Question #1: Consider Eratosthenes's experiment to measure the size of the Earth. Suppose the Earth were a smaller planet -- but the sun were still directly overhead in Syene at noon on the Summer Solstice, and it was still 500 miles from Syene to Alexandria. Would the shadow of the stick in Alexandria at noon on the Summer Solstice have been longer, shorter, or the same as it was on our Earth? Briefly explain your reasoning.

Answer:

time to fall is 3.914 seconds

Explanation:

given data

half distance time = 1.50 s

to find out

find the total time of its fall

solution

we consider here s is total distance

so equation of motion for distance

s = ut + 0.5 × at²   .........1

here s is distance and u is initial speed that is 0 and a is acceleration due to gravity = 9.8 and t is time

so for last 1.5 sec distance is 0.5 of its distance so equation will be

0.5 s = 0 + 0.5 × (9.8) × ( t - 1.5)²     ........................1

and

velocity will be

v = u + at

so velocity v = 0+ 9.8(t-1.5)    ..................2

so first we find time

0.5 × (9.8) × ( t - 1.5)² = 9.8(t-1.5)  + 0.5 ( 9.8)

solve and we get t

t = 3.37 s

so time to fall is 3.914 seconds

Answer:

Explanation:

Initial volume v₁ =46 x 10⁻⁶ m³

Initial temperature T₁ = 438.28 + 273 = 711.28 K

Initial pressure P₁ = nRT₁ / v₁

= 30.4 x8.3 x 711.28 / (46 x 10⁻⁶ )

= 3901.5 x 10⁶ Pa

Final temperature T₂ = 824 + 273 = 1097 K

Final volume V₂ =?

For isobaric process

v₁ / T₁ = V₂ / T₂

V₂ = V₁ X T₂ /T₁

= 46 X 10⁻⁶ X 1097/ 711.28

= 70.94 X 10⁻⁶ m³ = 70.94 cm³

b ) For isothermal change

P₁ V₁ = P₂V₂

P₂ = P₁V₁ / V₂

= 3901.5 X 10⁶ X 46 / 24.3

7385.55  X 10⁶ Pa.

Answer:

76 degree  south of west.

Explanation:

We shall represent velocities in vector form , considering east as  x axes and west as Y axes.

V₁ = 4.4 j

V₂ = 3.8, 30 degree  west of south

V₂ = -  3.8 sin 30 i - 3.8 cos 30 j

= - 1.9 i - 3.29 j

Change in velocity

= V₂ - V₁

= - 1.9 i - 3.29 j - 4.4 j

= - 1.9 i - 7.69 j

Acceleration

= change in velocity / time

(- 1.9 i - 7.69 j  ) / 60 ms⁻² .

Direction of acceleration θ

Tan θ = 7.69 / 1.9 =4.047

θ = 76 degree  south of west.

Answer:

18 miles east; 24 mph east

Explanation:

In order to find how far east of Wilmington is the ship after 1 hour, we just need to substitute t = 1 into the formula of the position.

The equation of the position is

[tex]s(t) = 12 t^2 +6[/tex]

where t is the time. Substituting t = 1,

[tex]s(1) = 12 (1)^2 + 6 = 12+6 = 18 mi[/tex]

So, the ship is 18 miles east of Wilmington.

To find the velocity of the boat, we just need to calculate the derivative of the position, so

[tex]v(t) = s'(t) = 24 t[/tex]

And by substituting t = 1, we find the velocity after 1 hour:

[tex]v(1) = 24 (1) = 24 mph[/tex]

And the direction is east.

Answer:

option (C)

Explanation:

Speed of car = 30 m/s

Let the time taken by the police car to catch the speeding car is t

The distance traveled by the speeding car in t + 1 second is equal to the distance traveled by the police car in time t

Distance traveled by the police car in time t

[tex]s=ut + 0.5 at^{2}[/tex]    .... (1)

Distance traveled by the speeding car in t + 1 second

s = 30 (t + 1) = 300

t + 1 = 10

t = 9 s

Put the value of t in equation (1), we get

300 = 0 + 0.5 x a x 9 x 9

a = 7.41 m/s^2

C. 7.41 meters per square second.

In this question, the car is travelling at constant speed, whereas the police car accelerates uniformly after some delay to catch the car, the respective kinematic formulas are shown below:

