A gallon of gasoline contains about 1.3 x 108joules
of energy. A 2000 kg car traveling at 20 m/s skids to astop.
Estimate how much gasoline it will take to bring the car backto the
original speed? To complicate matters further, consider thefact
that only about 15% of the energy extracted from gasolineactually
propels the car. The rest gets exhausted as heat andunburnt
fuel.

To calculate the final velocity of someone running with constant acceleration, we first find the acceleration using the distance and time, and then apply it to determine the final velocity. The final velocity after 10 seconds is 8 m/s.

To find the final velocity of the person running with constant acceleration, we use the kinematic equation v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time. The person starts from rest, so u = 0. We need to find a first, and we can do so using the equation s = ut + (1/2)at², where s is the distance traveled.

Insert the given values: 40m = 0 + (1/2)a(10s)², resulting in a = 0.8 m/s². Now apply the acceleration to the initial equation: v = 0 + 0.8 m/s² * 10s, which gives us v = 8 m/s. Therefore, the person's final velocity after 10 seconds is 8 m/s.

Answer:

460 Hz

Explanation:

the given resonating frequency of the device

f₁ =  920 Hz and f₂ =  1380 Hz

fundamental frequency of the device is

f₀ = n₂ - n₁

   = 1380 - 920

   = 460 Hz

expression of frequency of organic pipe open at both ends

[tex]f_0=n\dfrac{\nu}{2l}[/tex]

at n = 1

[tex]f_0=\dfrac{\nu}{2l} = 460 Hz[/tex]

the frequency ratios of the closed pipe

[tex]f_0:f_1:f_2: ...... =[1:2:3:.........]f_0[/tex]

                         =[1:2:3:.........]460 Hz

                         = 460 Hz : 920 Hz : 1380 Hz

so, the lowest frequency for the pipe open at both end is 460 Hz

Answer:

runway use is 3307.8 feet

Explanation:

given data

velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s

time = 28 seconds

weight = 28000 lbs

to find out

How many feet of runway was used

solution

we will use here first equation of motion for find acceleration

v = u + at     ..............1

here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time

put here value in equation 1

72.016 = 0 + a(28)

a = 2.572 m/s²

and

now apply third equation of motion

s = ut + 0.5×a×t²    .......................2

here s is distance and u is initial speed and t is time and a is acceleration

put here all value in equation 2

s = 0 + 0.5×2.572×28²  

s = 1008.24 m = 3307.8 ft

so  runway use is 3307.8 feet

To find the runway distance used by an aircraft accelerating to 140 knots in 28 seconds, we convert knots to feet per second, find the acceleration, and then apply the kinematic formula for distance to obtain approximately 3308 feet.

To calculate how many feet of runway an aircraft used to accelerate to 140 knots in 28 seconds, we need to convert the speed to consistent units and use the kinematic equations for uniformly accelerated motion. Knots need to be converted to feet per second (since the answer is required in feet). Since 1 knot = 1.68781 feet per second, 140 knots is equivalent to 236.293 feet per second. Now, we can use the formula for distance d when given initial velocity vi, final velocity vf, and acceleration a:

d = (vi + vf) / 2 * t

Here, as the plane is starting from a standstill, vi is 0, vf is 236.293 feet/second, and t is 28 seconds. We first need to find the acceleration:

a = (vf - vi) / t = 236.293 / 28 = 8.439 feet/second2

Now we can calculate the distance:

d = (0 + 236.293) / 2 * 28 = 3308.1 feet

The aircraft used approximately 3308 feet of runway.

Answer:

A) 29.5m/s

B) 25.9m/s

C) 47.9 m/s

Explanation:

This is a relative velocity problem, which means that the velocity perception will vary from each observer at a different reference point.

We can say that the velocity of the train respect the ground is 28m/s on east direction If I walk 1.5m/s respect the train, the velocity of the person respect to the ground is the sum of both velocities:

Vgp=Vgt+Vtp

where:

Vgp=velocity of the person respect the ground

Vgt=Velocity of the train respect the ground

Vtp=Velocity of the person respect the train

1) Vgp=28m/s+1.5m/s=29.5m/s

2) here the velocity of the person respect the train is negative because it is going backward

Vgp=28m/s-2.1m/s=25.9m/s

3) the velocity of the train respect with your friend will be the sum of both velocities(Vft).

Vfp=Vft+Vtp

Vfp=velocity of the person respect the friend

Vft=Velocity of the train respect the friend

Vtp=Velocity of the person respect the friend

Vfp=(22m/s+28m/s)+(-2.1m/s)=47.9 m/s

Answer:

He knocks her. V = 7.97m/s

Explanation:

Let A be Tarzan's starting position and B Jane's position (Tarzan's final position).

Since there is no air resistance, energy at position A must be equal to energy at B.

At position A, Tarzan's speed is zero and since the 45° of the vine is greater than the final 30°, Tarzan will have potential energy at point A.

[tex]E_{A}=m_{T}*g*L*(cos(30)-cos(45))[/tex]

At point B, it is the lowest point (between A and B), so it has no potential energy. It will only have kinetic energy (ideally zero, for Jane's sake, but we don't know).

[tex]E_{B}=\frac{m_{T}*V^{2}}{2}[/tex]

Because of energy conservation, we know that Ea=Eb, so:

[tex]m_{T}*g*L*(cos(30)-cos(45))=\frac{m_{T}*V^{2}}{2}[/tex]  Solving for V:

[tex]V=\sqrt{2*g*L*(cos(30)-cos(45))}=7.97m/s[/tex]

Answer:

(a) 4.04 ohm

(b) 6.93 A

(c) 8.53°

Explanation:

f = 12 kHz = 12000 Hz

Vo = 28 V

R = 4 ohm

L = 30 micro Henry = 30 x 10^-6 H

C = 8 micro Farad = 8 x 10^-6 F

(a) Let Z be the impedance

[tex]X_{L} = 2\pi fL=2\times3.14\times12000\times30\times10^{-6}= 2.26 ohm[/tex]

[tex]X_{c} = \frac{1}{2\pi fC}=\frac{1}{2\times3.14\times12000\times8\times10^{-6}}= 1.66 ohm[/tex]

[tex]Z = \sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{4^{2}+\left ( 2.26-1.66 \right )^{2}}[/tex]

Z = 4.04 Ohm

(b) Let Io be the amplitude of current

[tex]I_{o}=\frac{V_{o}}{Z}[/tex]

[tex]I_{o}=\frac{28}{4.04}[/tex]

Io = 6.93 A

(c) Let the phase difference is Ф

[tex]tan\phi = \frac{X_{L}-X_{C}}{R}[/tex]

[tex]tan\phi = \frac{2.26-1.66}{4}[/tex]

tan Ф =0.15

Ф = 8.53°

Answer:

Option (c) [tex]2\times 10^{-3}\ m=2\ millimeters[/tex]

Explanation:

Here, it is required to describe the given length in a particular unit. Firstly, we need to see the following conversions as :

1 meter = 0.001 kilometers

1 meter = 10⁻⁶ megameters

1 meter = 1000 millimeters

1 meter = 1000000 micrometers

From the given option, the correct one is (c) because, 1 meter = 1000 millimeters

So, [tex]2\times 10^{-3}\ m=2\ millimeters[/tex]

Hence, the correct option is (c). Hence, this is the required solution.

2.0×10^-3 of a meter is equivalent to 2.0 millimeters, hence option c is the correct answer.

The given length, 2.0×10^-3 of a meter, corresponds to a unit of length commonly used in the metric system. In this system, 1 meter is equal to 10^3 millimeters. Therefore, 2.0×10^-3 of a meter equates to 2.0 millimeters. This means the best answer from the provided options would be option c: 2.0 millimeters.

