another force with the same magnitude and in the opposite direction.

Hence, this law states that for each force acting on a body, ther is a force of equal intensity but in the opposite direction on the body that produced it.

In this sense if the car hit the concrete barrier with a force of 3250 N, the concrete barrier will exert a force of -3250 N on the car (where the negative sign indicates the force is acting in the opposite direction).

Answer:3250 , third law

Explanation:

Answer:

[tex]v_1 = 896.35 m/s[/tex]

Explanation:

As we know that bullet + pendulum system will move to the height of 0.650 m above the initial position

so here we can use energy conservation to find the speed just after the bullet hit the block

[tex]mgh = \frac{1}{2}mv^2[/tex]

[tex]v = \sqrt{2gh}[/tex]

[tex]v = \sqrt{2(9.81)(0.650)}[/tex]

[tex]v = 3.57 m/s[/tex]

Now we can use momentum conservation to find the initial speed of the bullet

[tex]m_1v_1 = (m_1 + m_2)v[/tex]

[tex]0.0100 v_1 = (2.50 + 0.01)(3.57)[/tex]

[tex]v_1 = 896.35 m/s[/tex]

Answer:

(a) Magnitude = [tex]1.14\times 10^5\ N/C[/tex]

Direction = toward the center of the disc

(b) Magnitude = [tex]8.90\times 10^4\ N/C[/tex]

Direction = toward the center of the disc

(c) Magnitude = [tex]1.46\times 10^5\ N/C[/tex]

Direction = toward the center of the disc

(d) See explanation

(e) See explanation.

Explanation:

Given:

Assume:

Part(a):

Using the formula for the electric field due to the uniform charged disc at a point P on its axis, we have

[tex]E = \dfrac{2 kQ}{r^2}\left ( 1-\dfrac{x}{\sqrt{x^2+r^2}} \right )\\\Rightarrow E = \dfrac{2 \times 9\times 10^9\times (-6.50\times 10^{-9})}{(0.0125)^2}\left ( 1-\dfrac{0.02}{\sqrt{(0.02)^2+(0.0125)^2}} \right )\\\Rightarrow E =-1.14\times 10^5\ N/C\\[/tex]

Hence, the magnitude of electric field at the given distance from the disc along its axis is [tex]1.14\times 10^5\ N/C[/tex] in the direction toward the center of the disc along the axis.

Part(b):

When all the charges on the disc are pushed away from the center of the disc to get uniformly distributed to its outer rim, the disc behaves like a uniformly charged ring.

Using the formula for the electric field due to the uniform charged ring at a point P on its axis, we have

[tex]E= \dfrac{kQx }{\left ( \sqrt{x^2+r^2} \right )^3}\\\Rightarrow E = \dfrac{ 9\times 10^9\times (-6.50\times 10^{-9})\times 0.02}{\left ( \sqrt{(0.02)^2+(0.0125)^2} \right )^3}\\&=-8.90\times 10^4\ N/C[/tex]

Hence, the magnitude of electric field at the given point from the rim along its axis is [tex]8.90\times 10^4\ N/C[/tex], and it is also toward the center of the along the axis.

Part (c):

When all the charge on the disc is pushed to the center of the disc, then the disc behaves as a point charge.

Using the formula of electric field due to a point charge at a fixed point, we have

[tex]E = \dfrac{kQ}{x^2}\\\Rightarrow E = \dfrac{9\times 10^9\times (-6.50\times 10^{-9})}{(0.02)^2}\\\Rightarrow E = -1.46\times 10^5[/tex]

Hence, the magnitude of electric field due to the point charge at the point P on the axis is [tex]1.46\times 10^5\ N/C[/tex]. which also points toward the center of the disc along the axis.

The direction of the electric field is found toward the negative x-axis because the electric field due to a point negative charge is radially inward.

Part (d):

The magnitude of electric field in part (a) is stronger than that in part (b) because the disc in part (a) is composed of infinite number of rings having radius varying from 0 cm to 2.5 cm whose components along the axis of electric field are added together which is greater than the electric field produced by a single ring in part (b).

Hence, the electric field in part (a) is greater than the electric field in part (b).  

Part (e):

The electric field in part (c) is the strongest of the three fields. This is because the field due to a point charge on the axis has the only component along the axis. However, the field components in part (a) and part (b) are distributed along two axes whose axial component is considerable and the other component gets canceled out. This causes the field strength to be weaker in part (a) and part (c).