Car

[tex]x_{C} = x_{o} + v_{C}\cdot t[/tex] (1)

Police car

[tex]x_{P} = x_{o} + \frac{1}{2}\cdot a_{P}\cdot (t-t')^{2}[/tex] (2)

Where:

If we know that [tex]x_{o} = 0\,m[/tex], [tex]x_{C} = x_{P} = 300\,m[/tex], [tex]t' = 1\,s[/tex] and [tex]v_{C} = 30\,\frac{m}{s}[/tex], then we have the following system of equations:

[tex]300 = 30\cdot t[/tex] (1)

[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (t-1)^{2}[/tex] (2)

By (1):

[tex]t = 10[/tex]

Then we find that acceleration of the police car must be:

[tex]300 = \frac{1}{2}\cdot a_{P}\cdot (10-1)^{2}[/tex]

[tex]a_{P} = 7.407\,\frac{m}{s^{2}}[/tex]

Therefore, the correct choice is C.

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Answer:

carrier power is 7.8 kW

Explanation:

given data

power = 10 kW

modulation percentage = 75 %

to find out

carrier power

solution

we will use here power transmitted equation that is

power = [tex]carrier power * ( 1+  \frac{modulation}{2})[/tex]   .................1

put here value in equation 1  we get carrier power

10 = [tex]carrier power *(1+ \frac{0.75}{2})[/tex]

carrier power = 7.8

so carrier power is 7.8 kW

Answer:

[tex]T = 10.43 s[/tex]

Explanation:

During deceleration we know that the deceleration is proportional to the velocity

so we have

[tex]a = - kv[/tex]

here we know that

[tex]\frac{dv}{dt} = - kv[/tex]

so we have

[tex]\frac{dv}{v} = -k dt[/tex]

now integrate both sides

[tex]\int \frac{dv}{v} = -\int kdt[/tex]

[tex]ln(\frac{v}{v_o}) = - kt[/tex]

[tex]ln(\frac{40}{70}) = - k(t)[/tex]

[tex]kt = 0.56[/tex]

Also we know that

[tex]a = \frac{vdv}{ds}[/tex]

[tex]-kv = \frac{vdv}{ds}[/tex]

[tex]\int dv = -\int kds[/tex]

[tex](v - v_o) = -ks[/tex]

[tex](40 - 70)mph = - k (480 ft)[/tex]

[tex]-30 mph = -k(0.091 miles)[/tex]

[tex]k = 329.67[/tex]

so from above equation

[tex]t = \frac{0.56}{329.67} = 1.7 \times 10^{-3} h[/tex]

[tex]t = 6.11 s[/tex]

initially it starts from rest and uniformly accelerate to maximum speed of 70 mph and covers a distance of 220 ft

so we have

d = 220 ft = 67 m = 0.042 miles[/tex]

now we know that

[tex]d = \frac{v_f + v_i}{2} t[/tex]

[tex]0.042 = \frac{70 + 0}{2} t[/tex]

[tex]t = 4.32 s[/tex]

so total time of motion is given as

[tex]T = 4.32 + 6.11 = 10.43 s[/tex]

Answer:

18.96 m

Explanation:

Height from person jump, h = 10 m

Let it takes time t to reach to the car.

Use second equation of motion

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

Here, a  g = 10 m/s^2 , u = 0, h = 10 m  

By substituting the values, we get

10 = 0 + 0.5 x 10 x t^2

t = 1.414 s

The speed of car, v = 30 mi/h = 13.41 m/s

Distance traveled by the car in time t , d = v x t = 13.41 x 1.414 = 18.96 m

So, the distance of car from the bridge is 18.96 m as the man jumps.

Answer:

 % of water boils away= 12.64 %

Explanation:

given,

volume of block  = 50 cm³ removed from temperature of furnace = 800°C

mass of water = 200 mL = 200 g

temperature of water  = 20° C

the density of iron = 7.874 g/cm³ ,

so the mass of iron(m₁)  = density × volume = 7.874 × 50 g = 393.7 g

the specific heat of iron C₁ = 0.450 J/g⁰C

the specific heat of water Cw= 4.18 J/g⁰C

latent heat of vaporization of water is L_v = 2260 k J/kg = 2260 J/g

loss of heat from iron is equal to the gain of heat for the water

[tex]m_1\times C_1\times \Delta T = M\times C_w\times \Delta T + m_2\times L_v[/tex]

[tex]393.7\times 0.45\times (800-100) = 200\times 4.18\times(100-20) + m_2\times 2260[/tex]

m₂ = 25.28 g

25.28 water will be vaporized

% of water boils away =[tex]\dfrac{25.28}{200}\times 100[/tex]

 % of water boils away= 12.64 %

The percentage of the water boils away when the iron block is placed into the water after furnace is 1264%.

