A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.0 s later. You may ignore air resistance. If the height of the building is 20 m, what must the initial speed be of the first ball if both balls are to hit the ground at the same time

Answer:

Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.

Energy can also be defined as the ability to do work.

Answer:

[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]

[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]

Explanation:

Firstly, there is supposed to be a diagram attached in order  to complete this question;

I have attached  the diagram below in order to solve this question.

From the data given;

The radius of the car R = 600 ft

Velocity of the car  B, [tex]V_B = 45 mi / hr[/tex]

We are to determine  the magnitude of the acceleration of A and the rate at which the speed of B is changing.

To start with the magnitude of acceleration A;

We all know that

1 mile = 5280 ft and an hour =  3600 seconds

Thus for ; 1 mile/hr ; we have :

5280 ft/ 3600 seconds

= 22/15 ft/sec

However;

for the velocity of the car B = 45 mi/hr; to ft/sec, we have:

= (45 × 22/15) ft/sec

= 66 ft/sec

A free body diagram is attached in the second diagram showing how we resolve the vector form

Now; to determine the magnitude of the acceleration of A; we have:

[tex]^ \to {a_A} = a_A sin 45^0 ^{\to} + a_A cos 45^0 \ j ^{\to} \\ \\ ^\to {a_B} = -(a_t)_B \ i ^ \to + (a_c )_B cos 45 ^0 \ j ^{\to}[/tex]

Where;

[tex](a_c)_B[/tex] = radial acceleration of B

[tex](a_t)_B[/tex] = tangential acceleration of B

From observation in the diagram; The acceleration of B is 0 from A

So;

[tex]a_B ^\to - a_A ^\to = a_{B/A} ^ \to[/tex]

[tex](-(a_t)_B - a_A sin 45^0 ) ^\to i+ ((a_t)_B-a_A \ cos \ 45^0) ^ \to j = 0[/tex]

[tex](a_c)_B = \dfrac{V_B^2}{R}[/tex]

[tex](a_c)_B = \dfrac{(66)^2}{600}[/tex]

[tex](a_c)_B = 7.26 ft/s^2[/tex]

Equating the coefficient of i and j now; we have :

[tex](a_t)_B = -a_A \ sin 45^0 --- (1)\\ \\ (a_c)_B = a_A cos \ 45^0 --- (2)\\ \\[/tex]

From equation (2)

replace [tex](a_c)_B[/tex] with 7.26 ft/s^2; we have

[tex]7.26 \ ft/s^2 = a_A cos \ 45^0 \\ \\ a_A = \dfrac{7.26 \ ft/s^2}{co s \ 45^0}[/tex]

[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]

Similarly;

From equation (1)

[tex](a_t)_B = -a_A \ sin 45^0[/tex]

replace [tex]a_A[/tex] with 10.267 ft/s^2

[tex](a_t)_B = -10.267 \ ft/s^2 * \ sin 45^0[/tex]

[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]

Answer:

Solution,

Mass=100 kg

Acceleration due to gravity(g)=24.79 m/s^2

Now,.

[tex]weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton[/tex]

hope this helps ..

Good luck on your assignment..

Answer:

a_total = 14.022 m/s²

Explanation:

The total acceleration of a uniform circular motion is given by the following formula:

[tex]a=\sqrt{a_c^2+a_T^2}[/tex]         (1)

ac: centripetal acceleration

aT: tangential acceleration

Then, you first calculate the centripetal acceleration by using the following formula:    

[tex]a_c=r\omega^2[/tex]

r: radius of the circular trajectory = 2.0m

w: final angular velocity  of the ball = 7.0 rad/s

[tex]a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}[/tex]        

Next, you calculate the tangential acceleration. aT is calculate by using:

[tex]a_T=r\alpha[/tex]    (2)

α: angular acceleration

The angular acceleration is:

[tex]\alpha=\frac{\omega_o-\omega}{t}[/tex]

wo: initial angular velocity = 13 rad/s

t: time = 15 s

Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:

[tex]a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}[/tex]

Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:

[tex]a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}[/tex]

The total acceleration of the ball is 14.022 m/s²

Answer:

the particle arrangement in liquid are close together with no regular arrangement

Answer:

0.303s

Explanation:

horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m

Using equation of linear motion

Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative

-0.45 = 0 - 0.5 × 9.81×t²

0.45 / (0.5 × 9.81) = t²

t = √0.0917 = 0.303 s

Answer:

The centripetal acceleration of the car will be 12.32 m/s² .