Answer:

18.3 m , 25.4°

Explanation:

d1 = 6 m, θ1 = 40°

d2 = 8 m, θ2 = 30°

d3 = 5 m, θ3 = 0°

Write the displacements in the vector form

[tex]d_{1}=6\left ( Cos40\widehat{i}+Sin40\widehat{j} \right )=4.6\widehat{i}+3.86\widehat{j}[/tex]

[tex]d_{2}=8\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )=6.93\widehat{i}+4\widehat{j}[/tex]

[tex]d_{3}=5\widehat{i}[/tex]

The total displacement is given by

[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]

[tex]d=\left ( 4.6+6.93+5 \right )\widehat{i}+\left ( 3.86+4 \right )\widehat{j}[/tex]

[tex]d=16.53\widehat{i}+7.86\widehat{j}[/tex]

magnitude of resultant displacement is given by

[tex]d ={\sqrt{16.53^{2}+7.86^{2}}}=18.3 m[/tex]

d = 18.3 m

Let θ be the angle of resultant displacement with + x axis

[tex]tan\theta =\frac{7.86}{16.53}=0.4755[/tex]

θ = 25.4°

Answer: Ok, first start writing our movement equation.

First our acceleration will be the gravity; so a= g

integrating we can obtain the velocity  v= g*t + v₀

where v₀ depends on if we trhow the egg upsides or downsides, being the positive direction downsides.

after, the position will be r = [tex]\frac{gt^{2} }{2}[/tex] +  v₀*t - 12m

where the -12m is because you are upside a building.

first case: straight down with speed 11m/s

so the velocity is positive

then, if we  do [tex]\frac{10*t^{2} }{2}[/tex] +  11*t - 12m = 0 and with bashkara obtain the positive root for the time t=1.6 seconds

putting this on our velocity equation you will get v = 10*1.6 + 11 = 27m/s is the velocity of the egg when it hits the ground.

for the other case, our equation will have the form of:

 [tex]\frac{10*t^{2} }{2}[/tex] -  11*t - 12m = 0

and the positive root of the time is: t = 3 seconds.

putting it in the velocity equation gives you v= 10*3 - 11 = 29m/s

which is bigger than the first case, so the egg hits the ground with more velocity, ence more energy.

Final answer:

Throwing the egg straight down with an initial speed of 11 m/s from a 12 m height will result in a higher impact speed, calculated using the kinematic equation for final velocity, which factors in both the initial speed and acceleration due to gravity.

Explanation:

To maximize the impact of the egg and break it the hardest in the egg drop contest, we need to consider the effect of throwing it downward with an initial speed. If the egg is thrown straight down with an initial velocity of 11 m/s from a 12 m high building, it will gain speed due to gravity in addition to the initial speed given. The final velocity just before impact can be calculated using the kinematic equation:

v2 = u2 + 2gh

where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (9.8 m/s2), and h is the height.

Substituting the given values:

v2 = (11 m/s)2 + 2(9.8 m/s2)(12 m)

v2 = 121 m2/s2 + 235.2 m2/s2

v2 = 356.2 m2/s2

v = √356.2 m2/s2 ≈ 18.9 m/s

Therefore, throwing the egg straight down at 11 m/s will result in a higher impact speed upon hitting the ground compared to throwing it upwards or dropping it without initial velocity.

Answer:

2. Change in velocity

Explanation:

The units that acceleration use are distance/time^2 [tex](m/s^2)[/tex] (in metric units).

The acceleration is the change of velocity over time, so

[tex]\frac{dv}{dt} =\frac{dx}{dt}*\frac{1}{dt}[/tex]

In another way, if the velocity is constant, the acceleration is zero, but if the velocity change over the time (as a car do when the red light change), the acceleration takes value!.

Velocity changes in displacement divided by time. Acceleration is the change in velocity divided by time.  

Answer: Option 2

Explanation:

          Acceleration is defined as the rate of velocity changes of an object in a given time period. This can express in the equation as follows,  

                                [tex]\bold{a = \frac{v}{t} m / s^{2}}[/tex]

          There two types in it as linear and radial acceleration. When velocity changes in the opposite direction to an object then it is called deceleration. But these are the same as a change in velocity.