The heat transfer is the transfer of thermal energy due to the temperature difference.The heat flows from the higher temperature to the lower temperature.

The heat transfer of a closed system is the addition of change in internal energy and the total amount of work done by it.

As the initial volume of the iron block is 50 cm³ and the density of the iron is 7.874 g/cm³. Thus the mass of the iron block is,

[tex]m=50\times7.874\\m=393.7\rm g[/tex]

The temperature of the furnace is 800 degrees Celsius  and the specific heat of the iron block is 0.45 J/g-C.

As the boiling point of the water is 100 degree Celsius. Thus the heat loss by the block of iron is,

[tex]Q_L=393.7\times0.45\times(800-100)\\Q_L=124015.5[/tex]

The latent heat of the water is 2260 J/g. Thus the heat gain by vaporized water is,

[tex]Q_v=2260\times m_v\\[/tex]

Now the heat gain by the water is equal to the heat loss by the iron block.

As the specific heat of the water is 4.18 J/g-C and the temperature of the  water is 20 degrees and volume of water is 200 ml.

Thus heat gain by water can be given as,

[tex]Q_G=Q_L=200\times4.18(100-20)+2260m_v\\124015.5=200\times4.18(100-20)+2260m_w\\m_v=25.28\rm g[/tex]

Thus the total amount of the water boils away is 25.28 grams.

The percentage of the water boils away is,

[tex]p=\dfrac{25.25}{200}\times100\\p=12.64[/tex]

Thus the percentage of the water boils away when the iron block is placed into the water after furnace is 1264%.

Learn more about the heat transfer here;

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Answer:1.26 m/s

Explanation:

Given

translation speed of ball =3.5 m/s

Moment of inertia of ball about com [tex]I=\frac{2}{5}mr^2[/tex]

Initial Energy

[tex]E_i=\frac{1}{2}mu^2+\frac{1}{2}I\omega _i^2(\omega =\frac{u}{r})[/tex]

Final Energy

[tex]E_f=\frac{1}{2}mv^2+\frac{1}{2}I\omega _f^2+mgh[/tex]

Equating energy as no energy loss take place

[tex]E_i=E_f[/tex]

[tex]\frac{1}{2}mu^2+\frac{1}{2}I\omega _i^2=\frac{1}{2}mv^2+\frac{1}{2}I\omega _f^2+mgh[/tex]

[tex]\frac{1}{2}mu^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \left ( \frac{u}{r}\right )^2=\frac{1}{2}mv^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \left ( \frac{v}{r}\right )^2+mgh[/tex]

m term get cancel

[tex]\left ( \frac{u^2}{2}\right )+\left ( \frac{2u^2}{10}\right )=\left ( \frac{v^2}{2}\right )+\left ( \frac{2v^2}{10}\right )+gh[/tex]

[tex]\frac{7}{10}u^2=\frac{7}{10}v^2+gh[/tex]

[tex]v^2=3.5^2-\frac{10}{7}\times 9.81\times 0.76[/tex]

[tex]v=\sqrt{1.6}=1.26 m/s[/tex]

Answer:

[tex]y=-4.9x10^{-16}m[/tex]

Explanation:

From the exercise we have initial velocity on the x-axis, the final x distance and acceleration of gravity.

[tex]v_{ox}=4.3x10^{7}m/s[/tex]

[tex]x=43cm=0.43m\\g=9.8m/s^{2}[/tex]

From the equation on moving particles we can find how long does it take the electron beam to strike the screen

[tex]x=x_{o}+v_{ox}t+\frac{1}{2}at^{2}[/tex]

Since [tex]x_{o}=0[/tex] and [tex]a_{x}=0[/tex]

[tex]0.43m=(4.3x10^{7}m/s)t[/tex]

Solving for t

[tex]t=1x10^{-8} s[/tex]

Now, from the equation of free-falling objects we can find how far does the electron beam fell

[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]

[tex]y=-\frac{1}{2}(9.8m/s^{2})(1x10^{-8} s)=-4.9x10^{-16}m[/tex]

The negative sign means that the electron beam fell from its initial point.

