A ball is thrown straight up and reaches a maximum height of 36 m above the point from which it was thrown. With what speed was the ball thrown?

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:

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:

[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 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:

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:

[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:

a) 0.568 kg

b) 474 kg/m³

Explanation:

Given:

Inner radius = 8.82 cm = 0.0882 m

Outer radius = 9.91 cm = 0.0991 m

Density of the liquid = 948.00 Kg/m³

a) The volume of the sphere = [tex]\frac{4\pi}{3}\times(0.0991^2-0.0882^2)[/tex]

or

volume of sphere = 0.0012 m³

also, volume of half sphere = [tex]\frac{\textup{Total volume}}{\textup{2}}[/tex]

or

volume of half sphere = [tex]\frac{\textup{0.0012}}{\textup{2}}[/tex]

or

Volume of half sphere =0.0006 m³

Now, from the Archimedes principle

Mass of the sphere = Weight of the volume of object submerged

or

Mass of the sphere = 0.0006× 948.00 = 0.568 kg

b) Now, density =  [tex]\frac{\textup{Mass}}{\textup{Volume}}[/tex]

or

Density = [tex]\frac{\textup{0.568}}{\textup{0.0012}}[/tex]

or

Density = 474 kg/m³

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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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:

x=4.06m

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem

Vf=7.6m/s

t=1.07

Vo=0

we can use the ecuation number one to find the acceleration

a=(Vf-Vo)/t

a=(7.6-0)/1.07=7.1m/s^2

then we can use the ecuation number 2 to find the distance

{Vf^{2}-Vo^2}/{2.a} =X

(7.6^2-0^2)/(2x7.1)=4.06m

Answer:

Inverted

Real

Explanation:

u = Object distance =  30 cm

v = Image distance

f = Focal length = 10 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}{10}-\frac{1}{30}\\\Rightarrow \frac{1}{v}=\frac{1}{15}\\\Rightarrow v=15\ cm[/tex]

As, the image distance is positive the image is real and forms on the other side of the lens

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-15}{30}\\\Rightarrow m=-0.5[/tex]

As, the magnification is negative the image is inverted

By applying the lens equation, we calculate that the image is formed 15 cm behind the lens. This is a real image as it forms on the opposite side of the lens, and in the case of a converging lens, it will be inverted.

First, we use the lens equation, which is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. In this case, the object distance 'do' is 30cm and the focal length 'f' is 10cm. Solving for 'di', we find that the image is located 15 cm behind the lens (i.e., on the opposite side from the object).

Since the image forms on the opposite side of the lens from where the object is, this indicates it's a real image. A positive image distance indicates a real image and a negative image distance indicates a virtual image.

For a converging lens, a real image is always inverted, and a virtual image is always upright. Therefore, in this case, the image would be inverted.

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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:

The acceleration and the distance are 25200 mi/h² and 0.1008 mi.

Explanation:

Given that,

Initial speed = 71 mi/h

Final speed = 50 mi/h

Time = 3.0 s

(a). We need to calculate the acceleration

Using equation of motion

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

[tex]a=\dfrac{v-u}{t}[/tex]

Put the value in the equation

[tex]a=\dfrac{(50-71)\times3600}{3}[/tex]

[tex]a=-25200\ mi/h^2[/tex]

Negative sign shows the deceleration.

(b). We need to calculate the distance

Using equation of motion

[tex]v^2=u^2+2as[/tex]

[tex](50)^2=(71)^2+2\times(-25200)\times s[/tex]

[tex]s=\dfrac{(50)^2-(71)^2}{-25200}[/tex]

[tex]s=0.1008\ mi[/tex]

Hence, The acceleration and the distance are 25200 mi/h² and 0.1008 mi.

The distance traveled by the car when the car is constantly deaccelerating at a rate of 25200 miles/h² is 0.0504 miles.

Given to us

Initial Velocity of the car, u = 71 miles/h

Final Velocity of the car, v = 50 miles/h

Time = 3.0 s  [tex]=\dfrac{3}{3600}[/tex] hour

According to the first equation of motion, acceleration can be written as,

[tex]a=\dfrac{v-u}{t}[/tex]

substituting the values we get,

[tex]a=\dfrac{50-71}{\dfrac{3}{3600}}[/tex]

[tex]a=-25,200\rm\ miles/h^2[/tex]

Thus, the acceleration of the car is -25,200 miles/h².