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

The maximum tension at which the small box doesn't slip off the large box can be calculated using the formula T_max = μs * mg. Substituting the given values into the formula, the largest tension is found to be 29.4 N for the small box not to slip off a 10 kg wood sled on frictionless snow.

The maximum tension, T_max, for which the small box remains stationary with respect to the larger box can be found using the concept of static friction. This static friction, f_s, can be expressed as the product of the static friction coefficient, μs, and the normal force on the small box, which is its weight, mg. So, f_s = μs * mg.

Given that the maximum tension, T_max, equals f_s, we can derive the formula as T_max = μs * mg. This is an expression for the maximum tension in terms of the given variables.

For the numerical part of the question, given that the mass of the box m = 6 kg, and the coefficient of kinetic friction μk = 0.50, we can substitute these values into the derived formula: T_max = μk * m * g = 0.50 * 6.0 kg * 9.8 m/s^2 = 29.4 N. This is the largest tension force for which the box doesn't slip.

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The number of "wah"s every second is the difference between the frequencies of the tuning fork and the note he is playing.

Answer:

The two waves produced have different frequencies.

Explanation:

Rohit is trying to tune an instrument. He strikes a tuning fork and then plays a note that almost matches it. Rohit hears the volume of the two waves beating, which sounds like "wah-wah-wah-wah."

The options can be the following

The two waves produced have different speeds.

The two waves produced have different amplitudes.

The two waves produced have different wavelengths.

The two waves produced have different frequencies.

A wave is a disturbance which  transfers energy through a medium without displacing the medium itself.

The two waves have different frequencies to be able to produce such a sound. That's called a beat frequency. the waves interfere constructively.

Answer:

Acceleration will increase.

Explanation:

The relation between force, mass and acceleration according to the Newton's second law of motion is given as:

F = ma

We are given that the driving force on the truck remains constant, so F is constant here. We can rewrite the above equation as:

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

Since, F is constant, the acceleration of the truck is inversely proportional to the mass.

There is a hole at the bottom of the truck through which the sand is being lost at a constant rate. Since, the sand is being lost, the overall mass of the truck is being reduced.

Since, the acceleration of the truck is inversely proportional to the mass, the reduced mass will result in an increased acceleration.

So, the acceleration of the truck will increase.

As the truck loses mass due to leaking sand, while a constant driving force is maintained, the acceleration of the truck increases according to Newton's second law of motion.

The question is about the changes in the acceleration of a truck as it loses mass due to leaking sand. From a physics perspective, the truck's acceleration will actually increase as it loses mass. This is because the force propelling the vehicle remains constant while its mass decreases. According to Newton's second law of motion, which states that force equals mass times acceleration (F=ma), if the force remains constant and mass decreases, the acceleration must increase to keep the equation balanced.

For example, with initial values, if the force propelling the truck is 30 newtons and the truck (with its sand) has a mass of 15 kilograms, the acceleration would be 2 m/s² (from F=ma, or 30N=15kg*a). Now suppose the sand leaks out and the truck’s mass drops to 10kg. If the driving force stays the same at 30N, when plugging these values back into F=ma, the new acceleration would be 3 m/s². Thus, the acceleration of the truck has increased as it has lost mass.

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Answer:

h₂ = 27.22m : height the pellet reaches

Explanation:

Pellet kinetics

Pellet moves with uniformly accelerated movement in  

freefall  

v f₁²=v₀²+2g*h₁ (formula 1)  : ball movement down

v f₂²=v₀²-2g*h₂ (formula (formula 2)  : ball movement up

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

h :height in m

Known data  

g=9.8 m/s²

h₁=12.7m

v f₁= 27.9 m/s

Initial velocity calculation (v₀)

We replace the data in formula 1:

27.9²=v₀²+2*9.8* 12.7

27.9²-2*9.8* 12.7=v₀²

529.49=v₀²

[tex]v_{o} =\sqrt{529.49}[/tex]

[tex]v_{o} =23.1\frac{m}{s}[/tex]

Calculation of the height that the pellet reaches when it goes up

Data:

[tex]v_{o} =23.1\frac{m}{s}[/tex]

v f₂=0 :When the pellet reaches the maximum height the final speed is zero

We replace the data in formula (2)

0²=23.1²-2*9.8*h₂

19.6h₂ =23.1²

h₂ =23.1²÷19.6

h₂ =27.22m

Answer:

A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed.

