A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.8 mm above the river, whereas the opposite side is a mere 1.3 mm above the river. The river itself is a raging torrent 53.0 mm wide.
A) How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?
B) What is the speed of the car just before it lands safely on the other side?

Answer:

Water > Box of books > Stone > Ball

Explanation:

We'll begin by calculating the potential energy of each object. This can be obtained as follow:

For stone:

Mass (m) = 15 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 3 m

Potential energy (PE) =?

PE = mgh

PE = 15 × 10 × 3

PE = 450 J

For water:

Mass (m) = 10 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 9 m

Potential energy (PE) =?

PE = mgh

PE = 10 × 10 × 9

PE = 900 J

For ball:

Mass (m) = 1 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 20 m

Potential energy (PE) =?

PE = mgh

PE = 1 × 10 × 20

PE = 200 J

For box of books:

Mass (m) = 25 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 2 m

Potential energy (PE) =?

PE = mgh

PE = 25 × 10 × 2

PE = 500 J

Summary:

Object >>>>>>>> Potential energy

Stone >>>>>>>>> 450 J

Water >>>>>>>>> 900 J

Ball >>>>>>>>>>> 200 J

Box of books >>> 500 J

Arranging from greatest to least, we have:

Object >>>>>>>> Potential energy

Water >>>>>>>>> 900 J

Box of books >>> 500 J

Stone >>>>>>>>> 450 J

Ball >>>>>>>>>>> 200 J

Water > Box of books > Stone > Ball

Answer:

b 1.39 m/s²

Explanation:

Given the following data;

Time = 12 seconds

Distance, S = 100 m

Since it's starting from rest, the initial velocity is equal to 0m/s.

To find the acceleration, we would use the second equation of motion;

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

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 0(12) + ½*a*12²

100 = 0 + 72

100 = 72a

Acceleration, a = 100/72

Acceleration, a = 1.389 ≈ 1.39 m/s²

Answer:

Speed of the melon = 0.25 m/s

we would normally don't see the melon moving due to friction with the resting surface.

Explanation:

We use conservation of momentum:

Pi = Pf

with Pi = 0.1 kg * 30 m/s = 3 kg m/s

and Pf = 0.1 kg * 20 m/s + 4.0 kg * V = 2 kg m/s + 4 * V

Then using the equality above, we solve for V (velocity of the melon)

3 kg m/s = 2 kg m/s + 4 V

1 kg m/s = 4 kg * V

Then V = 1 / 4  M/s = 0.25 m/s

So we would normally don't see the melon moving due to friction with the resting surface.

Answer:

7 would be C, a cell.

Explanation:

Hi.

7 would be C, a cell.

A cell is the basic unit of structure and function in all living things.

If it is living, it is made of cells.

Hope this helps.

Answer:

7. Cell

8. Organelle

Answer:

Explanation:

Force of friction = μ N , where μ is coefficient of friction , N is normal force on the body .

a )

Given,

Normal force N = 400 N

Force of friction = 300 N

μ = coefficient of static friction = ?

Putting the values ,

300 = 400 μ

μ = .75

b )

Normal force N = 400 N

Force of friction = 200 N

μ = coefficient of kinetic  friction = ?

Putting the values ,

200 = 400 μ

μ = .50

c ) see attached file .

You're supposed to divide the mass by the volume, which is going to equal to 5

Answer:

The balls velocity is 1 divided by 3

The velocity of the ball is 18.85 m/s.

The ball’s centripetal acceleration is 236.87 m/s².

The ball's centripetal force is  118.44 Newton.

Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.

Given parameters:

length of the string: l = 1.5 meters.

Time interval = 60 seconds.

Total number of complete rotation = 120.

Hence, the velocity of the ball = 120×2π×1.5/60 m/s

= 18.85 m/s.

The ball’s centripetal acceleration = (velocity)²/ radius

= (18.85)²/1.5 m/s²

= 236.87 m/s²

The ball's centripetal force = mass × centripetal acceleration

= 0.5 × 236.87 Newton

= 118.44 Newton

Learn centripetal acceleration here:

https://brainly.com/question/14465119

#SPJ2

Explanation:

The energy of the system before the collision must equal the energy after the collision.

After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is

[tex]\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J[/tex]

Before the collision only the bullet has kinetic energy.

So we can now determine the speed of the bullet using

[tex]\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s[/tex]

Answer:

making sense of a 3-d world from 2-d data

Explanation:

Explanation:

Cheetahs can go from 0 to 60 miles per hour in just 3.4 seconds and reach a top speed of 70 miles per hour. While they are the fastest land animal in the world, they can only maintain their speed for only 20 to 30 seconds.