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Answer:

μ= 0.408 : coefficient of kinetic friction

Explanation:

Kinematic equation for the box:

[tex]a=\frac{v_{f} -v_{i} }{t}[/tex] Formula( 1)

a= acceleration

v_i= initial speed =0

v_f= final speed= 20 m/s

t= time= 5 s

We replace data in the formula (1):

[tex]a=\frac{0-20}{5}[/tex]

[tex]a= -\frac{20}{5}[/tex]

a= - 4m/s²

Box kinetics: We apply Newton's laws in x-y:

∑Fx=ma : second law of Newton

-Ff= ma Equation (1)

Ff is the friction force

Ff=μ*N Equation (2)

μ is the coefficient of kinetic friction

N is the normal force

Normal force calculation

∑Fy=0  : Newton's first law

N-W=0   W is the weight of the box

N=W= m*g  : m is the mass of the box and g is the acceleration due to gravity

N=9.8*m

We replace N=9.8m in the equation (2)

Ff=μ*9.8*m

Coefficient of kinetic friction ( μ) calculation

We replace Ff=μ*9.8*m and a= -4m/s² in the equation (1):

-μ*9.8*m: -m*4 : We divide by -m on both sides of the equation

9.8*μ=4

μ=4 ÷ 9.8

μ= 0.408

Answer:

The charge on each plate of the capacitor is [tex]19.38 \mu C[/tex]

with one plate positive and one negative, i.e., [tex]\pm 19.38 \mu C[/tex]

Solution:

According to the question:

Critical value of Electric field, [tex]E_{c} = 3.0\times 10^{6} V/m[/tex]

Area of each plate of capacitor, [tex]A_{p} =73 cm^{2} = 73\times 10^{- 4} m^{2}[/tex]

Now, the amount of charge on the capacitor's plates can be calculated as:

Capacitance, C  = [tex]\frac{epsilon_{o}\times Area}{Distance, D}[/tex]        (1)

Also, Capacitance, C = [tex]\frac{charge, q}{Voltage, V}[/tex]

And

Electric field, E = [tex]\frac{Voltage, V}{D}[/tex]

So, from the above relations, we can write the eqn for charge, q as:

q = [tex]\epsilon_{o}\times E_{c}\times A_{p}[/tex]

q = [tex]8.85\times 10^{- 12}\times 3.0\times 10^{6}\times 73\times 10^{- 4}[/tex]

[tex]q = 19.38 \mu C[/tex]

Answer:[tex]\theta =41.409 ^{\circ}[/tex]

Explanation:

Given

Jack and Jill ran up the hill at 2.8 m/s

Horizontal component of Jill's velocity vector was 2.1 m/s

Let [tex]\theta [/tex]is the angle made by Jill's velocity with it's horizontal component

Therefore

[tex]2.8cos\theta =2.1[/tex]

[tex]cos\theta =\frac{2.1}{2.8}[/tex]

[tex]cos\theta =0.75[/tex]

[tex]\theta =41.409 ^{\circ}[/tex]

Vertical velocity is given by

[tex]V_y=2.8sin41.11=1.85 m/s[/tex]

The question can be solved using the Pythagorean theorem to calculate the vertical velocity component and apply trigonometry to find the angle of the hill. Remember the conversion from radians to degrees.

The question asks about the angle of the hill that Jack and Jill climbed, as well as the vertical component of Jill’s velocity. We know that Jill ran up the hill at a velocity of 2.8 m/s and the horizontal component of her velocity was 2.1 m/s. These two components form a right angle triangle where the hypotenuse is the total velocity (2.8 m/s), one side is the horizontal velocity (2.1 m/s), and the other side, which we're finding, is the vertical component of the velocity.

We can find the angle of the hill, θ, using the tangent function of trigonometry. Tan θ = opposite side / adjacent side. In this case, the 'opposite side' is the vertical velocity component we're after, and the 'adjacent side' is the horizontal component (2.1 m/s). To find the vertical velocity component, you can use the Pythagorean theorem, which states that: (Hypotenuse)² = (Adjacent Side)² + (Opposite Side)² or (2.8 m/s)² = (2.1 m/s)² + (vertical velocity)².

Once you solve for the vertical velocity through the Pythagorean theorem, you then insert it into the tangent equation to solve for the angle θ. Remember that when you use the arctangent function on a calculator to find the angle, the answer will likely be in radians, so you might need to convert them into degrees.

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Answer:

Work done,[tex]w=5.12\times10^{6}\ \rm J[/tex]

change in internal Energy ,[tex]\Delta U=5.12\times10^6\ \rm J[/tex]

Explanation:

Given:

Since It is given that the process is adiabatic process it means that there is no exchange of heat between the system and surroundings

[tex]T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}\\\\286\times V_i^{\gamma -1}=T_f \left( \dfrac{V_i}{5} \right )^{\gamma -1}\\T_f=840.76\ \rm K[/tex]

Let n be the number of moles of Helium given by

[tex]n=\dfrac{m}{M}\\n=\dfrac{3\times10^3}{4}\\n=0.75\times10^3[/tex]

Work done in Adiabatic process

Let W be the work done

[tex]W=\dfrac{nR(T_1-T_2)}{\gamma-1}\\W=\dfrac{0.75\times10^3\times8.314(286-840.76)}{1.67-1}\\W=-5.12\times 10^6\ \rm J[/tex]

The Internal Energy change in any Process is given by

Let [tex]\Delta U[/tex] be the change in internal Energy

[tex]\Delta U=nC_p\Delta T\\\Delta U=0.75\times10^3\times1.5R\times(840.76-286)\\\Delta U=5.12\times10^6\ \rm J[/tex]

Answer:

a) time taken = 2.66 s

b) v = 28.34

Explanation:

given,

rate of descending = 2.3 m/s

height of camera above ground = 41 m              

using equation of motion                      

[tex]h = u t + \dfrac{1}{2}gt^2[/tex]              

[tex]41 =2.3t + \dfrac{1}{2}\times 9.8\times t^2[/tex]

4.9 t² + 2.3 t - 41 =0                      

t = 2.66 ,-3.13                    

time taken = 2.66 s

b) v² = u² + 2 g h

v² = 0 + 2× 9.8 × 41

v = 28.34                            

Answer:

h = 50.49 m

Explanation:

Data provided:

Speed of skier, u = 2.0 m/s

Maximum safe speed of the skier, v = 30.0 m/s

Mass of the skier, m = 85.0

Total work = 4000 J

Height from the starting gate = h

Now, from the law of conservation of energy

Total energy at the gate = total energy at the time maximum speed is reached

[tex]\frac{1}{2}mu^2+mgh=4000J+\frac{1}{2}mv^2[/tex]

where, g is the acceleration due to the gravity

on substituting the values, we get

[tex]\frac{1}{2}\times85\times2.0^2+85\times9.81\times h=4000J+\frac{1}{2}\times85\times30^2[/tex]

or

170 + 833.85 × h = 4000 + 38250

or

h = 50.49 m

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = [tex]\frac{4}{3}\pi(\frac{d}{2})^3[/tex]

or

Volume of the balloon = [tex]\frac{4}{3}\pi(\frac{0.296}{2})^3[/tex]

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314  kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = [tex]\frac{3}{2}\times K_BT[/tex]

where,

Boltzmann constant, [tex]K_B=1.3807\times10^{-23}J/K[/tex]

Average kinetic energy = [tex]\frac{3}{2}\times1.3807\times10^{-23}\times292[/tex]

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = [tex]\frac{3RT}{m}[/tex]

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = [tex]\frac{3\times8.314\times292}{0.004}[/tex]

or

rms speed = 1349.35 m/s

Answer:0.835 s

Explanation:

Given

Red ball initial velocity([tex]u_r[/tex])=1.1 m/s

height of building(h)=28 m

after 0.5 sec blue ball is thrown with a velocity([tex]u_b[/tex])=24.3 m/s

Height of blue after 2 sec red ball is thrown

i.e. height of blue ball at t=1.5 sec after blue ball is thrown upward

[tex]h=24.3\times 1.5-\frac{9.81\times 1.5^2}{2}=25.414 m[/tex]

therefore blue ball is at height of 25.414+0.9=26.314 from ground

moment after the two ball is at same height

for red ball

[tex]14=1.1\times \left ( t+0.5\right )+\frac{9.81\times \left ( t+0.5\right )^2}{2}-----1[/tex]

for blue ball

[tex]13.1=24.3\times t-\frac{9.81\times t^2}{2}-----2[/tex]

add 1 & 2

we get

[tex]27.1=1.1t+0.55+24.3t+\frac{g\left ( t+0.25\right )}{2}[/tex]

27.1=25.4t+0.55+4.905t+1.226

t=0.835 s

Answer:

(a ) vf= 188.12m/s  : Final speed at 8.75 s

(b) d= 823.04 m   : Distance the rocket sled traveled

Explanation:

Rocket sled kinematics :The rocket sled moves with a uniformly accelerated movement, then we apply the following formulas:

d =vi*t+1/2a*t² Formula (1)

vf= vi+at            Formula(2)

Where:

vi: initial speed =0

a: acceleration=21.5 m/s²

t: time=8.75 s

vf: final speed in m/s

d:displacement in meters(m)

Calculation of displacement (d) and final speed (vf)

We replace data in formulas (1) and (2):

d= 0+1/2*21.5*8.75²

d= 823.04 m

vf= 0+21.5*8.75

vf= 188.12m/s

Answer:

(a) 13.6 eV

(b) 10.2 V

Explanation:

a) Ionization potential energy is defined as the minimum energy required to excite a neutral atom to its ionized state i.e basically  the minimum energy required to excite an electron from n=1 to infinity.

Energy of a level, n, in Hydrogen atom is, [tex]E_{n}=-\frac{13.6}{n^{2} }[/tex]

Now ionization potential can be calculated as

[tex]E_{\infty}- E_{0}[/tex]

Substitute all the value of energy and n in above equation.

[tex]=-\frac{13.6}{\infty^{2}}-(-\frac{13.6}{1^{2}})\\=13.6eV[/tex]

Therefore, the ionization potential is 13.6 eV.

b) This is the energy required to excite a atom from ground state to its excited state. When electrons jumps from ground state level(n=1) to 1st excited state(n=2) the corresponding energy is called 1st excitation potential energy and corresponding potential is called 1st excitation potential.

So, 1st excitation energy = E(n 2)- E(n = 1)

[tex]=-\frac{13.6}{2^{2}}-(-\frac{13.6}{1^{2}})\\=-3.4eV - (-13.6eV) \\=10.2eV[/tex]

Now we can find that 1st excitation energy is 10.2 eV which gives,

[tex]eV'=10.2eV\\V'=10.2V[/tex]

Therefore, the 1st excitation potential is 10.2V.

Answer:

Magnitude of resultant force is  1190.314 N

Direction of force is 13.352°

Explanation:

given data:

thrust force = 725 N

Angle = 32.4 degree

Let x is consider as positive direction

Resultant force in x direction is  

Rx = 725 + 513cos32.4 = 1158.14 N

and Resultant force perpendicular to x direction is:

Ry = 513sin32.4 = 274.88 N

Magnitude of resultant force is

[tex]R=\sqrt{R_x^2+R_y^2} = 1190.314N[/tex]

and resultant force direction is  

[tex]\theta=tan^{-1}\frac{R_y}{R_x} = 13.352\degree[/tex]

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11)   Now we will calculate the module of this vector:

[tex]|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82[/tex]

The speed of the boat with respect to the shore is 1.91 m/s.

From the information given, we have that;

A boat's speed in still water is 1.60 m/s

The boat is to travel directly across a river whose current has speed 1.05 m/s

We can see that the movement is in both horizontal and vertical directions.

Using the Pythagorean theorem, let use determine the resultant speed of the boat with respect to the shore, we have that;

Resultant speed² = √((boat's speed)² + (current's speed)²)

Substitute the value as given in the information, we have;

= (1.60)² + (1.05 )²)

Find the value of the squares, we get;

= (2.56 + 1.1025 )

Find the square root of both sides, we have;

=  √3.6625

Find the square root of the value, we have;

= 1.91 m/s.

