A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a ramp of 64.8 degrees at a speed of 25.4 m/s. What would be the largest number of buses he can clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 10.0 m long

Answer:

social

Explanation:

Your social environment includes the people you spend time with – your family, your friends, classmates and other people in your community.  Your social environment is healthier when you choose friends who show concern for their own health and yours.

Answer:

Distance, S = 130m

Explanation:

Given the following data;

Initial velocity = 2m/s

Final velocity = 28m/s

Acceleration = 3m/s²

To find the distance, we would use the third equation of motion.

V² = U² + 2aS

Substituting into the equation, we have;

28² = 2² + 2*3*S

784 = 4 + 6S

6S = 784 - 4

6S = 780

S = 780/6

Distance, S = 130m

The sky changes colors at sunset because the atmosphere ABSORBS some colors of light more than other colors.

The second choice ("The atmosphere bends light") is a correct statement, but it's not the reason that the sky changes colors at sunset.

Answer:

The atmosphere bends light.

Explanation:

I got the test, and passed! =)

Answer:

Explanation:

A ) At constant volume :

ΔEint = n Cv x ΔT , n is no of moles , Cv is specific heat at constant volume , ΔT is increase in temperature .

For helium Cv = 3/2 R = 1.5 x 8.3 J = 12.45 J

ΔEint = .7 x 12.45 x ( 564 - 300 )

= 2300.76 J .

W₁ = 0 because volume is constant so work done by gas is zero .

Q₁ = ΔEint = 2300.76 J

B )

At constant pressure

Q₂ = n Cp Δ T , Cp is specific heat at constant pressure .

For monoatomic gas ,

Cp = 5/2 R = 2.5 x 8.3 J = 20.75 J

Q₂ = .7 x 20.75 x 264 J

= 3834.6 J

W₂ = work done by gas

= PΔV = nRΔT

= .8  x 8.3 x 264

= 1752.96 J

ΔEint = Q₂ - W₂

= 3834.6 - 1752.96

= 2081.64 J.

 ΔEint, 1, Q1, and W1 for the process at constant volume. and ΔEint, 2, Q2, and W2 for the process at constant pressure is mathematically given as

a)

dE1= 2300.76 J .

W1=0 as Volume is constant

Q1= 2300.76 J as Q= dE1

b)

Q2= 3834.6 J

W2= 1752.96 J

dE2= 2081.64 J.  

a)

Generally, the equation for the Constant Volume   is mathematically given as

dE = n Cv x dT

Where

Cv = 3/2

R = 1.5 x 8.3 J

R= 12.45 J

Therefore

dE = 0.7 x 12.45 x ( 564 - 300 )

dE1= 2300.76 J .

W1=0 as Volume is constant

Q1= 2300.76 J as Q= dE1

b)

Generally

Q2 = n Cp dT

Where

Cp = 5/2

R = 2.5 x 8.3 J

R= 20.75 J

Hemce

Q2 = 0.7 x 20.75 x 264 J

Q2= 3834.6 J

For Work done

W=PdV

W= nRdT

Therefore

W= 0.8  x 8.3 x 264

W2= 1752.96 J

Hence

dE = Q₂ - W₂

dE= 3834.6 - 1752.96

dE2= 2081.64 J.  

For more information on Pressure

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to plow fields beacuse it's much easier to plow on tractor than on foot

Answer:

See the explanation below.

Explanation:

Energy is the ability of bodies to perform work and produce changes in themselves or other bodies.

There are several types of energy, but let's talk specifically about mechanical energy.

Mechanical energy is associated or subdivided into kinetic, potential, and elastic energies.

Kinetic energy

[tex]E_{kin}=\frac{1}{2}*m*v^{2}[/tex]

Potential energy

[tex]E_{pot}=m*g*h[/tex]

Elastic Energy

[tex]E_{elas}=\frac{1}{2} *k*x^{2}[/tex]

Answer:

Explanation:

It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .

At levelled road , for stoppage

Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .

At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .

At upward inclined  road , for stoppage

Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force  = (friction force + gravitational force ) x displacement .