Answer:

3.24×10¯² nN

Explanation:

From the question given above, the following data were obtained:

Mass of girl (M₁) = 61.6 kg

Mass of boy (M₂) = 71.2 kg

Distance apart (r) = 95 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force (F) =?

The force of attraction between the girl and the boy can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 61.6 × 71.2 / 95²

F = 6.67×10¯¹¹ × 61.6 × 71.2 / 9025

F = 3.24×10¯¹¹ N

Finally, we shall convert 3.24×10¯¹¹ N to nN. This can be obtained as follow:

1 N = 10⁹ nN

Therefore,

3.24×10¯¹¹ N = 3.24×10¯¹¹ N × 10⁹ nN / 1 N

3.24×10¯¹¹ N = 3.24×10¯² nN

Thus, the force attraction between the girl and the boy is 3.24×10¯² nN

Answer:

a)   v = 26.2 m / s, b) acceleration is radial, a = 27.4 m / s², c)   a = 2.8 g and

d) a = - 8.73 10⁻² m / s²,  τ = 1.09 10⁴ N m

Explanation:

a) For this exercise we can use the relationships between rotational and linear motion

           v = w r

let's reduce the magnitudes to the SI system

          w = 10 rpm (2pi rad / 1 rev) 1 min / 60s) = 1,047 rad / s

          r = 25.0 m

let's calculate

          v = 1.047 25.0

          v = 26.2 m / s

b) When the body is rotating at constant speed, the relationship must be perpendicular to the speed, therefore the direction of acceleration is radial, that is, towards the center of the circle and its magnitude is

            a = v² / r

            a = 26.2²/25

            a = 27.4 m / s²

c) Let's look for the relationship between the centripetal acceleration and the acceleration due to gravity

           a / g = 27.4 / 9.8

           a / g = 2.8

           a = 2.8 g

d) let's find the deceleration and torque to stop the centripette in 5 min

           t = 5 min (60 s / 1min) = 300 s

let's use the rotational kinematics relations

           w = w₀ + α t

initial angular velocity is wo = 1,047 rad / s and the final as is stop do w = 0

           α = - w₀ / t

           α = - 1,047 / 300

           α = -3.49 10⁻³ rad / s²

angular and linear are related

           a = α r

           a = -3.49 10⁻³ 25

           a = - 8.73 10⁻² m / s²

the negative sign indicates that the acceleration is stopping the movement

torque is

           τ = F r

The force can be found with Newton's second law

          F = m a

we substitute

         τ = m a r

         τ = 5000.0   8.73 10⁻²  25

         τ = 1.09 10⁴ N m

Answer:

its x and y component is 24.749m/s

Explanation:

Given

Speed of the dog = 35m/s

x component of the speed = xcos theta

y component of the speed = ycos theta

Given theta =45 degrees

x-component = 35cos45

x-component = 35(0.7071)

x-component = 24.749m/s

y-component = 35sin45

y-component = 35(0.7071)

y-component = 24.749m/s

Hence its x and y component is 24.749m/s

Answer:

The time of motion is 0.64 s.

Explanation:

Given;

mass of the apple, m = 107 g

height of fall, h = 2 m

The velocity of the apple when it hits the ground is calculated from the law of conservation of energy;

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8\times 2} \\\\v = 6.261 \ m / s[/tex]

The time of motion is calculated;

v = u + gt

6.261 = 0 + 9.8t

6.261 = 9.8t

t = 6.261 / 9.8

t = 0.64 s

Therefore, the time of motion is 0.64 s

The time taken for the apple to hit the ground is 0.64 s.

The time taken for the apple to hit the ground can be calculated using the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Solve for t.

Hence, The time taken for the apple to hit the ground is 0.64 s

Learn more about time taken here: https://brainly.com/question/4931057

Answer:

It should be Mass, Gravity and Height

Explanation:

Answer:

a)  v = 4.37 10⁶ m / s,  speed is much less than c

b)     v = 2.01 10⁸  m / s,  this value is 67% of the speed of light, , for which relativistic corrections should be used

Explanation:

The bohr model for the hydrogen atom and dendroids is a classical model with a quantization of the angular momentum

let's start by using Newton's second law with the electric force

         F = m a

Coulomb's law electric force

         F = [tex]k \frac{q_1q_2}{r^2}[/tex]

in this case in an atom the number of protons is equal to the atomic number and there is only one electron

        q₁ = Ze

        q₂ = e

acceleration is centripetal

         a = v² / r

we substitute

        [tex]k \frac{Z e^2}{r^2} = m \frac{v^2}{r}[/tex]

        v² =  [tex]k \frac{Ze^2}{m r}[/tex]

quantization is imposed without justification in this model,

       L = p x r = n [tex]\hbar[/tex]

       \hbar= h /2π

if we consider circular orbits, the speed and position are perpendicular

       m v r = n \hbar

       r = [tex]\frac{n \hbar}{m v}[/tex]

we substitute

     v² = [tex]k \frac{Z e^2}{m} \frac{m v}{n \hbar}[/tex]

     v = [tex]k \frac{Z e^2 }{ n \hbar}[/tex]

let's apply this equation

     \hbar= h / 2π

     \hbar= 6.626 10-34 / 2π

     \hbar= 1.05456 10⁻³⁴ J s

a) He1 ion,  the atomic number of helium is 2

      v =  [tex]\frac{9 \ 10^9 \ 2 ( 1.6 \ 10^{-19})^2 }{n \ 1.0546 \ 10^{-34}}[/tex]

       v =4.3695 10⁶ / n m / s

the ground state occurs for N = 1

        v = 4.37 10⁶ m / s

the relationship of this value to the speed of light is

        v / c = 4.37 10⁶/3 10⁸

        v / c = 1.46 10⁻²

speed is much less than c

b) the uranium ion with atomic number Z = 92

       v = [tex]\frac{9 \ 10^9 \ 92 ( 1.6 \ 10^{-19})^2 }{n \ 1.054 \ 10^{-34} }[/tex]

       v = 2.01 10⁸  m / s

       v/c = [tex]\frac{2.01 \ 10^8 }{3 \ 10^8}[/tex]

       v/c =  0.67

this value is 67% of the speed of light, for atoms with a higher atomic number the effects are increasingly important, for which relativistic corrections should be used

Answer:

Untrasound

Explanation:

Your welcome :)

Answer:

t = 4.34 s

Explanation:

To get this, we need to calculate the time of each part for both vehicles.

In the case of the bicycle, we can calculate the time duration with it's acceleration:

a = Vf - Vo/t ------> t = Vf - Vo / a

But Vo = 0 so time:

t = 21 / 14 = 1.5 s

Doing the same with the car:

t = 49 / 8 = 6.125 s

With these values of time, we can calculate the distance covered by both vehicles during acceleration:

X = Vo*t + at²/2

X = at² / 2

With the bicycle:

X = 14 * (1.5)² / 2 = 15.75 mi

With the car:

X = 8 * (1.5)² / 2 =  9 mi

And now, we can also get the speed of the car, after the time of 1.5 s has passed.

V = 8 * 1.5 = 12 mi/h

Now we can actually write an equation with both data of the vehicles in function of the distance. We are going to say that "t" would be the time taken by both vehicles, to meet each other.

Distance covered by the bicycle = distance covered by car

Distance covered by the bicycle would be:

15.75 mi + 21t

Distance covered by car:

9 + 12t + (8t²/2)

Equalling both expressions:

15.75 + 21t = 9 + 12t + 4t²

4t² - 9t - 6.75 = 0

Solving for t, using quadratic expressions we have:

t = 9 ± √(9)² + 4*4*6.75 / 2*4

t = 9 ±√189 / 8

t = 9 ± 13.75 / 8

t₁ = 2.84 s

t₂ = -0.594 s

We are taking the positive time.

Then, the time where both vehicles meet, which is the same time interval when the bicycle is ahead of the car will be:

t = 2.84 + 1.5

Hope this helps

Answer: 1. The law of consevation of energy sates that energy can neither be created nor destroyed. It can only be transformed or transfered from one form to another. The law of conservation of energy is found everywhere for example, Water falls from the sky, converting potential energy to kinetic energy.

2. Different forms of energy are related because energy cannot be created or destroyed. they can all be transformed into from one form to another.

Explanation:

Answer:

(4) Its kinetic energy is a minimum.

Explanation:

Stone experiments a parabolic motion, which is a combination of horizontal motion at constant speed and vertical uniform accelerated motion due to gravity, where effects from air viscosity and Earth's rotation are neglected. Meaning that stone represents a conservative system.

When stone reachest highest point, horizontal velocity remains unchanged and vertical velocity is zero. Acceleration remains constant and different of zero. Hence, gravitational potential energy is a maximum and kinetic energy is a minimum.

Correct answer is: (4) Its kinetic energy is a minimum.

The highest point, that stone reaches its kinetic energy is a minimum. therefore the option 4 is correct.

Projectile motion -The motion of an object, thrown (projected) into the air is known as projectile motion.

(1) Its acceleration is zero-

When an object is at its highest point the acceleration will be equal to the gravitational force (9.8 m/sec squared). If we take air resistance into the account it will be slightly greater than the 9.8 meters per second squared but not equal to zero in any case. Hence, statement 1 is incorrect.

(2) Its acceleration is a minimum, but not zero-

At the highest point, the object will be at the one place where air resistance does not affect the object, and thus acceleration is exactly equal to the acceleration due to gravity and at this position, it will be the maximum. Hence, statement 2 is incorrect.

(3) Its gravitational potential energy is a  minimum-

At the highest point, the object will stop for the moment and have zero velocity. Thus it will have zero kinetic energy. Therefore total energy will have in the form of gravitational potential energy and which is maximum at this point. Hence, statement 3 is incorrect.

(4) Its kinetic energy is a minimum-

At the highest point the object will stop for the moment and have zero velocity. Thus it will have zero kinetic energy. Hence, statement 4 is correct.

At the highest point, that stone reaches its kinetic energy is a minimum. therefore the option 4 is correct.

For more about the projectile motion, follow the link below-

https://brainly.com/question/11049671

Answer:

A. 2.63 s B. 12.38 m

Explanation:

A. For what time interval (in s) after the light turned green is the bicycle ahead of your car?

The time interval at which the bicycle is ahead of the car is the time it takes for the car to reach the bicycle's speed of 21.0 mi/h.

So, using v = u + at where u = initial speed of car = 0 mi/h, v = final speed of car = 21.0 mi/h, a = acceleration of car = 8.00 mi/h/s and t = time taken for acceleration.

So, v = u + at

t = (v - u)/a

substituting the values of the variables into the equation, we have

t = (21.0 mi/h - 0 mi/h)/8.00 mi/h/s

= 21.0 mi/h ÷ 8.00 mi/h/s

= 2.63 s

B. By what maximum distance does the bicycle lead the car?

To find this distance, we find the distance moved by both the car in this time of t = 2.63 s

So, using s = ut + 1/2at² where u = initial speed of car = 0 mi/h = 0 m/s, t = time = 2.63 s, a = acceleration of car = 8.00 mi/h/s = 8.00 × 1609 m/3600 s = 3.58 m/s/s = 3.58 m/s² and s = distance moved by car.

So, substituting the values of the variables into the equation, we have

s = ut + 1/2at²

s = 0 m/s × 2.63 s + 1/2 × 3.58 m/s² × (2.63 s)²

s = 0 m + 1/2 × 3.58 m/s² × (2.63 s)²

s =  1.79 m/s² × 6.9169 s²

s = 12.38 m

which is also the maximum distance with which the bicycle leads the car.

Answer:

The angular displacement is 37.5  radian.

Explanation:

Given that,

The radius of the car, r = 0.3 m

The acceleration of the car, [tex]\alpha =3\ rad/s^2[/tex]

The initial speed of the car, [tex]\omega_i=0[/tex]

Time, t = 5 s

The angular displacement can be calculated using second equation of motion i.e.

[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\\theta=\dfrac{1}{2}\alpha t^2\\\\\theta=\dfrac{1}{2}\times 3\times (5)^2\\\\\theta=37.5\ rad[/tex]

So, it will make 37.5 radians.

Answer:

Mixed in a smoothie it like it licked

Explanation:

Answer:

84kliometers

Explanation:

divide one hundred and sixty eight kilo meters by two hours

Answer:

1.5 * 10^-2 Tm^2

Explanation:

Electric Flux = B.A cos(theta)

B = 0.055 T

A = 0.32 m^2

theta = 30

Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2

Answer:

Specific heat (Cp) water (at 15°C/60°F): 4.187 kJ/kgK = 1.001 Btu(IT)/(lbm °F) or kcal/(kg K)

Answer:

10s

Explanation:

Horizontal velocity is the velocity of an object in an horizontal direction

The ball's horizontal velocity is approximately 33.078 ft./s

Reason:

The known parameter are;

The horizontal distance the footballer kicks the ball, d = 43 yards

The time after which the ball lands, Δt = 3.9 seconds

Required:

To find the velocity of the ball

Solution:

[tex]Velocity = \dfrac{Distance}{Time} = \dfrac{d}{\Delta t}[/tex]

Therefore;

[tex]Horizontal \ velocity \ of \ the \ ball, \ v_x= \dfrac{43 \ yard}{3.9 \ seconds} \approx 11.026 \ yd/s[/tex]

The ball's horizontal velocity, vₓ ≈ 11.026 yd/s

1 yard = 3 feet

[tex]11.026 \ yard = 11.026 \ yard \times \dfrac{3 \ feet}{yard} = 22.078 \ feet[/tex]

The ball's horizontal velocity, vₓ ≈ 33.078 ft./s

Learn more about horizontal velocity here:

https://brainly.com/question/14898646

Answer:

a. i. -4.65 m/s ii. -13.95 m/s b. 5.89 m c. 2.85 s

Explanation:

a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?