Answer:

- 3.46 m

Explanation:

initial position, xi = 3.37 m

displacement, Δx = - 6.83 m

Let the final position is xf.

So, displacement = final position - initial position

Δx = xf - xi

- 6.83 = xf - 3.37

xf = 3.37- 6.83

xf = - 3.46 m

Thus, the final position of the squirrel is - 3.46 m.

Answer:

h = 0.204 m

Explanation:

given data:

radius r = 0.131 m

mass m = 9.86 kg

density of copper = 8960 kg/m3

we knwo that density is given as

[tex]\rho = \frac{mass}{volume}[/tex]

[tex]volume = \pi * r^2 h[/tex]

[tex]density  =  \frac{ mass}{\pi * r^2 h}[/tex]

[tex]h = \frac{ mass}{\pi * r^2 * density}[/tex]

putting all value to get thickness value

[tex]h = \frac{ 98.6}{ \pi 0.131^2*8960}[/tex]

h = 0.204 m

Answer:

567.126 x 10⁻⁶ N

[tex]5.6 x 10^{-6} N[/tex]

Explanation:

Thinking process:

[tex]13pC = 13 x 10^{-12} C[/tex]

The electric field is given by E = 15000 N/C

Electric force on the charge

= charge x electric field

= 13 x 10⁻¹² x 15000

= 195 x 10⁻⁹ N.

The force acts in upward direction as force on positive charge acts in the direction in which electric field exists.

Volume of droplet = 4/3 π R³

R = 2.4 X 10⁻³ m

Volume V = 4/3 x 3.14 x ( 2.4 x 10⁻³)³

= 57.87 x 10⁻⁹ m³

density of water = 1000 kg / m³

mass of water droplet = density x volume

                                     = 1000 x 57. 87 x 10⁻⁹ kg

                                     = 57.87 x 10⁻⁶ kg .

                        Weight = mass x g

                                    = 57.87 x 10⁻⁶ x 9.8

                                     = 567.126 x 10⁻⁶ N.

Therefore, the weight is more than the electric force.

Answer:t=0.54 s

Explanation:

Given

Jacob is traveling 5 m/s in North direction

Jacob throw a ball with a in south direction with a velocity of 5 m/s

Ball is thrown in opposite direction of motion of car therefore it seems as if it is dropped from car as its net horizontal velocity is 5-5=0

Time taken by ball to reach ground

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

[tex]1.45=0+\frac{9.81\times t^2}{2}[/tex]

[tex]t^2=frac{2\times 1.45}{9.81}[/tex]

t=0.54 s

Motion of ball will be straight line

Answer:

a) 0.31 s

b) 19.77 m

Explanation:

We will need the following two formulas:

[tex]V_{f} = V_{0}+at\\\\X=V_{0}t + \frac{at^{2}}{2}[/tex]

We first use the final velocity formula to find the time that it takes to decelerate the paratrooper:

[tex]4.95\frac{m}{s}=27.6\frac{m}{s}-74\frac{m}{s^{2}}t\\\\-22.65\frac{m}{s}=-74\frac{m}{s^{2}}t\\\\t= \frac{22.65\frac{m}{s}}{74\frac{m}{s^{2}}}=0.31s[/tex]

Now that we have the time, we can use the distance formula to calculate the distance travelled by the paratrooper:

[tex]X=27.6\frac{m}{s}*0.31s - \frac{74\frac{m}{s^{2}}*(0.31 s)^{2}}{2}=19.77 m[/tex]

Answer:

1.93 m

Explanation:

Initial velocity of fish u = 11 m /s

Angle of jump = 34 degree.

Vertical component of its velocity = u sin34 = 11 x.5592 = 6.15 m /s

Considering its motion in vertical direction ,

u = 6.15 m/s

g = - 9.8 m /s

maximum height attained = h

From the formula

v² = u² - 2gh

0 = 6.15x 6.15 - 2 x 9.8 h

h = (6.15 x 6.15 )/ 2 x 9.8

= 1.93 m .