Hence displacement is less .

At downward slopping road ,  friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle  also .

At downward inclined  road , for stoppage

Kinetic energy of vehicle + work done by gravitational force  = Work done by frictional force = friction force  x displacement .

Hence displacement is more .

Hence displacement is more in the downward slopping.

Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.

It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .

At levelled road , for stoppage

Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .

At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .

At upward inclined  road , for stoppage

Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force  = (friction force + gravitational force ) x displacement .

Hence displacement is less .

At downward slopping road ,  friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle  also .

At downward inclined  road , for stoppage

Kinetic energy of vehicle + work done by gravitational force  = Work done by frictional force = friction force  x displacement .

Hence displacement is more in the downward slopping.

To learn more about displacement refer to the link:

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Answer:

mm hope that helps

Explanation:

What are the 3 forms of energy?

Forms of energy

Potential energy.

Kinetic energy.

Answer:

P = 85.72 W

Explanation:

Given that,

Initial speed, u = 0

Final speed, v = 22 m/s

Time, t = 0.48 s

Mass, m = 170 g = 0.17 kg

Let a be the acceleration of the rattlesnake.

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{22-0}{0.48}\\\\a=45.84\ m/s^2[/tex]

Let x is the displacement of a rattlesnake. It can be given by :

[tex]x=ut+\dfrac{1}{2}at^2\\\\x=0+\dfrac{1}{2}\times 45.84\times (0.48)^2\\\\x=5.28\ m[/tex]

The power of the rattlesnake is given by :

[tex]P=\dfrac{W}{t}\\\\P=\dfrac{m\times a\times x}{t}\\\\P=\dfrac{0.17\times 45.84\times 5.28}{0.48}\\\\P=85.72\ W[/tex]

So, the power of the rattlesnake is 85.72 W.

(a) It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.

Recall that centripetal acceleration has a magnitude a of

a = v ² / R

where

v = tangential speed

R = radius of the curve

so that

a = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²

Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have

∑ F = Fs = m a

where

Fs = magnitude of static friction

m = mass of the car

Then

Fs = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ 5400 N

(b) Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,

N - W = 0

where

N = magnitude of normal force

W = weight

so that

N = W = m g = (950 kg) (9.8 m/s²) = 9310 N

Friction is proportional to the normal force by a factor of µ, the coefficient of static friction:

Fs = µ N

Assuming 35 m/s is the maximum speed the car can travel without skidding, we find

µ = Fs / N = (5400 N) / (9310 N) ≈ 0.581395 ≈ 0.58

Answer:

39.26 s

Explanation:

From the question given above, the following data were obtainedb

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Distance travelled (s) = 1700 m

Time (t) =?

Next, we shall determine the final velocity of the plane. This can be obtained as follow:

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Distance travelled (s) = 1700 m

Final velocity (v) =?

v² = u² + 2as

v² = 6² + (2 × 1.9 × 1700)

v² = 36 + 6460

v² = 6496

Take the square root of both side

v = √6496

v = 80.6 m/s

Finally, we shall determine the time taken before the plane lifts off. This can be obtained as follow:

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Final velocity (v) = 80.6 m/s

Time (t) =?

v = u + at

80.6 = 6 + 1.9t

Collect like terms

80.6 – 6 = 1.9t

74.6 = 1.9t

Divide both side by 1.9

t = 74.6 / 1.9

t = 39.26 s

This, it will take 39.26 s before the plane lifts off.

Answer:

* he new moon phase when the position is       Sun - Moon - Earth,

* have of the Full Moon when the position is     Sun - Earth - Moon,

*All the phases of the moon are governed by the movement of the Moon around the Earth.

Explanation:

In the solar system, the planets revolve around the sun, which is much more massive, in the case of the Earth it is more massive than its satellite, therefore the Moon revolves around the Earth in a period of approximately 28 days.

It is said that the moon is in the new moon phase when the position is Sun - Moon - Earth, so the moon cannot be seen

It is in the phase of the Full Moon when the position is

                  Sun - Earth - Moon, in this case the moon can be observed by the light reflected from it.