We write the equation of the forces acting on the mass.

So, T - mg = ma where T = tension in vertical rope = (36.0 N/s)t, m = mass of box = 3.50 kg, g = acceleration due to gravity = 9.8 m/s² and a = acceleration of box = dv/dt where v = velocity of box and t = time.

So, T - mg = ma

T/m - g = a

dv/dt = T/m - g

dv/dt = (36.0 N/s)t/3.50 kg - 9.8 m/s²

dv/dt = (10.3 m/s²)t - 9.8 m/s²

dv = [(10.3 m/s²)t - 9.8 m/s²]dt

Integrating, we have

∫dv = ∫[(10.3 m/s³)t - 9.8 m/s²]dt

∫dv = ∫(10.3 m/s³)tdt - ∫(9.8 m/s²)dt

v = (10.3 m/s³)t²/2 - (9.8 m/s²)t + C

v = (5.15 m/s³)t² - (9.8 m/s²)t + C

when t = 0, v = 0 (since at t = 0, box is at rest)

So,

0 = (5.15 m/s³)(0)² - (9.8 m/s²)(0) + C

0 = 0 + 0 + C

C = 0

So, v = (5.15 m/s³)t² - (9.8 m/s²)t

i. What is the velocity of the box at t = 1.00 s,

v =  (5.15 m/s³)(1.00 s)² - (9.8 m/s²)(1.00 s)

v = 5.15 m/s - 9.8 m/s

v = -4.65 m/s

ii. What is the velocity of the box at t = 3.00 s,

v =  (5.15 m/s³)(3.00 s)² - (9.8 m/s²)(3.00 s)

v = 15.45 m/s - 29.4 m/s

v = -13.95 m/s

b. What is the maximum distance that the box descends below its initial position?

Since v = (5.15 m/s³)t² - (9.8 m/s²)t and dy/dt = v where y = vertical distance moved by mass and t = time, we need to find y.

dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t

dy = [(5.15 m/s³)t² - (9.8 m/s²)t]dt

Integrating, we have

∫dy = ∫[(5.15 m/s³)t² - (9.8 m/s²)t]dt

∫dy = ∫(5.15 m/s³)t²dt - ∫(9.8 m/s²)tdt

∫dy = ∫(5.15 m/s³)t³/3dt - ∫(9.8 m/s²)t²/2dt

y = (1.72 m/s³)t³ - (4.9 m/s²)t² + C'

when t = 0, y = 0.

So,

0 = (1.72 m/s³)(0)³ - (4.9 m/s²)(0)² + C'

0 = 0 + 0 + C'

C' = 0

y = (1.72 m/s³)t³ - (4.9 m/s²)t²

The maximum distance is obtained at the time when v = dy/dt = 0.

So,

dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t = 0

(5.15 m/s³)t² - (9.8 m/s²)t = 0

t[(5.15 m/s³)t - (9.8 m/s²)] = 0

t = 0 or [(5.15 m/s³)t - (9.8 m/s²)] = 0

t = 0 or (5.15 m/s³)t = (9.8 m/s²)

t = 0 or t = (9.8 m/s²)/(5.15 m/s³)

t = 0 or t = 1.9 s

Substituting t = 1.9 s into y, we have

y = (1.72 m/s³)t³ - (4.9 m/s²)t²

y = (1.72 m/s³)(1.9 s)³ - (4.9 m/s²)(1.9 s)²

y = (1.72 m/s³)(6.859 s³) - (4.9 m/s²)(3.61 s²)

y = 11.798 m - 17.689 m

y = -5.891 m

y ≅ - 5.89 m

So, the maximum distance that the box descends below its initial position is 5.89 m

c. At what value of t does the box return to its initial position?

The box returns to its original position when y = 0. So

y = (1.72 m/s³)t³ - (4.9 m/s²)t²

0 = (1.72 m/s³)t³ - (4.9 m/s²)t²

(1.72 m/s³)t³ - (4.9 m/s²)t² = 0

t²[(1.72 m/s³)t - (4.9 m/s²)] = 0

t² = 0 or (1.72 m/s³)t - (4.9 m/s²) = 0

t = √0 or (1.72 m/s³)t = (4.9 m/s²)

t = 0 or t = (4.9 m/s²)/(1.72 m/s³)

t = 0 or t = 2.85 s

So, the box returns to its original position when t = 2.85 s