Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second oasis. What distance separates the two oases?

Answer:3.6

Explanation:

Given

Architect decide to double the width

length increases by 50 percent

height increases by 20 percent

Let L,B,H be length,height and width of rectangular room

L'=1.5L

B'=2B

H'=1.2H

therefore new volume V'=L'\times B'\times H'

[tex]V'=1.5L\times 2B\times 1.2H[/tex]

V'=3.6LBH

new volume is 3.6 times the original

Answer:

xf = - 3.16 m

Explanation:

the squirrel was initially in the position xi = 3.65 m, then it had a displacement of Δx = -6.81 m.

The negative sign indicates that it moved in the opposite direction, so we must subtract this displacement Δx = -6.81 m to the initial position xi = 3.65 m, to find its final position.

3.65 m - 6.81 m = - 3.16 m

Answer:

Volume left, v = 67 mL

Explanation:

Given that,

Volume of soda contained in a can, V = 375 mL

We need to find the volume of soda left after 308 mL of soda is removed, V' = 308 mL

Let v is the left volume of soda. It can be calculated using simple calculations as :

v = V - V'

v = 375 mL - 308 mL

v = 67 mL

So, the left volume in the can is 67 mL. Hence, this is the required solution.

Answer:

h/(m*c)

Explanation:

Hi!

Lets denote the units of X as [X]

Since the dimentions of h are:

[tex][h] = \frac{ML^{2}}{T}[/tex]

If we divide [h] by the units of mass, we get:

[tex]\frac{[h]}{[m]} = \frac{L^{2}}{T}}[/tex]

Also we know that:

[tex][c] = \frac{L}{T}[/tex]

So:

[tex][\frac{h}{mc}] = \frac{L^{2}}{T}}*\frac{T}{L}=L[/tex]

Therefore

 h/(mc) has dimentiosn of length

To construct a quantity with the dimensions of length using Planck's constant (h), mass (m), and the speed of light (c), the formula L = h/(mc) can be used. This length scale is relevant in the realms of quantum mechanics and high-energy physics.

The student has asked how to construct a quantity with the dimensions of length using Planck's constant (h), a mass (m), and the speed of light (c). To achieve this, we can use the formula for the Planck length given by:

Lp = √hG/c³

Where G is Newton's gravitational constant. However, since we need to construct a length using just h, m, and c without G, we can rearrange the Planck length equation to:

L = h/(mc)

This gives a length L which is dependent on the mass m in addition to Planck's constant h and the speed of light c. This length scale is significant in quantum mechanics and high-energy physics, where extremely small distances are relevant.

Answer:

work done is = 0

Explanation:

given data

distance = 2 cm

potential = 1 V

charge with magnitude = 1 nC

to find out

work done by the electrostatic force

solution

we know that at equipotential surface is that surface which have equal potential at each every point that we say

work done will be

work done = ∫dw

∫dw = [tex]\int\limits^v1_v2 {q} \, dv[/tex]

here q is charge

so

net work done = q ( v2 - v1 )

and

so v2 = v1 = 0

so

work done is = 0

No work is done in moving a charge along an equipotential line because there is no difference in electric potential along that line, making the work done by the electrostatic force zero.

The question asks to calculate the work done by the electrostatic force when moving a charge along an equipotential line. By definition, the potential difference along an equipotential line is zero. Therefore, the work done W by the electric force to move a charge q in an electric potential V along an equipotential line is given by the equation W = qΔV, where ΔV is the change in electric potential. Since ΔV is zero along an equipotential line, the work done is also zero. No work is required to move a charge along an equipotential line because there is no change in electric potential energy.

Answer:

The net force on X is Fx=75.4N

The net force on Y is FY=25.17N

Explanation:

This is an electrostatic problem, we can calculate de force applying the formula:

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

the force because of charge at 1cm on the X axis, will only have an X component of force, so:

[tex]Fx1=8.98755*10^9*\frac{(9\µC)(7\µC)}{(5*10^{-2}m)^2}\\Fx1=226.48N[/tex]

For the force because of the charge of the Y axis we have to find the distances usign pitagoras, and the angle:

[tex]r=\sqrt{(-1*10^{-2}m)^2+(6*10^{-2}m)^2} \\r=6.08cm=0.0608m[/tex]

we can find the angle with:

[tex]\alpha = arctg(\frac{1cm}{6cm})=9.46^o[/tex]

We now can calculate the force of the X axis because of the second charge:

[tex]Fx2=8.98755*10^9*\frac{(-9\µC)(7\µC)}{(6.08*10^{-2}m)^2}*cos(9.46 )\\Fx2=-151.08N[/tex]

and for the Force on Y axis:

[tex]Fy2=8.98755*10^9*\frac{(-9\µC)(7\µC)}{(6.08*10^{-2}m)^2}*sin(9.46 )\\Fy2=-25.17N[/tex]

The net force on X axis is:

Fx=226.48N-151.08N=75.4N

Fy=-25.17N

Answer:

The magnitude of the resultant force is equal to 0.0216 N

The direction is along the negative y axis

Explanation:

According to the exercise data:

q1 = 9x10^-6 C (0,1)

q2 = -9x10^-6 C (0,-1)

q = 7x10^6 C (6,0)

the distance between load q1 and load q will be equal to:

[tex]r_{1} = \sqrt{(0-1)^{2}+(6-0)^{2} }= 6.08 m[/tex]

The same way to calculate the distance between q2 and q:

[tex]r_{2} =\sqrt{(0-(-1))^{2}+(6-0)^{2} } =6.08 m[/tex]

The force on q due to the load q1, is calculated with the following equation:

[tex]F_{1}=\frac{K*q_{1}*q }{r1^{2} } *(cos45i-sin45j)=\frac{8.987559x10^{9}*9x10^{-6} *7x10^{-6} }{6.08^{2} }*(cos45i-sin45j)=0.0153*(0.707i-0.707j)=0.0108N(i-j)[/tex]

The same way to force F2:

F2 = (8.987559x10^9*-9x10^-6*7x10^-6))/(6.08^2)*(cos45i-sin45j) = 0.0108 N(i-j)

The resultant force:

F = F1 + F2 = 0.0108 N *(-2j) = - 0.0216 N j

The magnitude is equal to 0.0216 N

The direction is along the negative y axis

Answer:

option D

Explanation:

the correct answer is option D

acceleration of a body under free fall due to the effect of gravity is the called acceleration due to gravity.

Acceleration due to gravity is a constant value.

According to the International System of Measurements the acceleration due to gravity is 9.81 m/s² when an object falls vertically.

acceleration due to gravity is denoted by 'g'.

Answer:

1.35×10⁻⁷ m

37.278 mi/My

Explanation:

Speed of the tectonic plate= 6 cm/yr

Converting to seconds

[tex]6=\frac{6}{365.25\times 24\times 60\times 60}[/tex]

So in one second it will move

[tex]\frac{6}{365.25\times 24\times 60\times 60}[/tex]

In 71 seconds

[tex]71\times \frac{6}{365.25\times 24\times 60\times 60}=1.35\times 10^{-5}\ cm[/tex]

The tectonic plate will move 1.35×10⁻⁵ cm or 1.35×10⁻⁷ m

Convert to mi/My

1 cm = 6.213×10⁻⁶ mi

1 M = 10⁶ years

[tex]6\times 6.213\times 10^{-6}\times 10^6=37.278\ mi/My[/tex]

Speed of the tectonic plate is 37.278 mi/My

Answer:

False.

Explanation:

From Kepler's Third Law of plenetary motion, we know that:

"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."

Or, as expressed in mathematical terms:

[tex]\frac{a^3}{T^2}=constant[/tex], where a is the semi-major axis of the orbit (the distance from the center), and T is the orbital period of the satellite.

From this expression we can clearly see that if the orbit's semi-major axis is doubled, orbital period will be [tex]\sqrt{8}[/tex] times longer to compensate the variation.

Answer:

4.2×10⁷ m/s

Explanation:

L = length observed by an observer

L₀ = Proper length

v = Velocity of the rocket ship

Length contraction formula

[tex]L=L_{0}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\Rightarrow 0.99=1{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\Rightarrow \frac{0.99}{1}={\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\Rightarrow 0.99^2=1-\frac{v^{2}}{c^{2}}\\\Rightarrow 0.99^2-1=-\frac{v^2}{c^2}\\\Rightarrow 0.0199\times c^2=v^2\\\Rightarrow v=\sqrt{0.0199}c=v\\\Rightarrow v =\sqrt{0.0199}\times 3\times 10^8\\\Rightarrow v=4.2\times 10^7\ m/s[/tex]

The speed of the ship would be 4.2×10⁷ m/s

Answer:

8271.92 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 18\times 18^2\\\Rightarrow s=2916\ m[/tex]

The height reached at 18 seconds is 2916 m

[tex]v=u+at\\\Rightarrow v=0+18\times 18\\\Rightarrow v=324\ m/s[/tex]

The velocity at 2916 m is 324 m/s

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-324^2}{2\times -9.8}\\\Rightarrow s=5355.92\ m[/tex]

Maximum height reached by the toy rocket is 2916+5355.92 = 8271.92 m

The maximum height above the ground that the rocket will achieve is 8,271.92m.

To get the maximum height, we need to calculate the height reached by the rocket.

Using the equation:

[tex]S=ut+\frac{1}{2}at^2\\S =0(t) + \frac{1}{2}\times18\times 18^2\\S=9 \times 18^2\\S=2,916m[/tex]

Get the velocity also using the equation of motion:

[tex]v=u+at\\v=0+18(18)\\v=324m/s[/tex]

Get the maximum height above the ground the rocket will achieve:

[tex]v^2=u^2+2as\\324^2=0^2+2(9.8)s\\104,976=19.6s\\s=\frac{104,976}{19.6}\\s= 5,355.92m[/tex]

The total maximum height reached by rocket is 2,916 + 5,355.92 = 8,271.92m.

Learn more here: https://brainly.com/question/24018491

Answer:

[tex]E=0.697*10^{6}V/m\\[/tex]

Explanation:

If the voltage is constant, the relation between the electric field E and the voltage V , for two plates separated by a distance d, is:

[tex]V=E*d\\[/tex]

d=6cm=0.06m

V=4.18×10^4 V

We solve to find E:

[tex]E=V/d=4.18*10^{4}/0.06=6.97*10^{5}=0.697*10^{6}V/m\\[/tex]

Answer:

Explanation:

This problem can be solved using the center of mass of the system.

As there are no external forces in the horizontal direction, and the external forces in the vertical direction, the gravitational force and the buoyancy,  cancel each other, we know:

[tex]\vec{F}_{net_{external}} = M \frac{d\vec{P}}{dt} = 0[/tex]

where M is the total mass of the system, and [tex]\vec{P}[/tex] is the linear momentum of the system. This implies:

[tex]\frac{d\vec{P}}{dt} = 0[/tex]

so

[tex]\vec{P} = constant[/tex]

the linear momentum of the system is:

[tex]\vec{P} = M \vec{r}_{cm}[/tex]

where [tex] \vec{r}_{cm}[/tex] is the position of the center of mass. As the mass in this problem is constant, this implies that the position of the center of mass is conserved.

For a system formed by i particles, each of mass [tex]m_i[/tex] and position [tex]\vec{r}_i[/tex], the center of mass can be found with the equation:

[tex]\vec{r}_{cm} = \frac{\sum\limits_i m_i \vec{r}_i }{M}[/tex]

where M is the total mass of the system, found with the equation:

[tex]M = \sum\limits_i m_i [/tex]

For this problem, the total mass of the system is :

[tex]M = m_{brother} + m_{sister} + m_{canoe}[/tex]

[tex]M = 144.1 \ kg + 81.6 \ kg + 48.7 \ kg[/tex]

[tex]M = 274.4 \ kg[/tex]

To find the positions, we need an frame of reference.  Putting the origin of coordinate at the position of the sister.

[tex]\vec{r}_{sister_i} = 0 \ \hat{i}[/tex]

The brother is 2.03 meters apart in the positive direction:

[tex]\vec{r}_{brother_i} = 2.03 m \ \hat{i}[/tex]

and the center of mass of the canoe must be at half this distance:

[tex]\vec{r}_{canoe_i} = 1.015 m \ \hat{i}[/tex]

taking all this, the center of mass of the system is at :

[tex]\vec{r}_{cm} = \frac{ 81.6 \ kg * 0 \ \hat{i} + 144.1 \ kg \ 2.03 m \ \hat{i} + 48.7 \ kg \ 1.015 m \ \hat{i} }{274.4 \ kg}[/tex]

[tex]\vec{r}_{cm} = 1.2462 \ m\ \hat{i} [/tex]

After the switch, we found that the center of mass has to be displaced. If its displaced a distance d, we find that the new positions must be:

[tex]\vec{r}_{sister_f} = 2.03 m \ \hat{i} + d \ \hat{i}[/tex]

[tex]\vec{r}_{brother_f} = 0 \ \hat{i} + d \ \hat{i}[/tex]

[tex]\vec{r}_{canoe_f} = 1.015 m \ \hat{i} + d \ \hat{i}[/tex]

where the brother and the sister had switched the original position. Now, the center of mass will be located at:

[tex]\vec{r}_{cm} = 1.2462 \ m\ \hat{i} [/tex]

And the equation is :

[tex]\vec{r}_{cm} = \frac{ 81.6 \ kg * (2.03 m \ \hat{i} + d  \ \hat{i} )  + 144.1 \ kg \ d\ \hat{i} + 48.7 \ kg \ ( 1.015 m \ \hat{i} + d \ \hat{i}) }{274.4 \ kg}[/tex]

[tex]\vec{r}_{cm} = 0.7838 \ m \ \hat{i} + \frac{ 81.6 \ kg  d  \ \hat{i} )  + 144.1 \ kg \ d\ \hat{i} + 48.7 \ kg \ d \ \hat{i}) }{274.4 \ kg}[/tex]

[tex]\vec{r}_{cm} = 0.7838 \ m \ \hat{i} + \frac{ 81.6 \ kg   + 144.1 \ kg   + 48.7 \ kg \ }{274.4 \ kg} \ d \ \hat{i}[/tex]

[tex]\vec{r}_{cm} = 0.7838 \ m \ \hat{i} +  d \ \hat{i}[/tex]

[tex]1.2462 \ m\ \hat{i} = 0.7838 \ m \ \hat{i} +  d \ \hat{i}[/tex]

[tex]d \hat{i} = 0.4627 \ m \ \hat{i} [/tex]

And this is how much closer the paddle is to the end of the canoe.

After the passengers switch places, the distance to the paddle (which does not move) is the end of the canoe is 46.27 m.

The center of mass of a object or for a system of object is the point, where the center of distribution of mass in space.

FHE brother (mass 144.1 kg) is canoeing on Utah Lake with his sister (mass 81.6 kg) and canoe has a mass of 48.7 kg.

Let the position of the sister at the origin, his brother is at 2.03 meters. The position of the center of the mass of the canoe is at half distance (1.015 m). Thus, the center of mass of the system is,

[tex]r_{COM}=\dfrac{144.1(2.03)+81.6(0)+48.7(1.015)}{144.1+81.6+48.7}\\r_{COM}=1.2462[/tex]

For the displacement, the equation can be given as,

[tex]r_{COM}=\dfrac{144.1(d)+81.6(2.03+d)+48.7(1.015+d)}{144.1+81.6+48.7}\\1.2462=\dfrac{144.1(d)+81.6(2.03+d)+48.7(1.015+d)}{144.1+81.6+48.7}\\d=0.4627\rm \; m[/tex]

Thus, after the passengers switch places, the distance to the paddle (which does not move) is the end of the canoe is 46.27 m.

Learn more about the center of mass here;

https://brainly.com/question/13499822

Answer:

a) 200m, 100m/s

b) 710.20m

c) -117.98 m/s

d) 26.24 s

Explanation:

To solve this  we have to use the formulas corresponding to a uniformly accelerated motion problem:

[tex]V=Vo+a*t[/tex] (1)

[tex]X=Xo+Vo*t+\frac{1}{2}*a*t^2\\[/tex] (2)

[tex]V^2=Vo^2+2*a*X[/tex] (3)

where:

Vo is initial velocity

Xo=intial position

V=final velocity

X=displacement

a)

[tex]X=0+0*4+\frac{1}{2}*25*4^2[/tex]

the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.

[tex]X=200m[/tex]

[tex]V=0+25*4[/tex]

[tex]V=100m/s[/tex]

b)

The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8[tex]\frac{m}{s^2}[/tex] because it is going downward.

[tex]0=100^2+2*(-9.8)*X\\X=510.20\\totalheight=200+510.20=710.20m[/tex]

c)

In order to find the velocity when it crashes, we can use the formula (3).

the initial velocity is 0 because in that moment is starting to fall.

[tex]V^2=0^2+2*(-9.8)*(710.20)\\V=-117.98 m/s[/tex]

the minus sign means that the object is going down.

d)

We can find the total amount of time adding the first 4 second and the time it takes to going down.

to calculate the time we can use the formula (2) setting the reference at 200m:

[tex]-200=0+100*t+\frac{1}{2}*(-9.8)*t^2[/tex]

solving this we have: time taken= 22.24 seconds

total time is:

total=22.24+4=26.24 seconds.

Answer:

a) [tex]v=230 m/s[/tex]

b) [tex]v=196.5 m/s[/tex]

Explanation:

a) The formula for average velocity is

[tex]v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }[/tex]

For the first Δt=4.7s

[tex]v=\frac{(1150-70)m}{(4.7)s} =230 m/s[/tex]

b) For the secont  Δt=5.85s we know that the displacement is 1150m. So, the average velocity is:

[tex]v=\frac{(1150)m}{(5.85)s}=196.5m/s[/tex]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius  = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

[tex]E=\dfrac{kq}{r^2}[/tex]

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

[tex]E=\dfrac{kq}{r^2}[/tex]

Put the value into the formula

[tex]E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}[/tex]

[tex]E=6.88\times10^{7}\ N/C[/tex]

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d).  For, r = 7.00 cm

The electric field is

[tex]E=\dfrac{kq}{r^2}[/tex]

Put the value into the formula

[tex]E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}[/tex]

[tex]E=5.43\times10^{6}\ N/C[/tex]

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

Answer:

(a) 82 Km

(b) 32°

Explanation:

First you should draw the vectors in the cartesian plane (please see the picture below).

As the car is driven east for a distance of 47 Km, your first vector should be drawn from the origin (0,0) and on the x axis

Then the car is driven north for a distance of 28 Km, so your second vector should be drawn from the origin and on the y axis.

And the car finally goes east of north for 27Km, so the third vector should be drawn from the origin east of north forming an 35° angle with x axis.

Then you should find the components of the vector in x and y:

For Vector 1 ([tex]V_{1}[/tex])

[tex]V_{1x}=47Km[/tex]

[tex]V_{1y}=0[/tex]

For Vector 2 ([tex]V_{2}[/tex])

[tex]V_{2x}=0[/tex]

[tex]V_{2y}=28Km[/tex]

For Vector 3 ([tex]V_{3}[/tex])

[tex]V_{3x}=27cos37^{o}[/tex]

[tex]V_{3x}=21.56[/tex]

[tex]V_{3y}=27sin37^{o}[/tex]

[tex]V_{3y}=16.25[/tex]

To find the magnitud of the car´s total displacement, (R) you should add up all the x and y components.

For the x component:

[tex]R_{x}=V_{1x}+V_{2x}+V_{3x}[/tex]

[tex]R_{x}=47+0+21.56[/tex]

[tex]R_{x}=68.56Km[/tex]

For the x component:

[tex]R_{y}=V_{1y}+V_{2y}+V_{3y}[/tex]

[tex]R_{y}=0+28+16.25[/tex]

[tex]R_{y}=44.25Km[/tex]

Now please see the second picture that is showing the components x and y as the sides of a right triangle, and we are going to use the Pythagorean theorem to find the resultant, R.

[tex]R=\sqrt{R_{x}^{2}+R_{y}^{2}[/tex]

[tex]R=\sqrt{68.56^{2}+44.25^{2}}[/tex]

[tex]R=82Km[/tex]

And to find the angle of the car´s total displacement (α), we use the same right triangle with the relationship between its legs.

[tex]tan(\alpha)=\frac{44.25}{68.56}[/tex]

[tex]tan(\alpha)=0.64[/tex]

[tex]\alpha=tan^{-1}(0.64)[/tex]

[tex]\alpha =32^{o}[/tex]

Answer:

The speed of light remains the same.

Explanation:

(a)  Time : second, hour minutes are example of time.

(b) Distance : meter, kilometers etc are some example of distance.

(c) The speed of light : It always remains constant. It is equal to [tex]3\times 10^8\ m/s[/tex].

(d) The weight of an object is given by the product of mass of an object and the acceleration due to gravity. As the value of g is not same everywhere, its weight varies.

So, the speed of light is always remains the same. Hence, the correct option is (c).                                                                

Answer:

682.32 m

Explanation:

Speed of the passing speeder = 120 km/hr = 120 × 0.2777 = 33.33 m/s

time after which police man starts = 2 seconds

Acceleration of the policeman = 4.0 m/s²

let the time taken to catch be 't' seconds

Now,

The total distance to be covered by the policeman

= Distance covered by the speeder in the 2 seconds + Distance further traveled by the speeder in time t

thus,

From Newton's equation of motion

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

where,  

s is the total distance traveled by the police man

u is the initial speed  = 0

a is the acceleration

t is the time

thus,

[tex]0\times t+\frac{1}{2}\times4t^2[/tex]= 33.33 × 2 + 33.33 × t

or

2t² = 66.66 + 33.33t

or

2t² - 33.33t - 66.66 = 0

on solving the above equation, we get

t = 18.47 seconds (negative value is ignored as time cannot be negative)

therefore,

the total distance covered = 33.33 × 2 + 33.33 × 18.47 = 682.32 m

Answer:

Mass of second object will be 15 kg

Explanation:

We have given mass of first object = 12 kg

Acceleration [tex]a=5m/sec^2[/tex]

According to second law of motion we know that force F = MA

So force [tex]F=12\times 5=60N[/tex]

As the same force is applied to the second object of acceleration [tex]a=4m/Sec^2[/tex]

So force = ma

[tex]60=m\times 4[/tex]

m = 15 kg

So mass of second object will be 15 kg

Explanation:

The total distance in a path is called distance.

The shortest distance between two points is called displacement.

a) Here, the distance travelled between the park to your friend's house and back is

Distance between park to friends house + Distance from friend's house to your house.

b) Displacement would be the shortest distance between the park and your house.

a) Distance walked between your house to library and back is

Distance between your house and library + Distance between your house and library

b) Displacement would be zero (0) as the distance between you initial point and final point is zero. Here, the initial and final points are the same

Final answer:

The distance traveled is the total length of the path taken, while displacement is the distance between the initial and final positions in a straight line. Displacement is a vector quantity because it has both magnitude and direction.

Explanation:

a. The distance traveled is the total length of the path taken. In this case, you traveled from the park to your friend's house, and then back to your own house. So, the distance traveled is the sum of these two distances.

b. Displacement is the distance between the initial and final positions, considering only the straight line between them. In this case, since you ended up back at your own house, the displacement is zero because you did not change your position.

c. Displacement is a vector quantity because it has both magnitude (the distance between the initial and final positions) and direction. In this case, the direction is the straight line between the initial and final positions.

Answer:

Option d)

Explanation:

As the Kinetic energy of the body ids due to the speed of the body and depends on it, it does not depend on the height of the body.

The kinetic energy is given by:

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

where

m = mass of the body

v = kinetic energy of the body

Now, as per the question:

Initial velocity is u

and final velocity, v = 2u

Thus

Initial KE = [tex]\frac{1}{2}mu^{2}[/tex]               (1)

FInal KE = [tex]\frac{1}{2}mv^{2}[/tex] = [tex]\frac{1}{2}m(2u)^{2}[/tex]          

FInal KE = 4[tex]\frac{1}{2}mu^{2}[/tex]               (2)  

Thus from eqn (1) and (2):

Final KE = 4(Initial KE)

The Kinetic Energy has changed by a factor of 4

Answer:

Vx(7,22s)=67,42m/s

Explanation:

V(t)=Vo+a(t) [tex]Vx(7,22s)=Vxo+5,21\frac{m}{s^{2} } *7,22s

Vx(7,22s)=29,8m/s +37,62m/s=67,42m/s[/tex]

you just consider the x component of your velocity, then you apply the Velocity formula of uniform accelerated rectilinear motion, and you get to the result, by replacing t (time) by the value given 7,22s.

Answer:60 m,20 m

Explanation:

Given

You ran 10 yard line to 50 yard line therefore total distance traveled is 40 yard

Finally you turned around and came back to 30 yard line

Therefore total distance is 40+20=60 m(From 50 yard line to 30 yard line it is 20 yards)

Displacement=30-10=20 yards

Answer:

De broglie wavelength = 4.17 pm

Explanation:

We have given mass of proton [tex]m=1.67\times 10^{-27}kg[/tex]

Speed of proton v [tex]v=9.50\times 10^{6}m/sec[/tex]

Plank's constant [tex]h=6.63\times 10^{-34}J-s[/tex]

We have to find de broglie wavelength

De broglie wavelength is given by [tex]\lambda =\frac{h}{mv}=\frac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 9.5\times 10^6}=4.17\times 10^{-12}m=4.17pm[/tex]

Answer:

Part a) Big bang occurred about 13.8 billion years ago. This time is arrived after see the shift in the cosmic background radiation that fills the universe. The result of the big bang was the creation of universe itself. Everything we know space, matter, time was created by the event thus giving birth to the universe and stars, galaxies,e.t.c.

Part b) No string theory is currently not proven as of yet. The basic problem in formulating the theory of everything is to mix 2 completely different theories known as quantum theory of matter and theory of relativity. But this has not been achieved as of yet. There are different theories that try to explain the both quantum and relativistic nature of the matter and string theory is one of the theories that has been proposed to explain the same.The theory is not completely developed as of yet as there are numerous inconsistencies with it such as it proposes 34 dimension's of nature which have not been observed as yet.Also there is an inherent nature of the theory that it deals with quantities with [tex]10^{34}m[/tex] scales of length making it impossible to verify the theory which still is mathematically incomplete.

Part c)  Strong Nuclear forces come into action in the nucleus of an atom and holds the nucleon's together. As we know that a pair of protons will repel due to nature of their charge the strong nuclear force holds them together.

Part d)  Cosmology is the branch of astronomy that deals with the origin and evolution of the universe. It basically is the study about the universe was formed and how it evolved during different stages.

Answer:

Part a)

Final speed of the car is

[tex]v_f = 4.17 m/s[/tex]

Part b)

Acceleration of the car is

[tex]a = -5.39 m/s^2[/tex]

Explanation:

As we know that car makes a skid of 62.5 m

here acceleration of the car is

[tex]a = - 5.25 m/s^2[/tex]

now we have

[tex]d = v_i t + \frac{1}{2}at^2[/tex]

[tex]62.5 = v_i (4.15) + \frac{1}{2}(-5.25)(4.15^2)[/tex]

[tex]v_i = 25.95 m/s[/tex]

Part a)

Speed of the car by which it will hit the tree

[tex]v_f = v_i + at[/tex]

[tex]v_f = 25.95 - (5.25)(4.15)[/tex]

[tex]v_f = 4.17 m/s[/tex]

Part b)

Now if car will stop after travelling same distance which same initial speed

Then we can use kinematics

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 25.95^2 = 2(a)(62.5)[/tex]

[tex]a = -5.39 m/s^2[/tex]

Acceleration of a object is the rate of change of velocity of the object per unit time.

Acceleration of a object is the rate of change of velocity of the object per unit time.

Given information-

The car slows uniformly with acceleration of −5.25 m/s squared for 4.15 s.

The skid made by the car is 62.5 m long.

The initial velocity of the car can be find out using the distance formula of motion as,

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

Put the values as,

[tex]62.5=u\times4.15+\dfrac{1}{2}\times(-5.25)\times(4.15)^2\\u=25.95[/tex]

Thus the value of initial velocity is 25.95 m/s.

The velocity formula using the equation of motion can be given as,

[tex]v=u+at\\v=25.95+(-5.25)\times4.15\\v=4.17\rm m/s[/tex]

Thus, the speed does the car then strike the tree 4.17 m/s.

To stop the car the final velocity of it must be 0. Thus,

[tex]2ad=v^2-u^2\\2\times a\times62.5=0-25.95^2\\a=-5.93\rm m/s^2[/tex]

The acceleration need to be, so that the car narrowly avoids a collision -5.39 m/s squared.

Hence,

Learn more about the acceleration here;

https://brainly.com/question/605631

The magnitude of [tex]\( q_2 \)[/tex] is [tex]\( 674.99 \, \mu \text{C} \)[/tex].

To find the magnitude of [tex]\( q_2 \)[/tex], we can use Coulomb's Law and the principle of superposition. Coulomb's Law states that the force between two point charges is given by:

[tex]\[ F = \frac{k \cdot |q_1 \cdot q|}{r_1^2} + \frac{k \cdot |q_2 \cdot q|}{r_2^2} \][/tex]

where ( k ) is Coulomb's constant [tex](\( 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \))[/tex], [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the charges, [tex]\( q \)[/tex] is the test charge, and [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex] are the distances from [tex]\( q \)[/tex] to the charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex], respectively.

Given the problem, we have [tex]\( F = 23 \, \text{N} \)[/tex], [tex]\( q_1 = -27 \times 10^{-6} \, \text{C} \)[/tex], [tex]\( q = 9 \times 10^{-6} \, \text{C} \)[/tex], [tex]\( r_1 = 0.21 \, \text{m} \)[/tex], [tex]\( r_2 = 0.35 \, \text{m} \)[/tex], and we need to find [tex]\( q_2 \)[/tex].

First, let's calculate the force due to [tex]\( q_1 \)[/tex]:

[tex]\[ F_1 = \frac{k \cdot |q_1 \cdot q|}{r_1^2} = \frac{8.99 \times 10^9 \cdot |-27 \times 10^{-6} \cdot 9 \times 10^{-6}|}{(0.21)^2} \][/tex]

[tex]\[ F_1 = \frac{8.99 \times 10^9 \cdot 243 \times 10^{-12}}{0.0441} = \frac{2185.57}{0.0441} = 49536.73 \, \text{N} \][/tex]

Now, the force due to [tex]\( q_2 \)[/tex]:

[tex]\[ F_2 = \frac{k \cdot |q_2 \cdot q|}{r_2^2} \][/tex]

We know the net force is in the positive y-direction, so [tex]\( F_2 \)[/tex] must be positive.

Now, sum the forces and equate to the given net force:

[tex]\[ F = F_1 + F_2 \][/tex]

[tex]\[ 23 = 49536.73 + F_2 \][/tex]

[tex]\[ F_2 = 23 - 49536.73 = -49513.73 \, \text{N} \][/tex]

Now, plug in [tex]\( F_2 \)[/tex] and solve for [tex]\( q_2 \)[/tex]:

[tex]\[ -49513.73 = \frac{8.99 \times 10^9 \cdot |q_2 \cdot 9 \times 10^{-6}|}{(0.35)^2} \][/tex]

[tex]\[ -49513.73 = \frac{8.09 \times 10^9 \cdot |q_2|}{0.1225} \][/tex]

[tex]\[ |q_2| = \frac{-49513.73 \cdot 0.1225}{8.99 \times 10^9} \][/tex]

[tex]\[ |q_2| = -674.99 \times 10^{-6} \][/tex]

Since [tex]\( q_2 \)[/tex] is positive, [tex]\( q_2 = 674.99 \times 10^{-6} \, \text{C} \)[/tex].

Thus, the magnitude of [tex]\( q_2 \)[/tex] is [tex]\( 674.99 \, \mu \text{C} \)[/tex].

Answer:

Insufficient data.

Explanation:

Hi!

The data you have is only potential diference, so you know that the variation in potential energy of the electron when moving from -1.5V to 3.0V, is 4.5 V.

But you cannot know the acceleration. For that you need to know the electric field, so you can calculate force.

Answer:

B)5.52x10^9 J

Explanation:

The equation for calculating kinetic energy is as follows

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

so what we have to do is calculate the kinetic energy of the space probe in each state and calculate the difference

E=[tex]\frac{1}{2} m.V2^{2}-\frac{1}{2} m.V1^{2}=E[/tex]

E=[tex]\frac{1}{2} m(V2^{2} -V1^{2})=E[/tex]

E=(1500)(5000^2-4200^2)/2

E=5.52x10^9 J