All the phases of the moon are governed by the movement of the Moon around the Earth.

Answer:

we all know that deforestation is when tress are cut down but this may cause cardio dioxide to go up into the atmosphere and this may cause the rise in sea level and temperates tend to fluctuate

Explanation:

I'm just that smart yah dig

Answer:

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

[tex]V_i = \pi R^2 h_i --- (1)[/tex]

where;

height h which is the height for the free surface in a rotating tank is expressed as:

[tex]h = \dfrac{\omega^2 r^2}{2g} + C[/tex]

at the bottom surface of the tank;

r = 0, h = 0

∴

[tex]h = \dfrac{\omega^2 r^2}{2g} + C[/tex]

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

[tex]h=\dfrac{\omega^2 r^2}{2g} --- (2)[/tex]

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

[tex]\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr[/tex]

replacing the value of h in equation (2); we have:

[tex]V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr[/tex]

[tex]V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr[/tex]

[tex]V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0[/tex]

[tex]V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)[/tex]

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then [tex]V_f = V_i[/tex]

Replacing equation (1) and (3)

[tex]\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i[/tex]

[tex]\omega^2 = \dfrac{4g \times h_i }{R^2}[/tex]

[tex]\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}[/tex]

[tex]\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }[/tex]

[tex]\omega = \sqrt{109.87 }[/tex]

[tex]\mathbf{\omega = 10.48 \ rad/s}[/tex]

Finally, the angular velocity of the tank at which the bottom of the tank is exposed  = 10.48 rad/s

Answer:

 h '= [ ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]²  h

Explanation:

Let's analyze this problem, first the two bodies travel together, second the superball bounces, third it collides with the marble and fourth the marble rises to a height h ’

let's start by finding the velocity of the two bodies just before the collision, let's use the concepts of energy

starting point. Starting point

         Em₀ = U = m g h

final point. Just before the crash

         Em_f = K = ½ m v²

as there is no friction the mechanical energy is conserved

         Em₀ = Em_f

         mg h = ½ m v²

         v = √2gh

this speed is the same for the two bodies.

Second point. The superball collides with the ground, this process is very fast, so we will assume that the marble has not collided, let's use the concept of conservation of moment

initial instant. Just when the superball starts contacting the ground

      p₀ = M v

this velocity is negative because it points down

final instant. Just as the superball comes up from the floor

      p_f = M v '

the other body does not move

      p₀ = p_f

     - m v = M v '

       v ’= -v

Therefore, the speed of the asuperbola is the same speed with which it arrived, but in the opposite direction, that is, upwards.

Let's use the subscript 1 for the marble and the subscript 2 for the superball

Third part. The superball and the marble collide

the system is formed by the two bodies, so that the forces during the collision are internal and the moment is conserved

initial instant. Moment of shock

        p₀ = M [tex]v_{1'}[/tex]+ m v_2

final instant. When the marble shoots out.

        P_f = Mv_{1f'}+ m v_{2f}

        p₀ = p_

        M v_{1'}+ m v_2 = M v_{1f'} + m v_{2f}

        M (v_{1'} - v_{1f'}) = -m (v_2 - v_{2f})

in this expression we look for the exit velocity of the marble (v2f), as they indicate that the collision is elastic the kinetic nerve is also conserved

       K₀ = K_f

       ½ M v_{1'}² + m v₂² = M v_{1f'}²  + ½ m v_{2f}²

        M (v_{1'}² - v_{1f'}²) = - m (v₂² - v_{2f}²)

Let's set the relation  (a + b) (a-b) = a² - b²

      M (v_{1'} + v_{1f'})  (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})

let's write our two equations

           M ( v_{1'} - v_{1f'}) = -m (v₂ - v_{2f})                 (1)

           M (v_{1'} + v_{1f'})  (v_{1'} - v_{1f'}) = -m (v₂ + v_{2f}) (v₂ - v_{2f})

if we divide these two expressions

           (v_{1'}+ v_{1f'}) = (v₂ + v_{2f} )

we substitute this result in equation 1 and solve

          v_{1f'}= (v₂ + v_{2f}) - v_{1'}

          M (v_{1'} - [(v₂ + v_{2f}) - v_{1'}] = -m (v₂ - v_{2f})