Explanation:

Final answer:

The velocity of a ball changes after bouncing off a wall because while its speed remains the same, its direction is reversed. Therefore, the ball's velocity and momentum experience a change, but the overall system momentum is conserved. So the correct option is A.

Explanation:

When considering whether the velocity of a ball changes after it bounces off a wall, it is important to understand that velocity is a vector quantity, meaning it has both magnitude (speed) and direction. If a ball is thrown against a wall and bounces back with the same speed but in the opposite direction, its velocity has indeed changed due to a change in direction. The correct answer here is A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed. The mass of the ball is irrelevant to the consideration of velocity in this context, as velocity does not depend on mass.

Additionally, during this process, the ball experiences a change in momentum. Momentum is also a vector, and the direction of the ball's momentum is reversed upon bouncing off the wall while maintaining the same magnitude, assuming an elastic collision with no energy loss. The change in the direction of the ball's velocity also indicates a reversal in momentum. However, the total system momentum, including the wall and the ball, remains conserved.

Final answer:

The magnitude of the force the 24-kg box applies to the 62-kg box on a frictionless surface can be calculated using Newton's second law (F=ma), considering both boxes as a single system to find the combined acceleration, then applying that acceleration to determine the force on the 62-kg box.

Explanation:

To find the magnitude of the force that the 24-kg box applies to the 62-kg box, we need to apply Newton's second law of motion which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F=ma). Since the 42-N pushing force is applied to the 24-kg box, this force is responsible for accelerating both the 24-kg and 62-kg boxes, which are considered to be one system in this context. Therefore, the acceleration of the system can be calculated using a = F_total/m_total, where F_total is the total force (42 N) and m_total is the combined mass of both boxes (24 kg + 62 kg).

Once the acceleration is known, the force exerted by the 24-kg box on the 62-kg box can be found by calculating the force needed to accelerate the 62-kg box at the previously determined acceleration. This is found using the same equation F=ma, where m is the mass of the 62-kg box and a is the acceleration of the system.

Explanation:

Given that, the height of an object tossed upward with an initial velocity of 136 feet per second is given by the formula :

[tex]h=-16t^2+136t[/tex]

Where

h is the height in feet

t is the time in seconds

Let t is the time required for the object to return to its point of departure. At this point, h = 0

[tex]-16t^2+136t=0[/tex]

t = 8.5 seconds

So, the time required for the object to return to its point of departure is 8.5 seconds.

Answer:

a) 90 miles is the distance between town and John's current location.

b) John's current position is the 104 mile marker.

Explanation:

See in the picture.

Answer:

weight  is 442.10 kg

Explanation:

given data

weight = 1400 kg

length = 6 m

to find out

how many kilogram 19 m beam support

solution

we know by inverse variation formula of weight  that is

weight = [tex]\frac{k}{l}[/tex]   .............1

here k is constant and l is length

so  k will be

1400 =  [tex]\frac{k}{6}[/tex]

k = 8400

so for 19 m weight will be by equation 1

weight = [tex]\frac{k}{l}[/tex]

weight = [tex]\frac{8400}{19}[/tex]

weight = 442.10 kg

The weight that a horizontal beam can support varies inversely as the length of the beam. Using this relationship, we can find the weight a 19-meter beam can support given that a 6-meter beam can support 1400 kg.

The weight W that a horizontal beam can support varies inversely as the length L of the beam. This can be expressed mathematically as W = k/L, where k is a constant. In this case, we are given that a 6-meter beam can support 1400 kg. Let's use this information to find the value of k.

Using the given information, we can write the equation as 1400 = k/6. To solve for k, we can multiply both sides of the equation by 6, giving us k = 1400 x 6 = 8400.

Now that we have the value of k, we can use it to find how many kilograms a 19-meter beam can support. Using the equation W = k/L, we can plug in the values: W = 8400/19. Solving this equation gives us W = 442.105 kg (rounded to three decimal places). Therefore, a 19-meter beam can support approximately 442.105 kilograms.

Answer:

d = 39.7 km

Explanation:

initial position of the boat is 45 km away at an angle of 15 degree East of North

so we will have

[tex]r_1 = 45 sin15 \hat i + 45 cos15 \hat j[/tex]

[tex]r_1 = 11.64 \hat i + 43.46\hat j[/tex]

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

[tex]r_2 = 30 cos15\hat i + 30 sin15 \hat j[/tex]

[tex]r_2 = 28.98\hat i + 7.76 \hat j[/tex]

now the displacement of the boat is given as

[tex]d = r_2 - r_1[/tex]

[tex]d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)[/tex]

[tex]d = 17.34 \hat i - 35.7 \hat j[/tex]

so the magnitude is given as

[tex]d = \sqrt{17.34^2 + 35.7^2}[/tex]

[tex]d = 39.7 km[/tex]

Still 9.8 m/s^2 downward.

Without air resistance, the acceleration doesn't depend on the initial velocity or the weight of the object. It only depends on what planet you're on.

Answer:longer

Explanation:

Answer:

Longer

Explanation:

Answer:

Part 2: 4.30 m/s

Part 3: 1503 m/s^2

Explanation:

First you already had -5.29 m/s

Part 2

With what velocity does it leave the ground ?

it is the same process as question 1, but in this case the height is 0.945 m

[tex]v =\sqrt{2gh}[/tex]

[tex]v =\sqrt{2*9.8*0.945} \\v =4.30\ m/s[/tex]

Part 3

This is a simple formula of :

[tex]v_{f}-v_{i}=at \\where \\v_{f}: final\ speed \\v_{i}: initial\ speed \\a: acceleration\\t: time[/tex]

Final speed: is the speed the ball leaves the ground = 4.30 m/s

Initial speed: is the speed the ball hits the ground = -5.29 m/s

4.30 - (-5.29) = 0.00638a

a = 1503 m/s^2

I was running at 25 km/h speed when i arrived the school. Thus, option c) 25 km/h is correct.

To determine how fast you were running when you arrived at school, we need to use the formula that relates initial velocity, acceleration, and time:

v = u + at

Where:

Given that:

We can substitute these values into the formula:

v = 10 km/h + (30 km/h² * 0.5 h)

This simplifies to:

v = 10 km/h + 15 km/h

Thus, v = 25 km/h.

Therefore, you were running at 25 km/h when you arrived at school. The correct answer is c) 25 km/h.

Answer:

B. Nebular Hypothesis

Explanation:

It is important to read the statement carefully so we can rule out options. The text is referring to solar system formation, it means we should discard all options which are not related to that. Big bang theory is a concept to explain the origin of the whole universe, not the solar system only, so it is discarded. Plate tectonics describes inner earth layers that are essential parts of the planet movemements like earthquakes and they are not related to solar system either. Discarded. Helioentric theory is the idea that defines the sun as the center of the solar system, but again it does not explain anything about its origin, so let me explain a litte in detail about the right choice: Nebular Hypothesis.

Nebular Hypothesis

This is the way astronomers accept the solar system was created as well as other other plantetary systems. Million of years after Big Bang, there were gases and small dust particles all over the universe and they started to join and concentrate into bigger and bigger amount of gases turning into enourmous clouds made of all the necessary elements to begin the formation of stars, planets and eventually the solar system.

Answer:

the equation is dimensionally consistent

Explanation:

To verify that the formula is dimensionally consistent, we must verify the two terms of the sum and verify that they are units of force. We achieve this by putting the units of each dimensional term of the equation and verifying that the answer is in units of force

μ=viscosity=Ns/m^2

D=diameter

V=velocity

ρ=density=Kg/m^3π

First term

3πμDV=[tex]\frac{N.s.m.m}{m^{2}.s }[/tex]=N

the first term is  dimensionally consistent

second term

(9π/16)ρV^2D^2=[tex]\frac{kg.m^2.m^2}{m^3.s^2} =\frac{kg.m}{s^2} =N[/tex]

.

as the two terms are in Newtons it means that the equation is dimensionally consistent

Final answer:

Richard will save approximately 13.68 minutes by driving at 73 mph for 135 miles instead of at the speed limit of 65 mph. The mathematical concept used involves calculating travel time at different speeds and finding the difference between these times.

Explanation:

The student's question involves calculating how much time Richard will save by driving at 73 mph instead of the speed limit of 65 mph for a distance of 135 miles. This is a mathematics problem that falls under the category of rate, time, and distance, which is often covered in high school math classes.

To solve this, we need to calculate the time taken to travel 135 miles at both speeds and then find the difference in time. At 65 mph, the time taken (T1) is 135 miles / 65 mph. At 73 mph, the time taken (T2) is 135 miles / 73 mph. The time saved is T1 - T2.

Using the formula time = distance / speed, we get:

Time at 65 mph: T1 = 135 mi / 65 mph = 2.077 hours

Time at 73 mph: T2 = 135 mi / 73 mph = 1.849 hours

The time saved is T1 - T2 = 2.077 hours - 1.849 hours = 0.228 hours, which is about 13.68 minutes.

Answer:

Final velocity will be 556.8 m/sec

Explanation:

We have given the average rate of change of velocity that is [tex]a=5.8\times 10^5m/sec[/tex]

Time [tex]t=9.6\times 10^{-4}sec[/tex]

Initial velocity of bullet will be zero that is u=0

Now according to first law of motion

v=u+at, here v is final velocity, u is initial velocity and t is time

So [tex]v=0+5.8\times 10^5\times 9.6\times 10^{-4}=556.8m/sec[/tex]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

                               [tex]A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex]

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

                                 [tex]A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )[/tex]

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

                                T = 2*p / w

                                T = 2*p / (2*p / 92.8)

Hence,                     T = 92.8 min

The time of the International Space Station  is mathematically given as

T = 5568sec

Generally, the equation for the altitude of the International Space Station  is mathematically given as

[tex]A(t) \left = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex]

Therefore

Using the equation

[tex]A(t) \left = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})[/tex] we decipher time to be

     T = 2*p / w

In conclusion

T = 2*p / (2*p / 92.8)

T = 5568sec

Read more about Time

https://brainly.com/question/4931057

Answer:

1.336 m

Explanation:

q = - 3.77 micro C at y = 0 m

Q = - 4.92 micro C at y = 2.86176 m

Let the electric field is zero at P which is at a distance y from origin.

Electric field at P due to q is equal to the electric field at P due to Q.

[tex]E_{q}=E_{Q}[/tex]

[tex]\frac{kq}{OP^{2}}=\frac{kQ}{AP^{2}}[/tex]

[tex]\frac{3.77}{y^{2}}=\frac{4.92}{\left ( 2.86176-y \right )^{2}}[/tex]

[tex]\frac{\left ( 2.86176-y \right )}{y}}=1.142[/tex]

2.86176 - y = 1.142 y

2.142 y = 2.86176

y = 1.336 m

Thus, the electric field is zero at y = 1.336 m.

Answer:

21 seconds

Explanation:

First, convert km/hr to m/s.

69 km/hr × (1000 m / km) × (1 hr / 3600 s) = 19.2 m/s

Distance = rate × time

400 m = 19.2 m/s × t

t = 20.9 s

Rounded to two significant figures, it takes the horse 21 seconds.

Answer:

The answer to your question is: 21 seconds

Explanation:

Data

speed = v = 69 km/h

distance = d = 400 m

time = t = ?

Formula

v = d / t

t = d / v

Process

Convert 400 m to km

                                 1 km ----------------- 1000 m

                                 x     -------------------   400 m

                                x = (400 x 1) / 1000

                                x = 0.4 km

Substitution

                      t = 0.4 / 69

                      t = 0.006 h

Covert time to minutes and seconds

                     60 min -------------------- 1  h

                     x            -------------------  0.006 h

                     x = (0.006 x 60) / 1

                     x = 0.36 min

Convert time to seconds

                     1 min ------------------  60 s

                   0.36 min --------------   x

                     x = (0.36 x 60) / 1

                     x = 21.6 seconds  ≈ 21 seconds

Answer: [tex]38\ m[/tex]

Explanation:

For this exercise you can use the Pythagorean theorem to find the magnitude of the resultant displacement.

Then:

[tex]d^2=\triangle x^2+\triangle y^2[/tex]

You can observe that the square of the displacement is equal to the sum of the square  of the horizontal displacement and the square of the vertical displacement.

Since the pirate walks 37.0 meters north and then turns and walks 8.5 meters east:

[tex]\triangle x=37.0\ m\\\triangle y=8.5\ m[/tex]

Substituting values and solving for "d", you get:

[tex]d=\sqrt{(37.0\ m)^2+(8.5\ m)^2}\\\\d=38\ m[/tex]