According to the third equation of motion,

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

Substituting the values we get,

[tex](50)^2-(71)^2=2(-25200)s[/tex]

[tex]s = 0.0504 \rm\ miles[/tex]

Thus, the distance car travel during the braking period is 0.0504 miles.

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Answer:

[tex]q=+8.34*10^{-7}C}[/tex]

Explanation:

The potential V due to a charge q,  at a distance r, is:

[tex]V=k\frac{q}{r}[/tex]

k=8.99×109 N·m^2/C^2      :Coulomb constant

We replace the values in order to find q:

[tex]q=\frac{V*r}{k}=\frac{500*15}{8.99*10^{9}}=8.34*10^{-7}C[/tex]

Answer:

i apolagize im late but yeah bois 700 points

Explanation:

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:

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:

3.06 seconds time passes before the watermelon has the same velocity

watermelon going at speed 59.9 m/s

watermelon traveling when it hits the ground at speed is 79.19 m/s

Explanation:

given data

height = 320 m

speed = 30 m/s

to find out

How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling

solution

we will use here equation of motion that is

v = u + at    ....................1

here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²

put the value and find time t

30 = 0 + 9.8 (t)

t = 3.06 s

so 3.06 seconds time passes before the watermelon has the same velocity

and

we know superman cover distance is = velocity × time

so distance = 30 × t

and distance formula for watermelon is

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

here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t

30 × t = 0 + 0.5×9.8×t²

t = 6.12 s

so  by equation 1

v = u + at

v = 0 + 9.8 ( 6.12)

v = 59.9 m/s

so watermelon going at speed 59.9 m/s

and

watermelon traveling speed formula is by equation of motion

v² - u² = 2as      ......................3

here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m

v² - 0 = 2(9.8) 320

v = 79.19 m/s

so watermelon traveling when it hits the ground at speed is 79.19 m/s

Answer:

a) 520m

b) 10.30 s

c) 100,95 m/s

Explanation:

a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.

This means the maximum height this rocket reached before falling  was 520 m.

b) As we are dealing with constant acceleration (due gravity) [tex]g=9.8 \frac{m}{s^{2}}[/tex] we can use the following formula:

[tex]y=y_{o}+V_{o} t-\frac{gt^{2}}{2}[/tex]   (1)

Where:

[tex]y_{o}=520 m[/tex]  is the initial height of the rocket (at the exact moment in which it stops due engines fail)

[tex]y=0[/tex]  is the final height of the rocket (when it finally hits the launch pad)

[tex]V_{o}=0[/tex] is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)

[tex]g=9.8m/s^{2}[/tex]  is the acceleration due gravity

[tex]t[/tex] is the time it takes to the rocket to hit the launch pad

Clearing [tex]t[/tex]:

[tex]0=520 m+0-\frac{9.8m/s^{2} t^{2}}{2}[/tex]   (2)

[tex]t^{2}=\frac{-520 m}{-4.9 m/s^{2}}[/tex]   (3)

[tex]t=\sqrt{106.12 s^{2}[/tex]   (4)

[tex]t=10.30 s[/tex]   (5)  This is the time

c) Now we need to find the final velocity [tex]V_{f}[/tex] for this rocket, and the following equation will be perfect to find it:

[tex]V_{f}=V_{o}-gt[/tex]  (6)

[tex]V_{f}=0-(9.8 m/s^{2})(10.30 s)[/tex]  (7)

[tex]V_{f}=-100.95 m/s[/tex]  (8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)

Answer:

(a) - 165.032 m/s

(b) 238.37 m/s

Explanation:

initial horizontal velocity, ux = 172 m/s

height, h = 1390 m

g = 9.8 m/s^2

Let it strikes the ground after time t.

Use second equation of motion in vertical direction

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

-1390 = 0 - 0.5 x 9.8 x t^2

t = 16.84 second

(a) Let vy be the vertical component of velocity as it strikes the ground

Use first equation of motion in vertical direction

vy = uy - gt

vy = 0 - 9.8 x 16.84

vy = - 165.032 m/s

Thus, the vertical component of velocity as it strikes the ground is 165.032 m/s downward direction.

(b)

The horizontal component of velocity remains constant throughout the motion.

vx = 172 m/s

vy = - 165.032 m/s

The resultant velocity is v.

[tex]v=\sqrt{172^{2}+165.032^{2}}[/tex]

v = 238.37 m/s

Thus, teh velocity with which it hits the ground is 238.37 m/s.

Answer:

Part (a) 10.15 m

Part (b) Rising

Explanation:

Given,

part (a)

Let 't' be the time taken to reach the ball to the cross bar,

In x-direction,

[tex]\therefore r\ =\ u_xt\\\Rightarrow t\ =\ \dfrac{r}{u_x}\ =\ \dfrac{r}{ucos\theta}\\\Rightarrow t\ =\ \dfrac{36.0}{23.6cos45^o}\\\Rightarrow t\ =\ 2.15\ sec[/tex]

Let y be the height attained by the ball at time t = 2.15 sec,

[tex]y\ =\ u_yt\ \ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\ usin\theta t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow y\ =\  23.6\times sin45^o\times 2.15\ -\ 0.5\times 9.81\ 2.15^2\\\Rightarrow y\ =\ 13.205\ m[/tex]

Now Let H be the height by which the ball is clear the crossbar.

[tex]\therefore H\ =\ y\ -\ h\ =\ 13.205\ -\ 3.05\ =\ 10.15\ m[/tex]

part (b)

At the maximum height the vertical velocity of the ball becomes zero.

i,e, [tex]v_y\ =\ 0[/tex]

Let h be the maximum height attained by the ball.

[tex]\therefore v_y^2\ =\ u_y^2\ -\ 2gh\\\Rightarrow 0\ =\ (usin\theta)^2\ -\ 2gh\\\Rightarrow h\ =\ \dfrac{(usin\theta)^2}{2g}\\\Rightarrow h\ =\ \dfrac{23.6\times sin45.0^o)^2}{2\times 9.81}\\\Rightarrow h\ =\ 14.19\ m[/tex]

Hence at the cross bar the ball attains the height 13.205 m but the maximum height is 14.19 m. Therefore the ball is rising when it reaches at the crossbar.

Along the three Cartesian coordinates. The Miller indices of the plane are [tex](1/2,1/4,1/3)[/tex].

A system known as Miller indices is used to explain how crystal planes and directions inside a crystalline substance are oriented. They serve as a tool to depict the crystal lattice's three-dimensional configuration of atoms or lattice points. In order to describe the surfaces and axes of crystals, Miller indices are frequently used in crystallography.

Miller indices are a crucial tool that crystallographers use to convey crystallographic data and comprehend the geometric arrangement of atoms in crystals. They support the characterization of crystal structures, the prediction of material characteristics, and the comprehension of the microscopic behavior of materials.

Given:

Intercepts  =  [tex]9.66\ A, 19.32\ A, 14.49\ A[/tex]

Lattice constant, [tex]a = 4.83\ A[/tex]

The reciprocal of intercepts is given as:

[tex]r = (1/9.66),1/19.32,1/14.49)[/tex]

The Miller indices are given as:

[tex]M = r/(1/a)\\M = ((1/9.66),1/19.32,1/14.49))/(1/4.83)\\M = (1/2,1/4,1/3)[/tex]

Hence, along the three Cartesian coordinates. The Miller indices of the plane are [tex](1/2,1/4,1/3)[/tex].

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The Miller indices for a plane intercepting a body-centered cubic lattice with intercepts of 9.66 Å, 19.32 Å, and 14.49 Å are (6,3,4). These are found by taking reciprocals of the intercepts, then multiplying by a common factor to get the smallest set of integers.

The student's question is about finding the Miller indices of a plane in a body-centered cubic lattice with given lattice constants and plane intercepts. Miller indices describe the orientation of a plane or set of planes in a crystal lattice.

To find these, we first take the reciprocals of the intercepts, which in this case gives us 1/2, 1/4, and 1/3. We then need to multiply these by a common factor to eliminate any fractions and get the smallest set of integers. Multiplying by 12 gives us the Miller indices of (6,3,4).

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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:

[tex] v_f = 15 \frac{m}{s}  [/tex]

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum [tex]\vec{L}[/tex] is

[tex]\vec{L}  = \vec{r} \times \vec{p}[/tex]

where [tex]\vec{r}[/tex] is the position and [tex]\vec{p}[/tex] the linear momentum.

We also know that the torque is

[tex]\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )[/tex]

[tex]\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p} [/tex]

[tex]\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F} [/tex]

but, as the linear momentum is [tex]\vec{p} = m \vec{v}[/tex] this means that is parallel to the velocity, and the first term must equal zero

[tex]\vec{v} \times \vec{p}=0[/tex]

so

[tex]\vec{\tau} =   \vec{r} \times \vec{F} [/tex]

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

[tex]\vec{\tau}_{rod} =   0 [/tex]

this means, for the angular momentum measure from the rod:

[tex]\frac{d\vec{L}_{rod}}{dt} =   0 [/tex]

that means :

[tex]\vec{L}_{rod} = constant[/tex]

So, the magnitude of initial angular momentum is :

[tex]| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)[/tex]

but the angle is 90°, so:

[tex]| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| [/tex]

[tex]| \vec{L}_{rod_i} | = r_i * m * v_i[/tex]

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

[tex]| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s} [/tex]

[tex]| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s} [/tex]

For our final angular momentum we have:

[tex]| \vec{L}_{rod_f} | = r_f * m * v_f[/tex]

and the radius is 0.250 m and the mass is 2.00 kg

[tex]| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f [/tex]

but, as the angular momentum is constant, this must be equal to the initial angular momentum

[tex] 7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f [/tex]

[tex] v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg} [/tex]

[tex] v_f = 15 \frac{m}{s}  [/tex]

Answer:

15 m/s

Explanation:

L = mvr

Li = (2.00 kg)(0.750 m)(5m/s) = 7.5 kgm^2/s

conservation of angular momentum --> Li=Lf

Lf = 7.5 kgm^2/s

7.5 kgm^2/s = (2.00 kg)(0.250 m)(vf)

vf = 15 m/s

Explanation:

Every rotating body experiences centrifugal force. Due to this force the body tends to bulge out around it mid point and gets flattened at the poles. Same is applicable to Earth as well. Since the Earth is rotating at a very high speed, its equator gets bulged out due to centrifugal force. Because of this bulged equator, Earth's pole to pole diameter and equatorial diameter has difference of around 42.76 km. It is flatter on the poles. This also proves that Earth is not a perfect sphere.

Answer and Explanation:

The reason for the not being perfectly spherical ad bulging out at the equator is that The centripetal force acting toward's the earth gravitational center tries to keep the Earth in perfect spherical shape.

Also the angular momentum of the orbiting planet influences the bulge,

The greater angular momentum results in more bulge while the lower value of it results in lesser bulge and more perfect spherical shape.

Also, a greater amount of force directed towards the center and acting on the object at the equator results in the bulges at the equator whereas at poles this force is not required and hence radius is lower in that region.

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%.

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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:

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]

To get a sharp image of the Sun, the paper should be placed at the focal length of the lens, which is 10 centimeters away from the lens. The image is in focus at this point because the light rays from the Sun are effectively parallel when they reach the lens, which then focuses these rays at its focal point.

Given the sun is so far away, the light it emits is nearly parallel by the time it reaches Earth. When using a lens to cast an image of the Sun, the point where the image is in focus, that is, the focal point, is also the focal length of the lens.

In this case, the student uses a lens with f = 10 cm, meaning the focal length of the lens is 10 centimeters. To get a sharp image, the paper on which the image is being projected should be placed 10 cm away from the lens, or at the focal length of the lens. This is because the light is in sharp focus at this distance, creating a clear image on the paper.

An important concept here is that the Sun is an astronomical unit away, so the light rays from the Sun are essentially parallel when they reach the lens. The lens then focuses these parallel rays to its focal point, forming a sharp image at a distance equal to its focal length.

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