           -M v₂ - M v_{2f1'} + 2M v_{1'} = m v₂ - m v_{2f}

          -M v_{2f} -m v_{2f} = m v₂ -M v₂ + 2M v_{1'}

          v_{2f} (M + m) = - v₂ (M-m) + 2 M v_{1'}

          v_{2f} = - [tex]( \frac{ M-m}{M +m } )[/tex]) v₂ + 2 [tex](\frac{M}{M+m})[/tex] v_{1'}

now we can substitute the velocity values ​​found in the first two parts

          [tex]v_{2f}[/tex] = - ( \frac{ M-m}{M +m  } ) √2gh + 2(\frac{M}{M+m}) √2gh

we simplify

          v_{2f} = [( \frac{ M-m}{M +m  } ) + 2 (\frac{M}{M+m})] [tex]\sqrt{2gh}[/tex]

let's call the quantity in brackets that only depends on the masses

          A = ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]

           v_{2f}= A \sqrt{2gh}

in general, the marble is much lighter than the superball, so its speed is much higher than the speed of the superball

finally with the conservation of energy we find the height that the marble reaches

Starting point

          Emo = K = ½ mv_{2f}²

Final point

          Emf = U = m g h'

          Em₀ = Em_f

          ½ m v_{2f}² = m g h ’

          h ’= ½ v_{2f}² / g

         h ’= ½ (A \sqrt{2gh})² / g

         h ’= A² h

         h '= [ ( \frac{ M-m}{M +m  } )+ 2 (\frac{M}{M+m})]²  h

Answer:

mechanical energy

Explanation:

when raised up has potential energy

Answer:

potential energy

Explanation:

potential energy

Answer:

40J

Explanation:

Given parameters:

Mass of box  = 10kg

distance moved  = 2m

Force applied  = 20N

Unknown:

Work done in pushing the box  = ?

Solution:

The work done in moving a body is a function of the force applied to move it through a particular distance.

 Work done  = Force x distance

So;

   Work done  = 20 x 2  = 40J

Answer:

The 59th element is Praseodymium it's symbol is Pr.

Answer:

y = -2.69 m

the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.

Explanation:

his problem must be solved with the missile launch equations.

Let's start by looking for the jumper's initial velocity

          R = v₀² sin 2θ / g

for the long jump the angle used is tea = 45º, in the exercise they indicate that the best record is R = 7.9m

          v₀² = R g / sin 2te

          v₀ = [tex]\sqrt{ \frac{7.9 \ 9.8}{1 }[/tex]

          v₀ = 8.80 m / s

Now suppose you jump with this speed to get to the other building, let's use trigonometry for the components of the speed

          sin 45 = [tex]v_{oy}[/tex] /v₀

          cos 45 = v₀ₓ / v₀

         v_{oy} = v₀ sin 45

          v₀ₓ = v₀ cos 45

          v_{oy} = 8.8 sin 45 = 6.22 m / s

          v₀ₓ = 8.8 cos 45 = 6.22 m / s

now let's calculate the sato with these speeds

           x = [tex]v_{ox}[/tex] t

the minimum jump is x = 10 m

           t = x / v₀ₓ

           t = 10 / 6.22

           t = 1.61 s

let's find the vertical distance for this time

           y = v_{oy} t - ½ g t²

where zero is placed on the jump building

           y = 6.22 1.61 - ½ 9.8 1.61²

            y = -2.69 m

Let's analyze this result, the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.

Answer:

1470J

Explanation:

Given parameters:

Mass of the person  = 50kg

height  = 3m

Unknown:

Work done  = ?