Answer:

Explanation:

When saw slices wood by exerting a force on the wood , wood also exerts a reaction force on the saw in opposite direction which is equal to the force of action that is 104 N.

So torque exerted by wood on the blade

= force x perpendicular distance from the axis of rotation

= 104 x .128

=13.312 Nm.

Since this torque opposes the movement of blade , it turns the blade slower.

Answer:

120 m/s^2

(a) 6000 m

(b) 1200 m/s

Explanation:

mass, m = 10 kg

initial velocity, u = 0

Force, F = 1200 N

time, t = 10 s

Let a be the acceleration of the block.

By use of Newton,s second law

Force = mass x acceleration

1200 = 10 x a

a = 120 m/s^2

(a) Let the block travels by a distance s.

Use second equation of motion

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

s = 0 + 0.5 x 120  x 10 x 10

s = 6000 m

(b) Let v be the final velocity of the block

Use first equation of motion

v = u + at

v = 0 + 120 x 10

v = 1200 m/s

Answer:

We should only agree with the first statement of the first student but not with the second part of his statement. While as the statement of the other student is completely wrong.

Explanation:

The earth and the moon are locked in a process known as tidal locking. Tidal locking occurs when any object which revolves around other object takes the same amount of time to rotate around it's own axis as it takes to revolve around the planet.

This is exactly the case with moon and the earth system ,the moon is tidally locked with earth thus we cannot see the other side of the moon but this side is not dark as claimed by the first student but this side of the moon is also illuminated by the sunlight as the face of moon that we are able to see.

The second student is wrong as we cannot see the other side of moon from earth.

Final answer:

The Moon is in synchronous rotation with Earth, which means the same side, the near side, always faces Earth. The far side, incorrectly called the 'dark side,' is not perpetually dark but simply not visible from Earth and receives sunlight just like the near side.

Explanation:

The student who claimed that we always see only one side of the Moon is correct; however, their assertion that the other side is the "dark side" and always dark is incorrect. This is because the Moon is in synchronous rotation with Earth—it rotates on its axis in the same amount of time it takes to orbit Earth. Because of this, only one hemisphere, the near side, is visible from Earth, while the other hemisphere, the far side, remains out of view.

The idea that the far side is the "dark side" is a misconception; it receives sunlight just as the near side does, only at different times during the Moon's orbit. The term "dark side" only means that it's the side not visible from Earth. As the Moon revolves around Earth, different parts experience day and night, similar to how the Sun rises and sets on Earth. Therefore, the "dark side" is not perpetually dark; it simply refers to the lunar face we do not see that is also subject to changing illumination by the Sun.

In summary, the correct understanding is that the same side of the Moon always faces Earth, not because it doesn't rotate, but because it has a synchronous orbit with Earth, and the use of the term "dark side" is a misnomer since all parts of the Moon experience both sunlight and darkness at different times.

Answer:

a) Traveling at 45 mph, the driving time is 58 h. Traveling at 72 mph, the driving time will be 36 h.

b) Traveling at 45 mph, the gasoline cost will be $225.7.

Traveling at 72 mpg, the gasoline cost will be $417.9

Explanation:

The average speed can be calculated as the distance traveled over time:

speed = distance / time

Then:

time = distance / speed

a)If you drive at an average speed of 45 mph during a 2600-mile trip, the driving time will be:

time = 2600 mi / 45 mi/h = 58 h

If you drive at 72 mph:

time = 2600 mi / 72 mi/h = 36 h  

b) For the 2600-mile trip, you will need ( 2600 mi * (1 gallon/ 35 mi)) 74 gallons if you travel at 45 mph.

If you travel at 72 mph, you will need (2600 mi * (1 gallon /19 mi)) 137 gallons.  

Traveling at 45 mph, the gasoline cost will be (74 gallons * ($3,05/gallon)) $225.7

Traveling at 72 mph, the gasoline cost will be (137 gallons * (3.05/gallon)) $417.9

Answer:

a) You should position the cannon at 981 m from the wall.

b) You could position the cannon either at 975 m or 7.8 m (not recomended).

Explanation:

Please see the attached figure for a graphical description of the problem.

In a parabolic motion, the position of the flying object is given by the vector position:

r =( x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)

where:

r = position vector

x0 = initial horizontal position

v0 = module of the initial velocity vector

α = angle of lanching

y0 = initial vertical position

t = time

g = gravity acceleration (-9.8 m/s²)

The vector "r" can be expressed as a sum of vectors:

r = rx + ry

where

rx = ( x0 + v0 t cos α ; 0)

ry = (0 ; y0 + v0 t sin α + 1/2 g t²)

rx and ry are the x-component and the y-component of "r" respectively (see figure).

a) We have to find the module of r1 in the figure. Note that the y-component of r1 is null.

r1 = ( x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)

Knowing the the y-component is 0, we can obtain the time of flight of the cannon ball.

0 = y0 + v0 t sin α + 1/2 g t²

If the origin of the reference system is located where the cannon is, the y0 and x0 = 0.

0 = v0 t sin α + 1/2 g t²

0 = t (v0 sin α + 1/2 g t)         (we discard the solution t = 0)    

0 = v0 sin α + 1/2 g t

t = -2v0 sin α / g

t = -2 * 100 m/s * sin 53° / (-9.8 m/s²) = 16.3 s  

Now, we can obtain the x-component of r1 and its module will be the distance from the wall at which the cannon sould be placed:

x = x0 + v0 t cos α

x = 0 m + 100m/s * 16.3 s * cos 53

x = 981 m

The vector r1 can be written as:

r1 = (981 m ; 0)

The module of r1 will be: [tex]x = \sqrt{(981 m)^{2} + (0 m)^{2}}[/tex]

Then, the cannon should be placed 981 m from the wall.

b) The procedure is the same as in point a) only that now the y-component of the vector r2 ( see figure) is not null:

r2y = (0 ; y0 + v0 t sin α + 1/2 g t² )

The module of this vector is 10 m, then, we can obtain the time and with that time we can calculate at which distance the cannon should be placed as in point a).

module of r2y = 10 m

10 m = v0 t sin α + 1/2 g t²

0 = 1/2 g t² + v0 t sin α - 10 m

Let´s replace with the data:

0 = 1/2 (-9.8 m/s² ) t² + 100 m/s * sin 53 * t - 10 m

0= -4.9 m/s² * t² + 79.9 m/s * t - 10 m

Solving the quadratic equation we obtain two values of "t"

t = 0.13 s and t = 16.2 s

Now, we can calculate the module of the vector r2x at each time:

r2x = ( x0 + v0 t cos α ; 0)

r2x = (0 m + 100m/s * 16.2 s * cos 53 ; 0)

r2x = (975 m; 0)

Module of r2x = 975 m

at t = 0.13 s

r2x = ( 0 m + 100m/s * 0.13 s * cos 53 ; 0)

r2x = (7.8 m ; 0)

module r2x = 7.8 m

You can place the cannon either at 975 m or at 7.8 m (see the red trajectory in the figure) although it could be dangerous to place it too close to the enemy fortress!

Answer:

a) d=65.7 m : The automobile overtake passes the truck at a distance of 65.7 m  from the traffic signal

b) vₐ=16.94m/s : Automobile  speed for t = 7.7s

Explanation:

When the car catches the truck, the two will have passed the same time (t) and will have traveled the same distance (d):

Automobile kinematics:

car moves with uniformly accelerated movement:

d = v₀*t + (1/2)a*t²

v₀ = 0 : initial speed  

d = (1/2)*a*t² Equation (1)

Truck kinematics:

Truck moves with constant speed:

d = v*t Equation (2)

Data:

a=2.20 m/s² : automobile acceleration

v= 8.50 m/s  ; truck speed

Equation (1) =Equation (2)

(1/2)*a*t²= v*t

(1/2)*2.2*t²= 8.5*t   We divide both sides of the equation by t:

1.1*t=8.5

t=8.5÷1.1

t=7.7 s

We replace t=7.7 s in the equation (2)

d=8.5*7.7

d=65.7 m

Automobile  speed for t = 7.7s

vₐ=v₀₊a*t

vₐ=o₊2,2*7.7

vₐ=16.94m/s

Answer:

The car stops after 32.58 m.

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 20 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6 m/s²

Time taken by the car to stop

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20}{-6}\\\Rightarrow t=3.33\ s[/tex]

Total Time taken by the car to stop is 0.5+3.33 = 3.83 s

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=20\times 3.83+\frac{1}{2}\times -6\times 3.83^2\\\Rightarrow s=32.58\ m[/tex]

The car stops after 32.58 m.

Distance between car and obstacle is 50-32.58 = 17.42 m

Answer:

t = 2.96 s

Explanation:

Since the two stones hit the water at same instant of time

so we will have

[tex]d =vt + \frac{1}{2}gt^2[/tex]

here we know that

d = 47 m

v = 1.4 m/s

[tex]g = 9.81 m/s^2[/tex]

[tex]d = 1.40 t + \frac{1}{2}(9.81) t^2[/tex]

now by solving above equation for  d = 47 m

t = 2.96 s

Answer:1 s

Explanation:

Given

Initial velocity in Y-direction [tex]v_1[/tex]=+4 m/s

Final Velocity in Y-direction [tex]v_2=-5.8 m/s[/tex]

Acceleration in Y-direction is 9.81 [tex]m/s^2[/tex]

Using equation of motion

v=u+at

[tex]-5.8=4+9.81\times t[/tex]

[tex]t=0.998 \approx 1 s[/tex]

The magnitude of the velocity with which the ball hits the crater floor is approximately 401.24 m/s.

Let's convert the height to meters:

Height = 23 km

= 23,000 m

Now we can use the kinematic equation to find the time it takes for the ball to reach the ground:

[tex]h=\frac{1}{2}gt^2[/tex]

Where: h = height (23,000 m)

g = acceleration due to gravity (3.5 m/s²)  

t = time

Solving for t:

[tex]t=\sqrt{\frac{2h}{g}}[/tex]

Plug in the values:

[tex]t=\sqrt{\frac{2 \times 23000}{3.5}}[/tex]

t=114.64 s

So, the time for the ball to reach the crater floor is approximately 114.64 seconds.

Now calculate the magnitude of the velocity with which the ball hits the crater floor.

We can use the following kinematic equation:

v=gt

Where:

v = final velocity

g = acceleration due to gravity (3.5 m/s²)

t = time (114.64 s)

v=3.5×114.64

v=401.24 m/s

Hence, the magnitude of the velocity with which the ball hits the crater floor is approximately 401.24 m/s.

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The time it would take for the ball to hit the crater floor would be approximately 360.8 seconds, and the velocity it would hit the floor with would be approximately 1262.8 m/s.

To calculate the time it takes for the ball to reach the crater floor, we use the equation of motion d = 1/2gt^2, where 'd' represents the distance, 'g' represents the acceleration due to gravity, and 't' represents time. Here, the distance is 23,000m (converted from km to m), and the acceleration due to gravity is 3.5 m/s^2. By solving for 't' in this equation, we get t = sqrt(2d/g) = sqrt((2*23000)/3.5) = approx. 360.8s.

To calculate the magnitude of the final velocity as the ball hits the crater floor, we use the equation v = gt, where 'v' is the final velocity and 't' is the amount of time it has fallen. Plugging in the values we have for 'g' and 't', we get v = 3.5 * 360.8s = approx. 1262.8 m/s

Therefore, the time it would take for the ball to hit the crater floor would be approximately 360.8 seconds, and the velocity it would hit the floor with would be approximately 1262.8 m/s.

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Using Ohm's law and properties of series and parallel circuits, it's possible to find the resistances. In series, resistances are directly added and for parallel, the reciprocal of total resistance is the sum of reciprocals of individual resistances. Applying these principles with given current and voltage, one can solve for resistances.

This question pertains to electricity and specifically the characteristics of resistors when they are connected in series or parallel. Using Ohm's Law, we know that the voltage (V) is the product of the current (I) and the resistance (R). Therefore, when the resistors are connected in series, the combined resistance (Rtotal) is the sum of the individual resistances, while the current remains the same. This gives us Rtotal = V/I = 12.0V / 1.12A.

When in parallel, however, the total resistance can be found differently. In a parallel circuit, the total resistance is given by 1/Rtotal = 1/R+ 1/R. As per the problem, we know that the total current of the circuit connected in parallel is 9.39 A, so we can use the equation Itotal = V/ Rtotal to find the total resistance in the parallel circuit.

Combining the information from both the circuits would allow us to solve two simultaneous equations to get the values of R and R.

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The two resistances connected in series have a total resistance of 10.71 Ω, and connected in parallel have a total resistance of 1.28 Ω. Solving the equations, we find the resistances to be 1.43 Ω and 9.28 Ω.

Let's solve this step-by-step:

Step 1: Series Connection

When two resistors are connected in series, the total resistance, [tex]R_{total[/tex], is the sum of the resistances:

[tex]R_{total[/tex] = R₁ + R₂

Using Ohm's Law (V = IR), we can find the total resistance in series:

V = 12.0 V;

I = 1.12 A;

[tex]R_{total[/tex] = V/I

[tex]R_{total[/tex] = 12.0 V / 1.12 A = 10.71 Ω

Step 2: Parallel Connection

When the same resistors are connected in parallel, the total resistance, [tex]R_{parallel[/tex], can be found using the formula:

1/[tex]R_{parallel[/tex] = 1/R₁ + 1/R₂

Again, using Ohm's Law, we first find [tex]R_{parallel[/tex]:

V = 12.0 V; [tex]I_{total[/tex] = 9.39 A;

[tex]R_{parallel[/tex] = V/[tex]I_{total[/tex] = 12.0 V / 9.39 A = 1.28 Ω

Step 3: Solving the Equations

We now have two equations:

R₁ + R₂ = 10.71 Ω

(R₁ * R₂) / (R₁ + R₂) = 1.28 Ω (since 1/[tex]R_{parallel[/tex] = 1/R₁ + 1/R₂ )

Let's solve these equations:

Substitute R₂ = 10.71 - R₁ into the parallel equation:

(R₁ * (10.71 - R₁)) / 10.71 = 1.28

R₁ * (10.71 - R₁) = 1.28 * 10.71

10.71R₁ - R₁² = 13.69

R₁² - 10.71R₁ + 13.69 = 0

Solving the quadratic equation using the quadratic formula:

R₁ = [10.71 ± √((10.71)² - 4*1*13.69)] / 2

Solving this, we get R₁ ≈ 1.43 Ω or R₁ ≈ 9.28 Ω

Then, R₂ = 10.71 - 1.43 = 9.28 Ω ,

R₂ = 10.71 - 9.28 = 1.43 Ω

Answer:

[tex]a=2.36\ m/s^2[/tex]

T=157.06 N

Explanation:

Given that

Mass of first block = 21.1 kg

Mass of second block = 12.9 kg

First mass is heavier than first that is why mass second first will go downward and mass second will go upward.

Given that pulley and string is mass  less that is why both mass will have same acceleration.So lets take their acceleration is 'a'.

So now from force equation

[tex]m_1g-m_2g=(m_1+m_2)a[/tex]

21.1 x 9.81 - 12.9 x 9.81 =(21.1+12.9) a

[tex]a=2.36\ m/s^2[/tex]

Lets tension in string is T

[tex]m_1g-T=m_1a[/tex]

[tex]T=m_1(g-a)[/tex]

T=21.1(9.81-2.36) N

T=157.06 N

The magnitude of their acceleration is 2.36m/s² while the tension in the rope is 157.06N.

From the information given, the following can be depicted:

The force equation can be used to calculate the magnitude of the acceleration which will go thus:

(21.1 × 9.81) - (12.9 × 9.81) = (21.1 + 12.9)a

(8.2 × 9.81) = 34a

a = (8.2 × 9.81) / 34

a = 2.36m/s².

Therefore, the tension will be:

T = m(g - a)

T = 21.1(9.81 - 2.36)N

T = 157.06N

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The speed of the ball when passing its original height on its way down is 20 m/s. The time it takes to reach the ground level is 3.19 seconds. The ball travels a total distance of 70 meters.

The motion of the tennis ball can be analyzed by using the principles of Physics, particularly the laws of motion and the concept of gravity.

(a) The ball will have the same speed when it passes its original height on its way down, which is 20 m/s. This is based on the principle of conservation of energy. Since air resistance is ignored, the speed when it passes the original height on its way down should be the same as its initial speed.

(b) Calculating the total time it takes the ball to reach the ground involves the formula for time in free fall, which is t = sqrt(2h/g), wherein h is the height and g is the gravity. Thus plug in the values: t = sqrt((2*50)/9.8) = 3.19 seconds.

(c) The total distance traveled by the ball includes it going up and falling back down. The ball rises to a height of 20 m/s before falling from that height plus an additional height corresponding to the height of the cliff, total distance is 70 meters.

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