Explanation:

Given that

radius ,R= 57 m

Velocity , V=26.5 m/s

We know that centripetal acceleration given as follows

[tex]a_c=\dfrac{V^2}{R}[/tex]

Now by putting the values in the above equation we get

[tex]a_c=\dfrac{26.5^2}{57}=12.32\ m/s^2[/tex]

Therefore the centripetal acceleration of the car will be 12.32 m/s² .

Answer:

17.656 m

Explanation:

Initial speed u = 20 m/s

angle of projection α = 60°

at the top of the trajectory, one fragment has a speed of zero and drops to the ground.

we should note that the top of the trajectory will coincide with halfway the horizontal range of the the projectile travel. This is because the projectile follows an upward arc up till it reaches its maximum height from the ground, before descending down by following a similar arc downwards.

To find the range of the projectile, we use the equation

R = [tex]\frac{u^{2}sin2\alpha }{g}[/tex]

where g = acceleration due to gravity = 9.81 m/s^2

Sin 2α = 2 x (sin α) x (cos α)

when α = 60°,

Sin 2α  = 2 x sin 60° x cos 60° = 2 x 0.866 x 0.5

Sin 2α  = 0.866

therefore,

R = [tex]\frac{20^{2}*0.866 }{9.81}[/tex] = 35.31 m

since the other fraction with zero velocity drops a top of trajectory, distance between the two fragments assuming level ground and zero air drag, will be 35.31/2 = 17.656 m

Answer:

B. 2nmv

Explanation:

Pressure is force over area.

P = F / A

Force is mass times acceleration.

F = ma

Acceleration is change in velocity over change in time.

a = Δv / Δt

Therefore:

F = m Δv / Δt

P = m Δv / (A Δt)

The total mass is nm.

The change in velocity is Δv = v − (-v) = 2v.

A = 1 and Δt = 1.

Plugging in:

P = (nm) (2v) / (1 × 1)

P = 2nmv

Answer:

a) Jack does more work uphill

b) Numerically, we can see that Jill applied the most power downhill

Explanation:

Jack's mass = 75 kg

Jill's mass = [tex]1.5x = 75[/tex]

Jill's mass = [tex]x = \frac{75}{1.5}[/tex] = 50 kg

distance up hill = 15 m

a) work done by Jack uphill = mgh

where g = acceleration due to gravity= 9.81 m/s^2

work = 75 x 9.81 x 15 = 11036.25 J

similarly,

Jill's work uphill = 50 x 9.81 x 15 = 7357.5 J

this shows that Jack does more work climbing up the hill

b) assuming Jack's time downhill to be t,

then Jill's time = [tex]\frac{t}{3}[/tex]

we recall that power is the rate in which work id done, i.e

P = [tex]\frac{work}{time}[/tex]

For Jack, power = [tex]\frac{11036.25}{t}[/tex]

For Jill, power =  [tex]\frac{3*7357.5}{t}[/tex] =  [tex]\frac{22072.5}{t}[/tex]

Numerically, we can see that Jill applied the most power downhill

Answer:

Change in momentum = 2.197 kgm/s

Explanation:

Momentum = MV

Initial momentum = MU

Final momentum = MV

Computation:

⇒ Change in momentum = MV - MU

⇒ Change in momentum = M (V - U)

⇒ Change in momentum = 0.13(-10.3 - 6.6)

⇒ Change in momentum = 0.13(16.9)

⇒ Change in momentum = 2.197 kgm/s

if the velocity of the car reduces from 70km/h to 50km/h then the speed of the car will be equal to the speed of the lorry...

thus the relative velocity will be 0

Answer:

[tex]d_2=3.16cm[/tex]

Explanation:

So, in order to solve this problem, we must start by building a diagram of the problem itself. (See attached picture) And together with the diagram, we must build a free body diagram, which will include the forces that are being applied on the given charged particle together with their directions.

In this case we only care about the x-direction of the force, since the y-forces cancel each other. So if we do a sum of forces on the x-direction, we get the following:

[tex]\sum{F_{x}}=0[/tex]

so:

[Tex]-F_{12}+F_{13x}+F_{14x}=0[/tex]

Since [tex]F_{13x}=F_{14x}[/tex] we can simplify the equation as:

[tex]-F_{12}+2F_{13x}=0[/tex]

we can now solve this for [tex]F_{12}[/tex] so we get:

[tex]F_{12}=2F_{13x}[/tex]

Now we can substitute with the electrostatic force formula, so we get:

[tex]k_{e}\frac{q_{1}q_{2}}{r_{12}^{2}}=2k_{e}\frac{q_{1}q_{3}}{r_{13}^{2}}cos \theta[/tex]

We can cancel [tex]k_{e}[/tex] and [tex]q_{1}[/tex]

so the simplified equation is:

[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{r_{13}^{2}}cos \theta[/tex]

From the given diagram we know that:

[tex]cos \theta = \frac{d_{1}}{r_{13}}[/tex]

so when solving for [tex]r_{13}[/tex] we get:

[tex]r_{13}=\frac{d_{1}}{cos\theta}[/tex]

and if we square both sides of the equation, we get:

[tex]r_{13}^{2}=\frac{d_{1}^{2}}{cos^{2}\theta}[/tex]

and we can substitute this into our equation:

[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{d_{1}^{2}}cos^{3} \theta[/tex]

so we can now solve this for [tex]r_{12}[/tex] so we get:

[tex]r_{12}=\sqrt{\frac{d_{1}^{2}q_{2}}{2q_{3}cos^{3}\theta}}[/tex]

which can be rewritten as:

[tex]r_{12}=d_{1}\sqrt{\frac{q_{2}}{2q_{3}cos^{3}\theta}}[/tex]

and now we can substitute values.

[tex]r_{12}=(3cm)\sqrt{\frac{7x10^{-19}C}{2(2x10^{-19}C)cos^{3}(20^{o})}}[/tex]

which solves to:

[tex]r_{12}=6.16cm[/tex]

now, we must find [tex]d_{2}[/tex] by using the following equation:

[tex]r_{12}=d_{1}+d_{2}[/tex]

when solving for [tex]d_{2}[/tex] we get:

[tex]d_{2}=r_{12}-d_{1}[/tex]

when substituting we get:

[tex]d_{2}=6.16cm-3cm[/tex]

so:

[tex]d_{2}=3.16cm[/tex]

Answer:

a

  [tex]KE = 7.17 *10^{7} \ J[/tex]

b

 [tex]t = 6411.09 \ s[/tex]

Explanation:

From the question we are told that

    The radius of the flywheel is  [tex]r = 1.50 \ m[/tex]

      The mass of the flywheel is [tex]m = 430 \ kg[/tex]

          The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]

      The power supplied by the motor is  [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]

     Generally the moment of inertia of the flywheel is  mathematically represented as

       [tex]I = \frac{1}{2} mr^2[/tex]

substituting values

       [tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]

       [tex]I = 483.75 \ kgm^2[/tex]

The kinetic energy that is been stored is  

       [tex]KE = \frac{1}{2} * I * w^2[/tex]

substituting values

        [tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]

        [tex]KE = 7.17 *10^{7} \ J[/tex]

Generally power is mathematically represented as

          [tex]P = \frac{KE}{t}[/tex]

=>      [tex]t = \frac{KE}{P}[/tex]

substituting the value

        [tex]t = \frac{7.17 *10^{7}}{11190}[/tex]

        [tex]t = 6411.09 \ s[/tex]

Answer:

Explanation:

Gravitational force between two objects having mass m₁ and m₂ at a distance R

F = G m₁ m₂ / R²

Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²

= 1.01 x 10⁻⁶ N

b )

Force between baby and Jupiter

F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²

= 1.31 x 10⁻⁶  N

c )

Ratio = 1.01 / 1.31

= .77

Answer:

I = 7.5*10^-10 kg m/s

Explanation:

In order to calculate the impulse you first take into account the following formula:

[tex]I=m\Delta v=m(v-v_o)[/tex]       (1)

m: mass of the pollen grain = 1.0*10^-7g = 1.0*10^-10 kg

v: final speed of the pollen grain = ?

vo: initial speed of the pollen grain = 0 m/s

Next, you calculate the final speed of the pollen grain by using the information about the acceleration and time. You use the following formula:

[tex]v=v_o+a t[/tex]       (2)

a: acceleration = 2.5*10^4 m/s^2

t: time = 0.30ms = 0.30*10^-3 s

[tex]v=0m/s+(2.5*10^4m/s^2)(0.30*10^{-3}s)=7.5\frac{m}{s}[/tex]

Next, you replcae this value of v in the equation (1) and calculate the impulse:

[tex]I=m(v-v_o)=(1.0*10^{-10}kg)(7.5m/s-0m/s)=7.5*10^{-10}kg.\frac{m}{s}[/tex]

The impulse delivered to the pollen grain is 7.5*10^-10 kg m/s

Answer:

I’m saying kinetic gravitational and electromagnetic and I will comment on this if I got it right

Explanation:.

Answer:

The inner diameter is 57.3 cm

Explanation:

The inner diameter of the hollow spherical iron shell can be found using the weight of the sphere ([tex]W_{s}[/tex]) and the weight of the water displaced ([tex]W_{w}[/tex]):

[tex] W_{s} = W_{w} [/tex]

[tex] m_{s}*g = m_{w}*g [/tex]            

[tex] D_{s}*V_{s} = D_{w}*V_{w} [/tex]    

Where D is the density and V is the volume

[tex] D_{s}*\frac{4}{3}\pi*(\frac{d_{o}^{3} - d_{i}^{3}}{2^{3}}) = \frac{4}{3}\pi*(\frac{d_{o}}{2})^{3} [/tex]    

Where [tex]d_{o}[/tex] is the outer diameter and [tex]d_{i}[/tex] is the inner diameter    

[tex] D_{s}*(d_{o}^{3} - d_{i}^{3}) = d_{o}^{3} [/tex]                    

[tex] D_{s}*d_{i}^{3} = d_{o}^{3}(D_{s} - 1) [/tex]          

[tex] 7.87*d_{i}^{3} = 60.0^{3}(7.87 - 1) [/tex]  

[tex] d_{i} = 57.3 cm [/tex]                  

Therefore, the inner diameter is 57.3 cm.    

I hope it helps you!    

Rock can melt at a depth of about below Earth's surface  100 km

On the other hand, interface physics offers a wide variety of spectroscopic & microscopic techniques to characterize that deposition & structure creation process upon that sub-nanometer size and, therefore, to pave the way for effective manufacturing techniques.

It is the study of both the chemical processes that take place at the meeting point of two surfaces, such as solid-liquid, solid-gas, sturdy, liquid-gas, etc. Surface engineering refers to a few surface chemistry applications.

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Answer:

(a) 7.67 m/s.

(b)  4.9 J

Explanation:

(a) From the law of conservation of energy,

P.E = K.E

mgh = 1/2(mv²)

therefore,

v = √(2gh)....................... Equation 1

Where v = speed of the ball before bounce, g = acceleration due to gravity, h = height from which the ball was dropped.

Given: h = 3 m, g = 9.8 m/s²

Substitute into equation 1

v = √(2×9.8×3)

v = √(58.8)

v = 7.67 m/s.

(b) Energy of the ball before the bounce = mgh = 0.5×9.8×3 = 14.7 J

Energy of the ball after the bounce = mgh' = 0.5(9.8)(2) = 9.8 J

Amount of energy converted to heat = 14.7-9.8 = 4.9 J

Answer:

Density depends on the temperature and the gap between particles of the liquid. In most of cases temperature is inversely proportional to density means if the temperature increases then the density decreases and the space between particles of that liquid is also inversely proportional to the density means if the intraparticle space increases then the density decreases.

Answer:

a) 8.33 x [tex]10^{-6}[/tex] Pa  or  8.22 x [tex]10^{-11}[/tex] atm

b) 1.66 x [tex]10^{-5}[/tex] Pa  or  1.63 x [tex]10^{-10}[/tex] atm

c) 2.77 x [tex]10^{-14}[/tex] kg/m^2-s

Explanation:

Intensity of light = 2500 W/m^2

area = 25 ft^2

a) average radiation pressure on a totally absorbing section of the floor[tex]Pav = \frac{I}{c}[/tex]

where I is the intensity of the light

c is the speed of light = [tex]3*10^{8} m/s[/tex]

[tex]Pav = \frac{2500}{3*10^{8} }[/tex] = 8.33 x [tex]10^{-6}[/tex] Pa

1 pa = [tex]9.87*10^{-6}[/tex]

8.33 x [tex]10^{-6}[/tex] Pa = 8.22 x [tex]10^{-11}[/tex] atm

b) average radiation for a totally radiating section of the floor

[tex]Pav = \frac{2I}{c}[/tex]

this means that the pressure for a totally radiating section is twice the average pressure of the totally absorbing section

therefore,

Pav = 2 x 8.33 x [tex]10^{-6}[/tex]  = 1.66 x [tex]10^{-5}[/tex] Pa

or

Pav in atm = 2 x 8.22 x [tex]10^{-11}[/tex] = 1.63 x [tex]10^{-10}[/tex] atm

c) average momentum per unit volume is

[tex]m = \frac{I}{c^{2} }[/tex]

[tex]m = \frac{2500}{(3*10^{8}) ^{2} }[/tex] = 2.77 x [tex]10^{-14}[/tex] kg/m^2-s

Answer:

The value of the potential energy of the particle at point B is 85 joules.

Explanation:

According to the Principle of Energy Conservation, the energy cannot be created nor destroyed, only transformed. The particle at point A has kinetic and potential energy and receives a work due to an external conservative force (Work-Energy Theorem), whose sum is equal to potential energy at point B. Mathematically speaking, the expression that describes the phenomenon is:

[tex]K_{A} + U_{A} + W_{A \rightarrow B} = U_{B}[/tex]

Where:

[tex]K_{A}[/tex] - Kinetic energy at point A, measured in joules.

[tex]U_{A}[/tex] - Potential energy at point A, measured in joules.

[tex]W_{A \rightarrow B}[/tex] - Work due to conservative force from A to B, measured in joules.

[tex]U_{B}[/tex] - Potential energy at point B, measured in joules.

The initial kinetic energy of the particle is:

[tex]K_{A} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v[/tex] - Velocity, measured in meters per second.

If [tex]m = 0.4\,kg[/tex] and [tex]v = 10\,\frac{m}{s}[/tex], then:

[tex]K_{A} = \frac{1}{2}\cdot (0.4\,kg)\cdot \left(10\,\frac{m}{s} \right)^{2}[/tex]

[tex]K_{A} = 20\,J[/tex]

Finally, the value of the potential energy at point B is:

[tex]U_{B} = 20\,J + 40\,J + 25\,J[/tex]

[tex]U_{B} = 85\,J[/tex]

The value of the potential energy of the particle at point B is 85 joules.

The potential energy of the particle at point B is 85 J.

Given to us:

Mass of the particle, [tex]m=0.40\ kg[/tex]

velocity of the particle, [tex]v= 10\ m/s[/tex]

potential energy of the particle, [tex]PE= 40\ J[/tex]

Workdone from pt. A to B, [tex]WD_{(A\rightarrow B)} = 25\ J[/tex]

Calculating the kinetic energy of the particle,

[tex]\begin{aligned}KE&= \frac{1}{2}mv^2 \\\\&=\frac{1}{2}\times0.40\times (10)^2\\\\&=20 J\\\end{aligned}[/tex]

According to the  Principle of Energy Conservation,

The energy cannot be created nor be destroyed, it can only be transformed from one form to another.Therefore,

Also,

Total Energy at point A ,

[tex]\begin{aligned}(TE)_A &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_A+ PE_A+UE_A+ WD_{(0\rightarrow A)}\\&=20+40+0+0\\&=60\ J\end{aligned}[/tex]

Total Energy at point B,

[tex]\begin{aligned}(TE)_B &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\end{aligned}[/tex]

As the total energy is conserved from point A to B and also an external work is done on the particle. we can write the above equation as,

[tex]\begin{aligned} TE_B&=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\\&=(KE_B+ PE_B+UE_B)+ WD_{(A\rightarrow B)}\\&= TE_A+ WD_{(A\rightarrow B)}\\&=60+25\\&=85\ J\end{aligned}[/tex]

Therefore, the total energy for the particle at point B is 85 J but as the particle is not moving neither work is done at point B, the total energy of the particle is the potential energy of the particle.

Hence, the potential energy of the particle at point B is 85 J.

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Answer:

Q = 5.9 nC (Approx)

Explanation:

Given:

Further distance = 10 cm

Electric potential(V) = 190 v

Potential difference(V1) = 140 v

Find:

Magnitude of the charge Q

Computation:

V = KQ / r

190 = KQ / r.............Eq1

V1  = KQ / (r+10)

140 = KQ / (r+10) ............Eq2

From Eq2 and Eq1

r = 28 cm = 0.28 m

So,

190 = KQ / r

190 = (9×10⁹)(Q) / 0.28

53.2 = (9×10⁹)(Q)

5.9111 = (10⁹)(Q)

Q = 5.9 nC (Approx)

Answer : The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:

[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]

Explanation :

Beta decay : It is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.

The released beta particle is also known as electron.

The beta decay reaction is:

[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]

The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:

[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]

Answer:

the blank is 39

Explanation: a p e x

Answer:

t = 17.68s

Explanation:

In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:

[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex]            (1)

y: height for a time t

yo: initial height = 1000m

vo: initial velocity = 0m/s

g: gravitational acceleration = 9.8m/s^2

t: time = 5.00 s

You replace the values of the parameters to get the values of the new height of the skydiver:

[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]

Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.

You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0

[tex]0=877.5-7.00t-4.9t^2[/tex]       (2)

The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:

[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]

You use the positive value of t1 because it has physical meaning.

Finally, you sum the times of both parts of the trajectory:

total time = 5.00s + 12.68s = 17.68s

The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s

Answer:

The magnitude of the acceleration of the car when [tex]s = 50\,m[/tex] is [tex]2\,\frac{m}{s^{2}}[/tex].

Explanation:

The acceleration can be obtained by using the following differential equation:

[tex]a = v \cdot \frac{dv}{ds}[/tex]

Where [tex]a[/tex], [tex]v[/tex] and [tex]s[/tex] are the acceleration, speed and distance masured in meters per square second, meters per second and meters, respectively.

Given that [tex]v = 0.2\cdot s[/tex], its first derivative is:

[tex]\frac{dv}{ds} = 0.2[/tex]

The following expression is obtained by replacing each term:

[tex]a = 0.2\cdot 0.2\cdot s[/tex]

[tex]a = 0.04\cdot s[/tex]

The magnitude of the acceleration of the car when [tex]s = 50\,m[/tex] is:

[tex]a = 0.04\cdot (50\,m)[/tex]

[tex]a = 2\,\frac{m}{s^{2}}[/tex]

Answer:

10.15m/s

Explanation:

The change in the frequency of sound (or any other wave) when the source of the sound and the receiver or observer of the sound move towards (or away from) each other is explained by the Doppler effect which is given by the following equation:

f₁ = [(v ± v₁) / (v ± v₂)] f            ----------------------(i)

Where;

f₁ = frequency received by the observer or receiver

v = speed of sound in air

v₁ = velocity of the observer

v₂ = velocity of the source

f = original frequency of the sound

From the question, the observer is the bicyclist and the source is the car driver. Therefore;

f₁ = frequency received by the observer (bicyclist) = 357Hz

v = speed of sound in air = 330m/s

v₁ = velocity of the observer(bicyclist) = (1 / 3) v₂ = 0.33v₂

v₂ = velocity of the source (driver)

f = original frequency of the sound = 365Hz

Note: The speed of the observer is positive if he moves towards the source and negative if he moves away from the source. Also, the speed of the source is positive if it moves away from the listener and negative otherwise.

From the question, the cyclist and the driver are moving in the same direction. But then, we do not know which one is at the front. Therefore, two scenarios are possible.

i. The bicyclist is at the front. In this case, v₁ and v₂ are negative.

Substitute these values into equation (i) as follows;

357 = [(330 - 0.33v₂) / (330 - v₂)] * 365

(357 / 365) = [(330 - 0.33v₂) / (330 - v₂)]

0.98 =  [(330 - 0.33v₂) / (330 - v₂)]

0.98 (330 - v₂) =  (330 - 0.33v₂)

323.4 - 0.98v₂ = 330 - 0.33v₂

323.4 - 330 = (0.98 - 0.33)v₂

-6.6 = 0.65v₂

v₂ = -10.15

The value of v₂ is not supposed to be negative since we already plugged in the right value polarity into the equation.

iI. The bicyclist is behind. In this case, v₁ and v₂ are positive.

Substitute these values into equation (i) as follows;

357 = [(330 + 0.33v₂) / (330 + v₂)] * 365

(357 / 365) = [(330 + 0.33v₂) / (330 + v₂)]

0.98 =  [(330 + 0.33v₂) / (330 + v₂)]

0.98 (330 + v₂) =  (330 + 0.33v₂)

323.4 + 0.98v₂ = 330 + 0.33v₂

323.4 - 330 = (0.33 - 0.98)v₂

-6.6 = -0.65v₂

v₂ = 10.15

The value of v₂ is positive and that is a valid solution.

Therefore, the speed of the car is 10.15m/s

Answer:

V = 8.01*10^-12 V

Explanation:

In order to calculate the electric potential produced by the dipole you use the following formula:

[tex]V=k\frac{p}{r^2}[/tex]        (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

p: dipole moment = 5.2×10−30 C⋅m

r: distance to the dipole = 2.8*10^-9m

You replace the values of the parameters:

[tex]V=(8.98*10^9Nm^2/C^2)\frac{5.2*10^{-30}Cm}{(2.8*10^{-9}m)}\\\\V=8.01*10^{-12}V[/tex]

The electric potential of the dipole is 8.01*10^-12V