Solution:

To solve this problem, we use the expression below:

     Work done  = mass x acceleration due gravity x height

So;

     Work done  = 50 x 9.8 x 3 = 1470J

Answer:

Explanation:

The formula holding the relationship between wavelength and frequency of a wave can be expressed as:

[tex]\lambda = \dfrac{v}{f}[/tex]

where;

[tex]\lambda =[/tex] wavelength

v = speed

f = frequency

Given that:

v = 192 m/s and f = 240Hz

Then;

[tex]\lambda = \dfrac{192 \ m/s}{240 \ Hz}[/tex]

[tex]\lambda = 0.800 \ m[/tex]

Now, to estimate the respective amplitude of the string, we need to approach it by using the concept of wave equation which is:

y = A sin kx

here;

A = amplitude of the standing wave

k = wave number

x = maximum displacement

y = distance from center

recall that:

[tex]k = \dfrac{2 \pi}{\lambda}[/tex]

∴

[tex]y = A sin \dfrac{2 \pi}{\lambda }x[/tex]

Now;

for A = 0.400 cm ; [tex]\lambda[/tex] = 0.800 m ; k = 40 cm

Then;

[tex]y =(0.400 \ cm ) \ sin \dfrac{2 \pi}{0.800 m }\times (40 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm } )[/tex]

y = 0.400 sin π

y = 0 cm

At distance 40 cm; the amplitude = 0 cm

Thus, it is a node.

For k = 20cm

Then:

[tex]y =(0.400 \ cm ) \ sin \dfrac{2 \pi}{0.800 m }\times (20 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm } )[/tex]

y = 0.400 sin π/2

y = 0.400 cm

At distance 20 cm; the amplitude = 0.400 cm

Thus, it is antinode.

For k = 10cm

Then:

[tex]y =(0.400 \ cm ) \ sin \dfrac{2 \pi}{0.800 m }\times (10 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm } )[/tex]

4y = 0.400 sin π/2

y = 0.283 cm

At distance 10 cm; the amplitude = 0.283 cm

b)

The required time taken to go through the displacement( i.e. from largest upward to downward) is the time required to cover half of the wavelength.

This is expressed as:

[tex]T = \dfrac{1}{2} \times \dfrac{1}{f}[/tex]

[tex]T= \dfrac{1}{2} \times \dfrac{1}{240 \ Hz}[/tex]

T = 0.00208

T = 2.08 × 10⁻³ s

Answer:

4.62 N-s

Explanation:

recall that the formula for impulse is given by

Impulse = Force x change in time

in our case, we are given

Force = 14 N

change in time = 0.33s

Simply substituting the above into the equation for impulse, we get

Impulse = Force x change in time

Impulse = 14 x 0.33

= 4.62 N-s

[tex]\\ \sf\longmapsto Impulse=Force(Time)[/tex]

[tex]\\ \sf\longmapsto Impulse=14(0.33)[/tex]

[tex]\\ \sf\longmapsto Impulse=4.62Ns[/tex]

Answer:

A. Gamma decay

Explanation:

A form of nuclear decay in which the atomic number is unchanged is a gamma decay.

The atom has undergone a gamma decay.

In a gamma decay, no changes occur to the mass and atomic number of the substance.

Answer:

Earth

lol... ....

Answer:

Hi lol

Explanation:

follow me lol

sarivigakarthi this is my account...

Answer:

Explanation:

Speed of sound = 331 m /s

speed of jet = 7 .00 Mach = 7 times speed of sound

= 7 x 331 = 2317 m /s

distance to be covered = 5000 x 1000 = 5 x 10⁶ m

Time taken = distance / speed of jet

= 5 x 10⁶ / 2317

= 2.158 x 10³ s

= 35.96 minutes .

Answer:

The formula of nitrogen trifluoride is NF 3

The formula of ammonia is NH3

The formula of magnesium hydroxide is Mg(OH)2

Explanation:

The formula of a compound refers to the correct representation of the compound using appropriate chemical symbols.

The formula of a compound is derived from the symbols of its constituent elements. The formula of the compound must correctly show the number of atoms of each element in one formula unit of the compound.

In the answer section, the correct names or formulas of some compounds have been shown

Answer:

A. The formula of nitrogen trifluoride is NF3.

B. The formula of ammonia is NH3.

D. The formula of calcium chloride is CaCI2.

H. The formula of magnesium hydroxide is Mg(OH)2.

Explanation: