At one point in space, the electric potential energy of a 15 nC charge is 42 μJ . Part A) What is the electric potential at this point?
Express your answer to two significant figures and include the appropriate units.

Part B) If a 20 nC charge were placed at this point, what would its electric potential energy be?
Express your answer to two significant figures and include the appropriate units.

Answer:

t=444.4s

Explanation:

m=1.5*10^7 kg

F=7.5*10^5 N

v=80km/h*(1h/3600s)*(1000m/1km)=22.22m/s

Second Newton's Law:

F=ma

a=F/m=7.5*10^5/(1.5*10^7)=0.05m/s^2

Kinematics equation:

vf=vo+at=at      

vo: initial velocity equal zero

t=vf/a=22.22/0.05=444.4s

To calculate the time it takes to increase the speed of a train from rest to 80 km/h, you can use Newton's second law of motion.

To calculate the time it takes to increase the speed of the train from rest to 80 km/h, we can use Newton's second law of motion. First, we need to calculate the force required to accelerate the train. Given the mass of the train, 1.5 x 10^7 kg, and the acceleration, we can use the formula: force = mass x acceleration

force = (1.5 x 10^7 kg) x (80 km/h to m/s conversion)

Next, we can use the formula: force = mass x acceleration to find the time it takes to accelerate the train:

time = force / (7.5 x 10^15 N)

Plugging in the values, we can calculate the time it takes to increase the speed of the train from rest to 80 km/h.

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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

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:

(C) zero (there is no net horizontal component of the E-field)

Explanation:

If we subdivide the bar into small pieces, each piece (dx) contains a charge (dq), the electric field of each piece is equivalent to the field of a punctual electric charge, and has a direction as shown in the attached figure. For each piece (dx) in the negative axis there is another symmetric piece (dx) in the positive axis, and as we see in the figure for symmetry the sum of their electric fields gives a resultant in the Y axis (because its components in X are cancelled by symmetry).

Then the resultant of the electric field will be only in Y.

(C) zero (there is no net horizontal component of the E-field)

The x-component of the electric field at point (0,a) due to a uniform continuous line charge on the x-axis is zero, due to the symmetrical distribution of charge and corresponding cancellation of horizontal electric field components.

The student is asking about the direction of the electric field at a point on the positive y-axis due to a uniform continuous line charge distributed along the x-axis. To find the direction of the x-component of the electric field at the point (0,a), we can consider the symmetry of the charge distribution. For any small element of charge on the positive side of the x-axis, there is an identical element of charge on the negative side at the same distance from the origin. The electric fields produced by these two elements at point (0,a) on the y-axis will have the same magnitude but opposite x-components. These x-components will cancel each other out, resulting in a net x-component of the electric field being zero. Therefore, the correct answer to the student's question is (C) zero (there is no net horizontal component of the electric field).

Answer:

-1.19 m

Explanation:

R1 = + 4 cm

R2 = - 15 cm

n = 1.5

distance of object, u = - 1 m

let the focal length of the lens is f and the distance of image is v.

use lens makers formula to find the focal length of the lens

[tex]\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]

By substituting the values, we get

[tex]\frac{1}{f}=\left ( 1.5-1 \right )\left ( \frac{1}{4}+\frac{1}{15} \right )[/tex]

[tex]\frac{1}{f}=\frac{19}{120}[/tex]   .... (1)

By using the lens equation

[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

[tex]\frac{19}{120}=\frac{1}{v}+\frac{1}{1}[/tex]    from equation (1)

[tex]\frac{1}{v}=\frac{19-120}{120}[/tex]

v = -1.19 m

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.

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:

5 g/cm³

Explanation:

Density of an alloy = 5000 kg/m³

The Density of a material is the ratio of the mass by the volume.

The units here are of the metric system

Mass is the resistance a body has which opposes motion when force is applied.

Volume is the amount of material in an object

Convert kg/m³ to g/cm³

1 kg = 1000 g

1 m³ = 10⁻⁶ cm

[tex]5000\ kg/m^3=5000\times 1000\times 10^{-6}=5\ g/cm^3[/tex]

∴ 5000 kg/m³ = 5 g/cm³

Answer:

Explanation:

The ball is going down with velocity. It must have momentum . It is stopped by

catcher so that its momentum becomes zero . There is change in momentum . So force is applied on the ball  by the gloves  .

The rate of change of momentum gives the magnitude of force. This force must be in upward direction to stop the ball. So force is in positive direction .

Let us measure the force applied on the ball .

Final velocity after the fall by 66 m

V = √ 2gh

= √ 2x9.8 x 66

35.97 m /s

Momentum = m v

0.145 x 35.97

= 5.2156 kgms⁻¹

Change in momentum

= 5.2156 - 0

= 5.2156

Rate of change of momentum

=  Change of momentum / time = 5.2156 / .0118

= 442 N

if R be the force exerted by gloves to stop the ball

R - mg represents the net force which stops the ball so

R - mg = 442

R = 442 + mg

= 442 + .145 x 908

443.43 N

According to Newton's second law of motion rate of change of momentum is equal to the applied force. The force exerted by the glove of the player on the ball will be 497.4 N.

According to Newton's second law of motion rate of change of momentum is equal to the applied force.

Momentum is given by the product of mass and velocity. it is denoted by P.it leads to the impulsive force.

P = mv

ΔP = mΔv

[tex] \rm{ F = \frac{\delP}{\delt} } [/tex]

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

[tex]\rm{V = \sqrt{2\times9.81\times60} [/tex]

v = 34.31 m/sec

ΔV = v -u

ΔV = 34.31-0

ΔV = 34.31 m /sec.

Δt = 0.0118 sec

[tex]\rm{\frac{\delv}{\delt} } [/tex]= 34.31

F = [tex]m\rm{\frac{\delv}{\delt} }[/tex]

F = [tex] 0.145\times3430 [/tex]

F = 497.4 N

Thus the force exerted by the glove of the player on the ball will be 497.4 N.

To learn more about newtons second law of motion refers to the link;

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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:[tex]1.066\times 10^7 m/s[/tex]

Explanation:

Given

Charge per unit area on each plate([tex]\sigma [/tex])=[tex]2.2\times 10^{-7}[/tex]

Plate separation(y)=0.013 m

and velocity is given by

[tex]v^2-u^2=2ay[/tex]

where a=acceleration is given by

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

e=charge on electron

E=electric field

m=mass of electron

[tex]E=\frac{\sigma }{\epsilon _0}[/tex]

[tex]a=\frac{e\sigma }{m\epsilon _0}[/tex]

substituting values

[tex]v=sqrt{\frac{2e\sigma y}{m\epsilon _0}}[/tex]

[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 2.2\times 10^{-7}\times 0.013}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}[/tex]

[tex]v=1.066\times 10^7 m/s[/tex]

The velocity of an electron just before it reaches the positive plate of a parallel plate capacitor can be calculated by equating the converted potential energy to kinetic energy, and then solving for velocity using the electric field, the charge and mass of the electron.

Calculating the Final Velocity of an Electron in a Parallel Plate Capacitor

To find out how fast an electron is moving just before it reaches the positive plate of a parallel plate capacitor, we can use the concepts of electric fields and potential energy. The electric field E between the plates can be found using the charge per unit area σ (sigma) and the vacuum permittivity ε₀ (epsilon nought), given by E = σ / ε₀. Once the electric field is known, we can determine the force on the electron as F = qE, where q is the charge of the electron.

Since the electron starts from rest, the potential energy at the negative plate is converted entirely into kinetic energy just before it hits the positive plate. We can use the electron's charge and the potential difference (V) between the plates to find this energy: qV. We know that the kinetic energy is ½mv², where m is the mass of the electron and v is the final velocity.

Setting the potential energy equal to the kinetic energy, we get qV = ½mv². Solving for the final velocity, v, we find that v = √(2qV/m). Since the potential difference V can be calculated as the product of electric field E and the distance d between the plates, we can substitute V with Ed. The final equation for velocity is therefore v = √(2qEd/m). This will give us the velocity of the electron just before it reaches the positive plate.

Answer:

Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.                      

while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.

Both spacecraft and the astronauts both are in a free-fall condition.

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:

Option d)

Solution:

As per the question:

Work done by farm hand, [tex]W_{FH} = 972J[/tex]

Force exerted, F' = 310 N

Angle, [tex]\theta = 23^{\circ}[/tex]

Now,

The component of force acting horizontally is F'cos[tex]\theta[/tex]

Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.

Thus

[tex]W_{FH} = \vec{F'}.\vec{d}[/tex]

[tex]972 = 310\times dcos23^{\circ}[/tex]

d = 3.406 m = 3.4 m

The distance the wagon travels is 3.4 m. So the correct option is d.

To find the distance the wagon travels, we need to use the work-energy principle. Since the force is applied at an angle, we can find the horizontal component of the force by multiplying the force by the cosine of the angle. The work done is equal to the force multiplied by the distance, so we can rearrange the equation to solve for distance. Therefore, the distance the wagon travels is equal to the work done divided by the force component, or 972 J / (310 N * cos(23°)) = 3.4 m.

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Answer:

10⁻⁶ C

Explanation:

Let the charge be Q

Gain of energy in electric field

= potential x charge

= 25 x 10³ x Q

Kinetic energy of droplet

= .5 x .63 x 10⁻³ x 9 x 9

= 25.515 x 10⁻³ J

So , equating the energy gained

25 x 10³ x Q = 25.515 x 10⁻³

Q = 10⁻⁶ C

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:

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

Answer:

Image is virtual and formed on the same side as the object. 2 m from the lens.

The size of the image is 7.97 cm

Image is upright as the magnification is positive and smaller than the object.

Explanation:

u = Object distance =  6 m

v = Image distance

f = Focal length = -3 m (concave lens)

[tex]h_u[/tex]= Object height = 24 cm

Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{-3}-\frac{1}{6}\\\Rightarrow \frac{1}{v}=\frac{-1}{2} \\\Rightarrow v=\frac{-2}{1}=-2\ m[/tex]

Image is virtual and formed on the same side as the object. 2 m from the lens.

Magnification

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-2}{6}\\\Rightarrow m=0.33[/tex]

[tex]m=\frac{h_v}{h_u}\\\Rightarrow 0.33=\frac{h_v}{0.24}\\\Rightarrow h_v=0.33\times 0.24=0.0797\ m[/tex]

The size of the image is 7.97 cm

Image is upright as the magnification is positive and smaller than the object.

Answer:

(a) The constants required describing the rod's density are B=2.6 and C=1.325.

(b) The mass of the road can be found using [tex]A\int_0^{12}\left(B+Cx)dx[/tex]

Explanation:

(a) Since the density variation is linear and the coordinate x begins at the low-density end of the rod, we have a density given by

[tex]2.6\frac{g}{cm^3}+\frac{18.5\frac{g}{cm^3}-2.6\frac{g}{cm^3}}{12 cm}x = 2.6\frac{g}{cm^3}+1.325x\frac{g}{cm^2}[/tex]

recalling that the coordinate x is measured in centimeters.

(b) The mass of the rod can be found by having into account the density, which is x-dependent, and the volume differential for the rod:

[tex]m=\int\rho dv=\int\left(B+Cx\right)Adx=5\int_0^{12}\left(2.6+1.325x\right)dx=126.6[/tex],

hence, the mass of the rod is 126.6 g.

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:

19 degrees is the answer

Explanation:

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: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:

Charge, q = 1.402 C

Explanation:

Given that,

Current from lightning bolt, [tex]I=7.01\times 10^4\ A[/tex]

Time, t = [tex]t=20\ \mu s=2\times 10^{-5}\ s[/tex]

Let q is the charge transferred in this event. We know that the total charge divided by time is called current. Mathematically, it is given by :

[tex]I=\dfrac{q}{t}[/tex]

[tex]q=I\times t[/tex]

[tex]q=7.01\times 10^4\times 2\times 10^{-5}[/tex]

q = 1.402 C

So, 1.402 coulomb of charge is transferred in this event. Hence, this is the required solution.

Answer:

b) 6.03 seconds

c) 43.164 m/s

Explanation:

t = Time taken

u = Initial velocity = 16 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

[tex]v=u+at\\\Rightarrow 0=16-9.81\times t\\\Rightarrow \frac{-16}{-9.81}=t\\\Rightarrow t=1.63 \s[/tex]

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=16\times 1.63+\frac{1}{2}\times -9.81\times 1.63^2\\\Rightarrow s=13.05\ m[/tex]

So, the stone would travel 13.05 m up

So, total height of the stone would fall is 13.05+80 = 93.05 m

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 93.05=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{93.05\times 2}{9.81}}\\\Rightarrow t=4.4\ s[/tex]

b) The stone will land 1.63+4.4 = 6.03 seconds later

[tex]v=u+at\\\Rightarrow v=0+9.81\times 4.4\\\Rightarrow v=43.164\ m/s[/tex]

c) The stone's velocity when it lands is 43.164 m/s

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 correct response from Kato is: 'When θi = 90°, sin^2 θi = sin(2θi) = 1, which is its maximum value, so h is maximum.

When a tennis ball is thrown into the air at an angle, its height can be calculated using the equation:

h = (vi^2 * sin^2(θi)) / (2g)

where:

The term sin^2(θi) represents the square of the sine of the launch angle.

When the launch angle is 90°, sin^2(θi) equals 1, which is its maximum value. This means that the height the tennis ball reaches is maximum when the launch angle is 90°.

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Kato's second response, "When θi = 90°, sin^2(θi) is maximum, so h is maximum," is the correct explanation.

To determine the maximum height of a tennis ball, we can use the equation:

h = (vi^2 * sin^2(θi)) / (2g)

where h is the maximum height, vi is the initial velocity, θi is the launch angle, and g is the acceleration due to gravity.

Let's analyze Kato's responses:

Therefore, Kato's second response, "When θi = 90°, sin^2(θi) is maximum, so h is maximum," is the correct explanation.

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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:

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.

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Answer:

horizontal direction force move wagon at  18.79 N

Explanation:

given data

force F = 20 N

angle = 20 degree

to find out

What part of the force moves the wagon

solution

we know here as per attach figure

boy pull a wagon at force 20 N at angle 20 degree

so there are 2 component

x in horizontal direction i.e  F cos20

and y in vertical direction i.e F sin20

so we can say

horizontal direction force is move the wagon that is

horizontal direction force = F cos 20

horizontal direction force = 20× cos20

horizontal direction force = 18.79 N

so horizontal direction force move wagon at  18.79 N