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Answer:  b) TRUE and d) TRUE

Explanation:  a) FALSE the electric field lines are used to represent the charges postives and negatives.

b) TRUE it is the definition of  electric field lines , they are tangent to the electric field vector.

c) FALSE  the electric field lines only for conductor are perpendicular to the surface in other any situation there is a tangencial electric field components.

d) TRUE since it is a way to describe and imagine the effect of vectorial fields.

e) FALSE

Answer:

[tex]F=39,68N[/tex]

Explanation:

Data:

Mass [tex]m=25 Kg[/tex]

Coefficient of kinetic friction [tex]\mu=0.12[/tex]

Angle = [tex]29^{0}[/tex]

Acceleration = [tex]0.12 \frac{m}{s^{2} }[/tex]

Solution:

By Newton's first law we know that for the x-axis:

[tex]F_{rope_x}-F_f=F_R[/tex] Where [tex]F_R[/tex] is the resulting force, and [tex]F_f[/tex] is the friction force.

And for the y-axis:

[tex]F_{rope_y}+N=W[/tex], where N is the normal force, and W is the weight of the sled.  

We know that the resulting force's acceleration is [tex]0.12 \frac{m}{s^{2} }[/tex], and by using Newton's second law, we obtain:

[tex]F=m.a[/tex]

[tex]F_R=25Kg. 0.12\frac{m}{s^2} \\ F_R=3N[/tex] .

Now, the horizontal component of the force in the rope will be given by

[tex]F_{rope_x}=F_{rope}.cos(29^0)=F_R+F_f[/tex], since the resulting force is completely on the x-axis, and the friction opposes to the speed of the sled.

To obtain the friction force, we must know the normal force:

[tex]F_f=\mu. N[/tex]

Clearing N in the y-axis equation:

[tex]N=W-F_{rope_y}=W-F_{rope}.sin(29^0)[/tex]

So we can express the x-axis equation as follows:

[tex]F.cos(29^0)=F_R+\mu.(W-F_{rope}.sin(29^0))[/tex]

Finally, solving for F we get

[tex]F = (F_R + \mu. m.g) / (cos (29^0) + \mu.sin (29^0))[/tex]

[tex]F=39,68N[/tex]

Answer:

[tex]C_{eq}=1.97\ \mu F[/tex]

Explanation:

Given that,

Capacitance 1, [tex]C_1=0.5\ \mu F[/tex]

Capacitance 2, [tex]C_2=11\ \mu F[/tex]

Capacitance 3, [tex]C_3=1.5\ \mu F[/tex]

C₁ and C₂ are connected in series. Their equivalent is given by :

[tex]\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}[/tex]

[tex]\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}[/tex]

[tex]C'=0.47\ \mu F[/tex]

Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :

[tex]C_{eq}=C'+C_3[/tex]

[tex]C_{eq}=0.47+1.5[/tex]

[tex]C_{eq}=1.97\ \mu F[/tex]

So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.

Answer:

The initial velocity of the ball should be 50 m/s.

Explanation:

Since the trip of the ball shall consist of upward ascend and the downward descend and since the ascend and the descend of the ball is symmetrical we infer that the upward ascend of the ball shall last for a time of 5 seconds.

Now since the motion of ball is uniformly accelerated we can find the initial speed of the ball using first equation of kinematics as

[tex]v=u+gt[/tex]

where,

'v' is final velocity of the ball

'u' is initial velocity of the ball

'g' is acceleration due to gravity

't' is the time of motion

Now we know that the ball will continue to ascend until it's velocity becomes zero hence to using the above equation we can write

[tex]0=u-9.81\times 5\\\\\therefore u=9.81\times 5=49.05m/s\approx 50m/s[/tex]

Answer:[tex]0.0686 m^3[/tex]

Explanation:

Given

No of moles=2

[tex]T=25^{\circ}\approx 298 K[/tex]

[tex]P_1=210 KPa[/tex]

[tex]T_2=48 ^{\circ}\approx 321 k[/tex]

[tex]P_2=77.7 KPa[/tex]

PV=nRT

Substitute

[tex]77.7\times V_2=2\times 8.314\times 321[/tex]

[tex]V_2=0.0686 m^3[/tex]

Answer:

Depends on the specific relation of acceleration as function of time, but it would always look like decresing or increasing negatively.

Explanation:

Acceleration is the derivative of velocity with respect to time. That means that it is the change. The fact that is negative means that it is decreasing, or increasing in the negative direction (i.e. going backwards faster).

Attached is the graph with constant negative acceleration, using kinematics relations, it would be a linear equation with negative slope.

Answer:a straight line with negative slope

Explanation:

Answer:

[tex]F = 1361.1 N[/tex]

Explanation:

As we know that a charge is split into two parts

so we have

[tex]q_1 + q_2 = 7\mu C[/tex]

now we have

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we know

[tex]F = \frac{kq_1(7\mu C - q_1)}{r^2}[/tex]

here we have

r = 9 mm

now to obtain the maximum value of the force between two charges

[tex]\frac{dF}{dq_1} = 0[/tex]

so we have

[tex]7 \mu C - q_1 - q_1 = 0[/tex]

so we have

[tex]q_1 = q_2 = 3.5 \mu C[/tex]

so the maximum force is given as

[tex]F = \frac{(9\times 10^9)(3.5 \mu C)(3.5 \mu C)}{0.009^2}[/tex]

[tex]F = 1361.1 N[/tex]

Answer:

Its dissipated by the brake.

Explanation:

Traditional brakes use friction to stop the wheels (or axis). This friction its dissipated in the form of heat. There are others mechanism to brake that don't dissipated the energy, they stored it. In electric cars (or hybrids), there are regenerative brakes, that store the kinetic energy as electrical energy.

As the driver slams the brakes, the kinetic energy of the car gets converted into other forms of energy until the car comes to a stop, at which point it becomes zero.

When the driver of a car slams on her brakes to avoid a collision with a deer crossing the highway, the kinetic energy of the car decreases until the car comes to rest. This is due to the principle of energy conservation. Initially, when the car is moving, it has kinetic energy. However, when the brakes are applied, the kinetic energy gets converted into other forms of energy such as heat energy (due to friction between the brakes and the wheels) and potential energy (if the car is moving uphill). The kinetic energy keeps decreasing until the car comes to a complete stop. At that point, the kinetic energy of the car is zero because kinetic energy is associated with motion and the car is no longer moving.

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Answer:

a) 2034 J

b) -1471 J

c) -509 J

Explanation:

The trunk has a mass of 50 kg, so its weight is

f = m * a

f = 50 * 9.81 = 490 N

If the incline is of 30 degrees, the force tangential to the incline is:

ft = f * sin(a)

ft = 490 * sin(30) = 245 N

And the normal force is:

fn = f * cos(a)

fn = 490 * cos(30) = 424 N

The frictional force is:

ff  = μ * fn

ff = 0.2 * 424 = 84.8 N

To push the trunk up one must apply a force slightly greater than the opposing forces, the opposing are the tangential component of the weight and the friction force

fp = fn + ff

fp = 245 + 84.8 = 339 N

The work of the applied force is:

L = f * d

Lp = fp * d

Lp = 339 * 6 = 2034 J

The work of the weight is done by the tangential component:

Lw = -245 * 6 = -1471 J (it is negative because the weight was opposed to the direction of movement)

The work of the friction force is

Lf = -84.8 * 6 = -509 J

Answer:

18.86° , it will bend towards normal.

Explanation:

For refraction,

Using Snell's law as:

[tex]n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}[/tex]

Where,  

[tex]{\theta_i}[/tex]  is the angle of incidence  ( 25.0° )

[tex]{\theta_r}[/tex] is the angle of refraction  ( ? )

[tex]{n_r}[/tex] is the refractive index of the refraction medium  (n=1.7)

[tex]{n_i}[/tex] is the refractive index of the incidence medium ( n=1.3)

Hence,  

[tex]1.3\times {sin\ 25.0^0}={1.7}\times{sin\theta_r}[/tex]

Angle of refraction = [tex]sin^{-1}0.3232[/tex] = 18.86°

Since, the light ray is travelling from a material with a refractive index of 1.3 into a material with a refractive index of 1.7 or lighter to denser medium, it will bend towards normal.

The diagram is shown below: