A 120-cm-thick layer of oil floats on a 140-cm-thick layer of water. What is the pressure at the bottom of the water layer?

Final answer:

The average velocity for the trip from the front door to the bench is 0.143 m/s due east, and the average speed for the same trip is 1.286 m/s.

Explanation:

To calculate the average velocity for the entire trip, we need to consider the displacement (final position relative to the starting position) and the total time taken for the trip. The displacement is 10.0 m east (50.0 m due east - 40.0 m back west), and the total time is 28.0 s + 42.0 s, which equals 70.0 s. Therefore, the average velocity is displacement divided by time, calculated as follows:

Average Velocity = Displacement/Total Time = 10.0 m / 70.0 s = 0.143 m/s due east.

To calculate the average speed, we consider the total distance traveled and the total time. The distance is 50.0 m east + 40.0 m west = 90.0 m, and the time is the same 70.0 s. Thus, the average speed is:

Average Speed = Total Distance/Total Time = 90.0 m / 70.0 s = 1.286 m/s.

Answer: 0.506 m

Explanation: To solve this we use the relationship for the harmonic in the string which are given by the following expression:

λ=2*L/3

λ=2*0.76 m/3= 0.506 m

Answer:

D. Community members, teachers

Explanation:

In the HighScope Curriculum teachers work in collaboration with family members, thus encouraging greater learning in students. They do this by providing information about the curriculum, inviting family members to participate in the activities carried out in the classroom, workshops for parents are also held. This allows discussing children's progress and sharing ideas to extend classroom learning to home.

Answer:

7.468 kN

Explanation:

Here the force of 7468 Newton is given.

Some of the prefixes of the SI units are

kilo = 10³

Mega = 10⁶

Giga = 10⁹

The number is 7468.0

Here, the only solution where the number of significant figures is kilo

1 kilonewton = 1000 Newton

[tex]1\ Newton=\frac{1}{1000}\ kilonewton[/tex]

[tex]\\\Rightarrow 7468\ Newton=\frac{7468}{1000}\ kilonewton\\ =7.468\ kilonewton[/tex]

So 7468 N = 7.468 kN

Answer:

The minimum distance is 40 m.

(a) is correct option.

Explanation:

Given that,

Mass of vehicle = 1500 kg

Coefficient friction = 0.80

Speed = 25 m/s

We need to calculate the acceleration

Using frictional force

[tex]f=\mu mg[/tex]

Put the value into the formula

[tex]ma=\mu mg[/tex]

[tex]a = \mu g[/tex]

We need to calculate the distance

Using equation of motion

[tex]v^2=u^2+2as[/tex]

Put the value into the formula

[tex]0=25^2-2times\mu g\times s[/tex]

[tex]s=\dfrac{25^2}{2\times0.80\times9.8}[/tex]

[tex]s=39.8 = 40 m[/tex]

Hence, The minimum distance is 40 m.

Answer:

a) [tex]v_{3} =8.43 m/s[/tex]

b) [tex]v_{2}=2.15m/s[/tex]

c) ΔK=[tex]-28.18x10^4J[/tex]

d)ΔK=[tex]-10.33x10^4J[/tex]

Explanation:

From the exercise we know that there is a collision of a sports car and a truck.

So, the sport car is going to be our object number 1 and the truck object number 2.

[tex]m_{1}=1050kg\\v_{1}=-13m/s\\m_{2}=6320kg\\v_{2}=12m/s[/tex]

Since the two vehicles remain locked together after the collision the final mass is:

[tex]m_{3}=7370kg[/tex]

a) To find the velocity of the two vehicles just after the collision we must use linear's momentum principle

[tex]p_{1}=p_{2}[/tex]

[tex]m_{1}v_{1}+ m_{2}v_{2}=m_{3}v_{3}[/tex]

[tex]v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3}}=\frac{(1050kg)(-13m/s)+(6320kg)(12m/s)}{7370kg}[/tex]

[tex]v_{3}=8.43m/s[/tex]

b) To find the speed the truck should have had so both vehicles stopped in the collision we need to use the same principle used before

[tex]m_{1}v_{1}+ m_{2}v_{2}=0[/tex]

[tex]v_{2}=\frac{-m_{1}v_{1}}{m_{2} }=\frac{-(1050kg)(-13m/s)}{(6320kg)}=2.15m/s[/tex]

c) To find the change in kinetic energy we need to do the following steps:

ΔK=[tex]k_{2}-k_{1}=\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )[/tex]

ΔK=[tex]\frac{1}{2}(7370)(8.43)^{2}-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(12)^{2} )=-28.18x10^{4}J[/tex]

d) The change in kinetic energy where the two vehicles stopped in the collision is:

ΔK=[tex]k_{2}-k_{1}=0-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )[/tex]

ΔK=[tex]-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(2.15)^{2} )=-10.33x10^4J[/tex]

Answer:

a) The taxi is 107 s in motion

b) The average velocity is 26.2 m/s

Explanation:

First, the car travels with an acceleration of 2.00 m/s². The equations for position and velocity that apply for the car are:

x = x0 + v0 t + 1/2 a t²

v = v0 + a t

where

x = position at time t

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = speed

Let´s calculate how much distance and for how long the taxi travels until it reaches a speed of 29.0 m/s:

Using the equation for velocity:

v = v0 + a t

v - v0 / a = t

(29.0 m/s - 0 m/s) / 2 m/s² = t

t = 14.5 s

Then, in the equation for position:

x = x0 + v0 t + 1/2 a t²

x = 0 + 0 + 1/2 * 2.00 m/s² * (14.5 s)²

x = 210 m

Then, the vehicle travels at constant speed for 87 s. The distance traveled will be:

x = v * t

x = 29.0 m/s * 87.0 s = 2.52 x 10³ m

Lastly the car stops (v = 0) in 5 s. In this case, the car has a constant negative acceleration:

Using the equation for velocity:

v = v0 + a t

if v=0 in 5 s, then:

0 = 29.0 m/s + a * 5.00 s

a = -29.0 m/s / 5.00 s  

a = -5.80 m/s²

Using now the equation for the position, we can calculate how far has the taxi traveled until it came to stop:

x = x0 + v0 t + 1/2 a t²

x = 0 + 29.0 m/s * 5.00 s -1/2 * 5.80 m/s² * (5.00s)²

x = 72.5 m

a) The taxi has been in motion for:

Total time = 14.5 s + 87.0 s + 5.00s = 107 s

Note that we have always used x0 = 0, then, we have calculated the displacement for each part of the trip.

Adding all the displacements, we will get the total displacement:

Total displacement = 210 m + 2.52 x 10³ m + 72.5 m = 2.80 x 10³ m

Average speed = total displacement / total time

Average speed = 2.80 x 10³ m / 107 s = 26.2 m/s

Answer:

(a). The displacement is 3.11 m.

(b). The average velocity is 2.42 km/hr.

Explanation:

Given that,

A bird watcher meanders through the woods, walking 1.46 km due east, 0.123 km due south, and 4.24 km in a direction 52.4° north of west.

(a). We need to calculate the displacement

Using Pythagorean theorem

[tex]D=\sqrt{(OA)^2+(AB)^2}[/tex]

[tex]D=\sqrt{(1.46-4.24\sin52.4)^2+(0.123-4.24\cos52.4)^2}[/tex]

[tex]D=3.11\ km[/tex]

(b). We need to calculate the average velocity

Using formula of average velocity

[tex]v_{avg}=\dfrac{D}{T}[/tex]

Where, D = displacement

T = time

Put the value into the formula

[tex]v_{avg}=\dfrac{3.11}{1.285}[/tex]

[tex]v_{avg}=2.42\ km/hr[/tex]

Hence, (a). The displacement is 3.11 m.

(b). The average velocity is 2.42 km/hr.

Answer:

260 g

Explanation:

Given:

Let n be the number of such silicon dioxide that would give the desired mass of sand.

According to the question, the surface area of all the sand particles will be equal to the surface area of the cube.

[tex]\therefore n\times \textrm{Area of a sand particle}=\textrm{Surface area of a cube }\\\Rightarrow n\times 4\pi r^2= 6a^2\\\Rightarrow n = \dfrac{6a^2}{4\pi r^2}[/tex]

Let the total mass of sand required be M.

[tex]\textrm{Total mass of sand} = \textrm{Mass of all the required sand particle}\\\Rightarrow M = n\times \textrm{mass of one sand particle}\\\Rightarrow M = n\times \textrm{Density of sand particle}\times \textrm{Volume of a sand particle}\\\Rightarrow M = n\times\rho \times \dfrac{4}{3}\pi r^3\\[/tex]

[tex]\Rightarrow M = \dfrac{6a^2}{4\pi r^2}\times\rho \times \dfrac{4}{3}\pi r^3\\\Rightarrow M = 2\times\rho \times a^2r\\\Rightarrow M = 2\times2600 \times (1)^2\times 5\times 10^{-5}\\\Rightarrow M =0.260\ kg\\\Rightarrow M =260\ g\\[/tex]

Hence, the total mass of the sand required will be equal to the 260 g.

Answer:

Distance, d = 101.388 meters

Explanation:

It is given that,

The average velocity of plane, v = 3650 km/h = 1013.88 m/s

The time for which eye blinks, [tex]t=100\ ms=0.1\ s[/tex]

Let d is the distance covered by the jet. It can be calculated as :

[tex]d=v\times t[/tex]

[tex]d=1013.88\ m/s\times 0.1\ s[/tex]        

d = 101.388 meters

So, the distance covered by a fighter jet during a pilot's blink is 101.388 meters. Hence, this is the required solution.                                                                              

Final answer:

During a blink that lasts 0.1 seconds, a fighter jet traveling at an average velocity of 3650 km/h will cover approximately 101.39 meters.

Explanation:

To calculate the distance a fighter jet travels during a pilot's blink, we can use the formula distance = velocity × time. The pilot's blink lasts 100 milliseconds (ms) which is 0.1 seconds because 1000 ms equals 1 second. The jet's velocity is given as 3650 kilometers per hour (km/h).

First, we'll need to convert the jet's velocity to meters per second (m/s) since the time of the blink is given in seconds. 1 km equals 1000 meters, and there are 3600 seconds in an hour, so:

Velocity in m/s = (Velocity in km/h) × (1000 m/km) / (3600 s/h) = (3650) × (1000) / (3600) = 1013.89 m/s

Now, we will multiply the velocity in m/s by the time in seconds to get the distance:

Distance = Velocity × Time = (1013.89 m/s) × (0.1 s) = 101.39 meters

So, during the blink of an eye, which is 0.1 seconds, a fighter jet traveling at an average velocity of 3650 km/h will cover approximately 101.39 meters.

Answer:

a) [tex]R_s = 0.092[/tex]

b) [tex]R_p = 0.085[/tex]

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

[tex]R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2[/tex]

using snell's law to calculate θ t

[tex]sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}[/tex]

[tex]cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}[/tex]

a) [tex]R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2[/tex]

[tex]R_s = 0.092[/tex]

b) [tex]R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2[/tex]

[tex]R_p = 0.085[/tex]

Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h  east. c) 16,5 km NE of the desired position

Explanation:

Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where

see fig 1

Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then

90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,

this because we have an isosceles triangle, then the cathetus length is  

hypotenuse/[tex]\sqrt{2}[/tex]

using Pythagoras, here the hypotenuse is 90, then the cathetus are of length

90/[tex]\sqrt{2}[/tex] km/h= 63,6396 km/h.  

Now the total speed of the plane is

690km/h south + 63,6396 km/h north +63,6396 km/h east,

this is 626,3604 km/h south + 63,6396 km/h east,  here north is as if we had -south.

then using again Pythagoras we get the magnitude of the total speed it is

[tex]\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h [/tex],

the direction is calculated with respect to the south using trigonometry, we know the

sin x= cathetus opposed / hypotenuse,

then

x= [tex]sin^{-1}'frac{63,6396}{629,5851}[/tex]=5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form

626,6396 km/h south + 63,6396km/h  east.

Finally since the detour is caused by the west speed component plus the slow down caused by the north component of the wind speed, we get

Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,

since 11 min=11/60 hours=0.1833 hours.

Then the total detour from the expected position, the one it should have without the influence of the wind, we get  

Xdetour=[/tex]\sqrt{2*  11,6672x^{2} }[/tex]  = 16,5km at 45 degrees from east pointing north

The situation is sketched as follows  see fig 2

Answer:

13.65 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 9.5 m

a = Acceleration = 9.81 m/s²

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 9.5-0^2}\\\Rightarrow v=13.65\ m/s[/tex]

The flower pot is moving at a speed of 13.65 m/s when it crashes the sidewalk.

Answer:

initial velocity is 3.62 m/s

Explanation:

given data

high = 1.32 m

length = 1.88 m

to find out

initial velocity

solution

we consider here top height point a and b point at ground and c point  at distance 1.88 m away from b on ground

and x is horizontal component and y is vertical component

so at point A initial velocity is Va = Vx i

and at point c velocity = Vc = Vx i + Vy j

first we calculate time taken when it come down by distance formula

distance = 1/2 ×gt²   ..............1

1.32 = 1/2 ×(9.8)t²

t = 0.519 sec

so velocity x = distance / time

velocity x = [tex]\frac{1.88}{0.519}[/tex]

velocity x = 3.622 m/s

so initial velocity is 3.62 m/s

Answer:

3.62 m/s

Explanation:

h = 1.32 m

d = 1.88 m

Let u be the initial horizontal velocity and the time taken by the board to reach the ground is t.

Use second equation of motion in vertical direction

[tex]s=ut+0.5at^{2}[/tex]

1.32 = 0 + 0.5 x 9.8 x t^2

t = 0.52 second

The horizontal distance = horizontal velocity x time

1.88 = u x 0.52

u = 3.62 m/s

Thus, the nitial speed of the skate board is 3.62 m/s.

Answers:

a) [tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]

b) [tex]\rho_{liq}= 1.48 g/cm^{3}[/tex]

c) When we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.

Explanation:

a) According to Archimedes’ Principle:

A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.

In this case we have a wooden cylinder floating (partially immersed) in water. This object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium.  This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero.

Hence:

[tex]W_(cylinder)=B[/tex] (1)

Where:

[tex]W_(cylinder)=m.g[/tex] is the weight of the wooden cylinder, where [tex]m[/tex] is its mass and [tex]g[/tex] gravity.

[tex]B[/tex] is the Buoyant force, which is the force the fluid (water in this situation) exert in the submerged cylinder, and is directed upwards.

We can rewrite (1) as follows:

[tex]m_{cylinder}g=m_{water}g[/tex] (2)

On the other hand, we know density [tex]\rho[/tex] establishes a relationship between the mass of a body andthe volume it occupies. Mathematically is expressed as:

[tex]\rho=\frac{m}{V}[/tex] (3)

isolating the mass:

[tex]m=\rho V[/tex]    (4)

Now we can express (2) in terms of the density and the volume of cylinder and water:

[tex]\rho_{cylinder} V_{cylinder} g=\rho_{water} V_{water} g[/tex] (5)

In this case [tex]V_{water}[/tex] is the volume of water displaced by the wooden cylinder (remembering Archimedes's Principle).

At this point we have to establish the total volume of the cylinder and the volume of water displaced by the sumerged part:

[tex]V_{cylinder}=\pi r^{2} h[/tex] (6)

Where [tex]r[/tex] is the radius and [tex]h=30 cm[/tex] the total height of the cylinder.

[tex]V_{water}=\pi r^{2} (h-h_{top})[/tex] (7)

Where [tex]h_{top}=13.5 cm[/tex] is the height of the top of the cylinder above the surface of water and [tex](h-h_{top})[/tex] is the height of the sumerged part of the cylinder.

Substituting (6) and (7) in (5):

[tex]\rho_{cylinder} \pi r^{2} h g=\rho_{water} \pi r^{2} (h-h_{top}) g[/tex] (8)

Clearing [tex]\rho_{cylinder}[/tex]:

[tex]\rho_{cylinder}=\frac{\rho_{water}(h-h_{top})}{h}[/tex] (9)

Simplifying;

[tex]\rho_{cylinder}=\rho_{water}(1-\frac{h_{top}}{h}[/tex] (10)

Knowing [tex]\rho_{water}=1g/cm^{3}[/tex]:

[tex]\rho_{cylinder}=1g/cm^{3}(1-\frac{13.5 cm}{30cm})[/tex] (11)

[tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex] (12) This is the density of the wooden cylinder

b) Now we have a different situation, we have the same wooden cylinder, which density was already calculated ([tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]), but the density of the liquid [tex]\rho_{liq}[/tex] is unknown.

Applying again the Archimedes principle:

[tex]\rho_{cylinder} V_{cylinder} g = \rho_{liq} V_{liq} g[/tex] (13)

Isolating [tex]\rho_{liq}[/tex]:

[tex]\rho_{liq}= \frac{\rho_{cylinder} V_{cylinder}}{V_{liq}}[/tex] (14)

Where:

[tex]V_{cylinder}=\pi r^{2} h[/tex]

[tex]V_{liq}=\pi r^{2} (h-h_{top})[/tex]

Then:

[tex]\rho_{liq}= \frac{\rho_{cylinder} \pi r^{2} h}{\pi r^{2} (h-h_{top})}[/tex] (15)

[tex]\rho_{liq}= \frac{\rho_{cylinder} h}{h-h_{top}}[/tex] (16)

[tex]\rho_{liq}= \frac{0.55 g/cm^{3} (30 cm)} {30 cm - 18.9 cm}[/tex] (17)

[tex]\rho_{liq}= 1.48 g/cm^{3}[/tex] (18) This is the density of the liquid

c) As we can see, it was not necessary to know the radius of the cylinder (we did not need to knoe its length and width), we only needed to know the part that was sumerged and the part that was above the surface of the liquid.

This is because in this case, when we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.

Answer and Explanation:

The motion of the viper is circular and it completes a semi- circle forming an arc and then retraces its path back.

Thus the force experienced by the bug is centripetal force thus centripetal acceleration and in the absence of this force the bug will get dislodged.

This centripetal force is mainly provided by the static friction between the blades and the bug.

When the wipers are moving with high velocity when turned on, larger centripetal force is required to keep the bug moving with the wiper on the arc  than at low setting.

Thus there are more chances for the bug to be dislodged at higher setting than at low setting.

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

[tex]1\ m=\frac{1}{1609.344}\ miles[/tex]

1 hour = 60×60 seconds

[tex]1\ s=\frac{1}{3600}\ hours[/tex]

[tex]28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h[/tex]

Top speed of the cheetah is 64.2 mi/h

Equation of motion

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2[/tex]

Acceleration of the cheetah is 8.68 m/s²

2)

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s[/tex]

It takes a cheetah 3.31 seconds to reach its top speed.

3)

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m[/tex]

It travels 47.5 m in that time

4) When s = 120 m

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s[/tex]

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

The cheetah's top speed of 28.7 m/s is approximately 64.2 mi/h. It takes a cheetah 3.31 seconds to reach its top speed, traveling a distance of 47.4 meters during this acceleration. To reach a stationary rabbit 120 meters away, it would take the cheetah a total of 5.84 seconds.

The question involves converting speeds from meters per second to miles per hour, finding the time taken to achieve a certain speed, calculating the distance traveled in that time, and determining the time required to reach a target.

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound.  The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

[tex]d=v\times t[/tex]

[tex]d=340\ m/s\times 5\ s[/tex]  

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.    

Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

[tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2[/tex]

Where:

x = Final distance

xo = Initial point

Vo = Initial velocity

a = Acceleration

t = time

We have the following values:

x = 1.66m      

xo = 0m (the kangaroo starts from the floor)

Vo = 0 m/s (each jump starts from the floor and from a resting position)

a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)

t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

[tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2[/tex]

[tex]1.66 = 0 + 0t + \frac{1}{2}9.8t^2[/tex]

[tex]1.66 = 4.9t^2[/tex]

[tex]\frac{1.66}{4.9}  = t^2[/tex]

[tex]\sqrt{0.339} = t\\ t = 0.582s[/tex]

It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

t = 1.164s

The kangaroo was 1.164s in the air before returning to Earth

Answer:

Explanation:

Given

Distance to grandmother's house=100 mi

it is given that during return trip Julie spend equal time driving with speed 30 mph and 70 mph

Let Julie travel x mi with 30 mph and 100-x with 70 mph

[tex]\frac{x}{30}=\frac{100-x}{70}[/tex]

x=30 mi

Therefore

Julie's Average speed on the way to Grandmother's house[tex]=\frac{100}{\frac{50}{30}+\frac{50}{70}}[/tex]

=42 mph

On return trip

[tex]=\frac{100}{2\frac{30}{30}}=50 mph[/tex]

Final answer:

Julie's average speed to Grandmother's house is 42.02 mph. The average speed for the return trip cannot be calculated without the total time of travel or the division of time at each speed. Her average speed for the entire round trip is approximately 26.67 mph, and the average velocity is 0 mph since the starting and ending points are the same.

Explanation:

To calculate Julie's average speed on the way to Grandmother's house, we need to use the formula for average speed, which is total distance traveled divided by the total time taken. Since Julie drives half the distance at 30.0 mph and the other half at 70.0 mph for a total distance of 100 miles, we can calculate the time taken for each half. At 30 mph, for 50 miles, the time taken is ≈ 1.67 hours, and at 70 mph, for 50 miles, the time taken is ≈ 0.71 hours. The total time is ≈ 2.38 hours. Therefore, average speed is 100 miles ÷ 2.38 hours = 42.02 mph.

On the return trip, she drives half the time at 30.0 mph and half the time at 70.0 mph. However, without additional information, we cannot calculate the average speed for the return trip because we need the total time or the portion of time spent at each speed. The calculation is different from the first trip because this time it is dependent on time, not distance. For her entire trip, if Julie returned home 7 hours and 30 minutes after she left, and assuming the same 100-mile distance back, her average speed for the entire trip is the total distance (200 miles) divided by the total time (7.5 hours), which is ≈ 26.67 mph. However, since she returns to the starting point, her displacement is zero, and thus her average velocity for the entire trip is 0 mph, similar to the example in Figure 2.10.

Answer with Explanation:

The force that is exerted on a object is proportional to the rate of change of momentum. Mathematically

[tex]F=\frac{\Delta p}{\Delta t}[/tex]

As we can see from the above equation the force that is exerted on the object is inversely proportional to the time in which this momentum change is brought meaning if the time interval in which the momentum change is brought about is larger smaller the force that acts on the object.

When the driver of a car crashes into a barrier the airbags in the car are deployed instantly as due to inertia of motion the driver continues to move in the original direction of motion thus hitting the air bag in the process. Since the air in the airbag is compressible it increases the time in which this momentum becomes zero, reducing the force that is exerted on the person, thus protecting the person.

Airbags reduce the force on a driver's head by increasing the time over which the head decelerates, thus applying the principle of impulse to protect during a car crash. They achieve this in tandem with seatbelts and car structures designed to crumple and absorb impact forces.

Airbags protect drivers during a crash by utilizing the principle of impulse, which is the product of the net force acting on an object and the time over which the force acts. The airbag increases the time over which the driver's head decelerates to a stop, in turn decreasing the force that acts on the head. By extending the stopping time from a fraction of a second (if hitting the dashboard) to a longer time (momentum is stopped by the airbag), the average force experienced by the driver's head is considerably lessened. This reduction in force reduces the likelihood of severe injuries. The airbag also has vents to deflate during the crash, preventing bounce-back injuries.

Further, the airbag works with seatbelts that may have variable tension to distribute forces across the body more evenly, thus reducing stress on any single part of the body. Modern vehicles include these safety features not just for comfort but for crucial safety improvements, allowing for a longer time of force application during a crash and therefore less force at any instant. The same concept applies to the car's structure with plastic components designed to crumple and prolong collision time, further reducing forces experienced by the passengers.

The skateboarder's horizontal distance traveled after leaving the ramp can be calculated from her initial velocity and time in the air during projectile motion.

The skateboarder's motion can be analyzed using kinematic equations. When she leaves the ramp, she will follow a projectile motion trajectory. The horizontal distance she travels can be calculated using her initial vertical velocity and the time she is in the air.

The skateboarder touches down approximately 4.87 meters from the end of the 1.0-meter-high, 30° ramp after being launched at a speed of 7.6 m/s.

First, we'll find the horizontal and vertical components of the skateboarder's initial velocity. The angle of the ramp is 30°, and her speed is 7.6 m/s:

Horizontal component ([tex]V_x[/tex]) = 7.6 * cos(30°)

or, [tex]V_x[/tex] = 7.6 * 0.866

or, [tex]V_x[/tex] = 6.58 m/s

Vertical component ([tex]V_y[/tex]) = 7.6 * sin(30°)

or, [tex]V_y[/tex] = 7.6 * 0.5

or, [tex]V_y[/tex] = 3.8 m/s

solving for [tex]{V_f}_y[/tex] in the horizontal and vertical components of velocity as the skater leaves the ramp :

[tex]V_f}_y^2 = V_y^2 - 2* g * \Delta x[/tex]

or, [tex]{V_f}_y[/tex] = [tex]\sqrt{V_y^2 - 2 * g * \Delta x }[/tex]

or, [tex]{V_f}_y[/tex] = [tex]\sqrt{|(3.8)^2 - 2 * 9.8 * 1.0|}[/tex]

or, [tex]{V_f}_y[/tex] = [tex]\sqrt{|14.44 - 19.6|}[/tex]

or, [tex]{V_f}_y[/tex] = [tex]\sqrt{|-5.16|}[/tex]

or, [tex]{V_f}_y[/tex] = √(5.16)

or, [tex]{V_f}_y[/tex] = 2.27 m/s

In the x-axis, there is no acceleration so, [tex]a_x[/tex] = 0 m/s²

[tex]{V_f}_x^{2}[/tex] = [tex]V_x^2[/tex]+ 2 * [tex]a_x[/tex] * Δx

or, [tex]{V_f}_x^{2}[/tex] = [tex]V_x^2[/tex] + 2 * 0 * Δx

or, [tex]{V_f}_x^{2}[/tex] = [tex]V_x^2[/tex]

or, [tex]{V_f}_x[/tex] = [tex]V_x[/tex]

or, [tex]{V_f}_x[/tex] = 6.58 m/s

The skateboarder reaches a height of 1.0 m. We will use the following kinematic equation to find the time (T) it takes for her to fall this distance:

y = [tex]{V_f}_y[/tex] * t + 0.5 * (-g) * t²

Here, y = -1.0 m (since she’s falling), [tex]{V_f}_y[/tex] = 2.27 m/s, and g = 9.8 m/s². Plugging in the values:

-1.0 = 2.27 * T - 0.5 * 9.8 * T²

or, 4.9 * T² - 2.27 * T - 1.0 = 0

Solving this quadratic equation (4.9 * T² - 2.27 * T - 1.0 = 0) for T gives:

T ≈ 0.74 seconds

Next, we calculate the horizontal distance traveled using this time and the horizontal velocity:

Distance = [tex]{V_f}_x[/tex] * T = 6.58 m/s * 0.74 s

Distance = 4.87 meters

Therefore, the skateboarder touches down approximately 4.87 meters from the end of the ramp.

Answer:

gravitational force between two objects is 8.37 N

Explanation:

given data

mass of object  = 85000 kg

distance between two object r  = 2.40 m

to find out

gravitational force between two objects

solution

we know that gravitational constant G is  6.67 × [tex]10^{-11}[/tex] m³/ s²-kg

so

gravitational force between two objects formula is

F  = [tex]G*\frac{m1m2}{r^2}[/tex]

here E is gravitational force and G is gravitational constant put here all value and m1 and m2 are mass of object

F  = [tex]6.67*10^{-11}*\frac{85000^2}{2.40^2}[/tex]

F = 8.37 N

gravitational force between two objects is 8.37 N

Answer:

the greatest height at which the ball can reach is 19.62 m.

Explanation:

given,

time taken by the ball to reach the ground = 4 s

time at which ball reach at maximum height = 4/2 = 2 s

velocity of the ball at the top most point = 0 m/s

we know,

v = u + a t

0 = u + (-9.81 ) × 2

u = 19.62 m/s

maximum height achieved

[tex]s = u t + \dfrac{1}{2}at^2[/tex]

[tex]s = 19.62\times 2 + \dfrac{1}{2}(-9.81)\times 2^2[/tex]

s = 19.62 m

hence, the greatest height at which the ball can reach is 19.62 m.

Answer:

Wavelength = 489.52 nm

Explanation:

Given that the wavelength of the light = 633 nm

The time taken by the light in unknown liquid = 1.38 ns

Also,

1 ns = 10⁻⁹ s

So, t = 1.38 × 10⁻⁹ s

Also,

Distance = 32.0 cm = 0.32 m ( 1 cm = 0.01 m)

So, speed of the light in the liquid = Distance / Time = 0.32 / 1.38 × 10⁻⁹ m/s = 2.32 × 10⁸  m/s

Frequency of the light does not change when light travels from one medium to another. So,

[tex]\frac {V_{air}}{\lambda_{air}}=\frac {V_{liquid}}{\lambda_{liquid}}[/tex]

[tex]{V_{air}}=3\times 10^8\ m/s[/tex]

[tex]{\lambda_{air}}=633\ nm[/tex]

[tex]{V_{liquid}}=2.32\times 10^8\ m/s[/tex]

[tex]{\lambda_{liquid}}=?\ nm[/tex]

So,

[tex]\frac {3\times 10^8\ m/s}{633\ nm}=\frac {2.32\times 10^8\ m/s}{\lambda_{liquid}}[/tex]

Wavelength = 489.52 nm

Wavelength is the distance between two points of the two consecutive waves.

The wavelength of the laser beam in the liquid is 489.52 nm.

Wavelength is the distance between two points of the two consecutive waves.

Given information-

The helium-neon laser beam has a wavelength in air of 633 nm.

The time taken by the helium laser beam to to travel through 32.0 cm or 0.32 m of an unknown liquid is  1.38 ns or [tex]1.38\times10^{-9}\rm s[/tex].

Speed of the light is the ratio of distance traveled by it in the time taken. Thus the speed of the given light is,

[tex]v=\dfrac{0.32}{1.38}\\v=2.32\times10^8 \rm m/s[/tex]

Frequency of the wave is the ratio of speed and wavelength of the wave.

As the frequency of the wave is equal for each medium. Thus,

[tex]f=\dfrac{v_{liq}}{\lambda_{liq}} =\dfrac{v_{air}}{\lambda_{air}}[/tex]

As the speed of the air is [tex]3\times10^8[/tex] m/s.

Thus put the values in the above ratio as,

[tex]\dfrac{2.32\times10^8}{\lambda_{liq}} =\dfrac{3\times10^8}{633}\\\lambda_{liq}=489.52 \rm nm[/tex]

Thus the wavelength of the laser beam in the liquid is 489.52 nm.

Learn more about the wavelength here;

https://brainly.com/question/10728818

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

Answer:

[tex]E=1.56\times 10^{23}\ N/C[/tex]

Explanation:

Given that,

Electric force applied to the electron, [tex]F=25000\ N[/tex]

Charge on electron, [tex]q=1.6\times 10^{-19}\ C[/tex]

We need to find the electric force acting on the electron. The electric field is given by :

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

[tex]E=\dfrac{25000}{1.6\times 10^{-19}}[/tex]

[tex]E=1.56\times 10^{23}\ N/C[/tex]

So, the electric field acting on the electron is [tex]1.56\times 10^{23}\ N/C[/tex]. Hence, this is the required solution

Explanation:

Charge, [tex]q=-30\ \mu C=-30\times 10^{-6}\ C[/tex]

Electric force, [tex]F=27\ mN=27\times 10^{-3}\ N[/tex]

We need to find the magnitude and direction of electric field at that location. The relation between the electric field and electric force is given by :

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

[tex]E=\dfrac{27\times 10^{-3}\ N}{-30\times 10^{-6}\ C}[/tex]

[tex]E=-900\ N/C[/tex]

For a negative charge, the direction of electric field is inward. The direction of electric force and electric field is same. So, the direction of electric field in this case is in upward direction. Hence, this is the required solution.

The magnitude of the electric field is 900 N/C, and the direction is downward. Therefore, the electric field at that location is  -900 N/C  (negative value indicating downward).

To determine the magnitude and direction of the electric field at the location where the point charge experiences an electrostatic force, we can use the relationship between the force  [tex]\mathbf{F}[/tex]  experienced by a charge  q  in an electric field [tex]\mathbf{E}[/tex] :

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

Given:

- The charge  q  is -30 µC (microcoulombs).

- The electrostatic force  [tex]\mathbf{F}[/tex]  is 27 mN (millinewtons) upward.

First, we convert the given quantities to standard SI units:

- Charge  [tex]q = -30 \, \mu \text{C} = -30 \times 10^{-6} \, \text{C}[/tex]

- Force  [tex]\mathbf{F} = 27 \, \text{mN} = 27 \times 10^{-3} \, \text{N}[/tex]

Next, we use the equation  [tex]\mathbf{F} = q \mathbf{E}[/tex] to solve for the electric field [tex]\mathbf{E}[/tex] :

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

Substituting the given values:

[tex]\mathbf{E} = \frac{27 \times 10^{-3} \, \text{N}}{-30 \times 10^{-6} \, \text{C}} \\\\\mathbf{E} = \frac{27 \times 10^{-3}}{-30 \times 10^{-6}} \, \text{N/C} \\\\\mathbf{E} = \frac{27}{-30} \times 10^{3} \, \text{N/C} \\\\\mathbf{E} = -0.9 \times 10^{3} \, \text{N/C} \\\\\mathbf{E} = -900 \, \text{N/C}[/tex]

The negative sign indicates the direction of the electric field. Since the force on a negative charge is upward, the electric field must be directed downward (opposite to the direction of the force on a negative charge).

Answer:

A) x and y components of the electric field  (Ep) at the origin.

Epx = -1620.5 N/C

Epy = -1620.5 N/C

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

   Ep= 2291.7 N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence  

1nC= 10⁻⁹C

Data

K= 9x10⁹N*m²/C²

q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C

d₁=d₂=0.190m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁ : Electric Field at point P  (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.

Ep₂: Electric Field at point  P (x=0, y=0) due to charge q₂. As the charge q₂  is positive (q₂+) ,the field leaves the charge

Ep: Total field at point P due to charges q₁ and q₂.

Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂

Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C

Look at the attached graphic :

Epx = Ep₁= -1620.5 N/C

Epy = Ep₂= -1620.5 N/C

A) x and y components of the electric field  (Ep) at the origin.

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

[tex]E_{p} =\sqrt{1620.5^{2}+1620.5^{2} } = 2291.7 \frac{N}{C}[/tex]

Answer:

The total force exerted on the Y axis is: -52.07μC

Explanation:

This is an electrostatic problem, so we will use the formulas from the Coulomb's law:

[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]

We are interested only of the effect of the force on the Y axis. We can notice that the charge placed on the x=4cm will exers a force only on the Y axis so:

[tex]Fy1=9*10^9*\frac{6.05*10^{-9}*(-1.97)*10^{-9}}{(3.02*10^{-2})^2}\\[/tex]

Fy1=-117.61μC

For the charge placed on the origin we have to calculate the distance and the angle:

[tex]r=\sqrt{(4*10^{-2}m)^2 +(3.02*10^{-2}m)^2} \\r=5cm=0.05m[/tex]

we can find the angle with:

[tex]alpha = arctg(\frac{3.02cm}{4cm})=37^o[/tex]

The for the Force on Y axis is:

[tex]Fy2=9*10^9*\frac{6.05*10^{-9}*(5)*10^{-9}}{(3.02*10^{-2})^2}*sin(37^o)\\[/tex]

Fy2=65.54μC

The total force exerted on the Y axis is:

Fy=Fy1+Fy2=-52.07μC