The figure below shows a motorcycle leaving the end of a ramp with a speed of 39.0 m/s and following the curved path shown. At the peak of the path, a maximum height h above the top of the ramp, the motorcycle's speed is 37.1 m/s. What is the maximum height h? Ignore friction and air resistance. (Enter your answer in m.)

_____m

Answer:

0.248

Explanation:

Initial speed u = 13.6

Final speed v = 0

Distance s = 38.1

We have umg = ma

We make u subject of the formula

u = a/g

V² = u² + 2as

a = v²-u²/2s

We substitute the values into the above

a = 0-(13.6)²/2*38.1

a = 184.96/76.2

a = 2.427m/sec

Remember that

u = a/g

u = 2.427/9.8

= 0.2476

This is approximately 0.248

This is the minimum coefficient of friction required to keep the crate from sliding.

Answer:

by reducing friction.....

Answer:

Any floating object displaces a volume of water equal in weight to the object's MASS. ... If you place water and an ice cube in a cup so that the cup is entirely full to the ... If you take a one pound bottle of water and freeze it, it will still weigh one ... Fresh, liquid water has a density of 1 gram per cubic centimeter (1g = 1cm^3, ...

Answer: I think false

Explanation:The velocity at which an object is sent moving and the mass of the object both play a hand in the level of kinetic energy that object produces. Mass and kinetic energy have a positive relationship, which means that as mass increases, kinetic energy increases, if all other factors are held constant.

Answer:

h over 2 dg

Explanation:

brainliest!!!!!!!

Answer:

A)

The impulse that reaches the cell phone first is the Longitudinal wave

B)  0.0206  seconds

Explanation:

length of Iron rail = 7.5 m

speed of sound in Iron = 5950 m/s

speed of sound in Air = 343 m/s

A) Determine which pulse reaches the cell phone first

The impulse that reaches the cell phone first is the Longitudinal wave

 Time for longitudinal pulse to be detected =  7.5 / 5950 = 0.00126 s

  Time for pulse through air to be detected = 7.5 / 343 = 0.02186  s

B) separation in time between the arrivals of the two pulses

   ΔT = 0.02186 - 0.00126  = 0.0206  seconds

Answer:

d velocity will be the one according to me

Answer:

D

Explanation:

Telescopes detect electromagnetic waves from space and it travels back to the telescope lens in order for you to see.

Answer:

D

Explanation:

Ap ex science exam lol

Answer:

a. Vc = 5.06 m/s

b. Vp =  22.18 m/s

Explanation:

The acceleration of the pulley-mass system is as follows:

a = [tex]\frac{mg}{m + M}[/tex]

Solving for acceleration, we get:  

a = [tex]\frac{6.17 *9.8}{6.17 + 55.6}[/tex]

a = 0.97

So, for the part a:

Calculate the velocity of the crewman by using the following equation:

Vc = [tex]\sqrt{Vi^{2} + 2ay}[/tex]

Substituting the values into the equation, we get:

Vc = [tex]\sqrt{1.50^{2} + 2*0.97*13.2}[/tex]

Vc = 5.06 m/s

Now, for part b:

Calculate the final velocity of the pulley by using the following expression:

Vp = [tex]\sqrt{Vi^{2} +2gy }[/tex]

Just plugging in the values.

Vp = [tex]\sqrt{5.06^{2} +2*9.8*23.8 }[/tex]

Vp =  22.18 m/s

Answer:

RThe answer is Reflection....

Answer:

The sugars produced by photosynthesis can be stored, transported throughout the tree, and converted into energy which is used to power all cellular processes. Respiration occurs when glucose (sugar produced during photosynthesis) combines with oxygen to produce useable cellular energy.

Explanation:

I think this is correct lol.

Answer: it is thin near the center and thick at the edges

Explanation: took the test on Plato :)

Answer: When the car speed triples, momentum also triples but Kinetic energy increases 9 times or by 9 fold.

Explanation:

The momentum of a car (an object) is

p= mv

where

m is =the mass of the object( in this case car)

v is its= velocity

While the kinetic energy is is given by the formulae

K=1/2mv²

To determine how momentum and kinetic energy of the  car changes when the speed of the object triples, We have that the new velocity,

v¹= 3v

So that  the momentum  change becomes

p¹=mv¹=m (3v)= 3mv

mv=p

therefore p¹= 3p

we can see that the momentum also triples.

And the kinetic energy change  becomes

K¹=1/2m(v¹)²= 1/2m (3v)²

= 1/2m9v²= 1/2 x m x 9 x v²=9 x1/2mv²

1/2mv²=K

K¹= Kinetic energy = 9k

but Kinetic energy increases 9 times

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance [tex]D_{A}[/tex] beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed [tex]v_{A}[/tex]. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed [tex]v_{B}[/tex], which is greater than [tex]v_{A}[/tex].

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]

              Part B: [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

[tex]x=x_{0}+vt[/tex]

where

[tex]x_{0}[/tex] is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is [tex]D_{A}[/tex].

The equation will be:

[tex]x_{A}=D_{A}+v_{A}t[/tex]

Car B started at the starting line. So, its equation is

[tex]x_{B}=v_{B}t[/tex]

Part A: When they meet, both car are at "the same position":

[tex]D_{A}+v_{A}t=v_{B}t[/tex]

[tex]v_{B}t-v_{A}t=D_{A}[/tex]

[tex]t(v_{B}-v_{A})=D_{A}[/tex]

[tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]

Car B meet with Car A after [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex] units of time.

Part B: With the meeting time, we can determine the position they will be:

[tex]x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )[/tex]

[tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]

Since Car B started at the starting line, the distance Car B will be when it passes Car A is [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex] units of distance.

The distance traveled by the car A and car B  should be equal to the as they meet at the same position.

The time car B will catch the car A after is,

[tex]\dfrac{D_A}{v_B-v_A}[/tex]

The distance is the product of the speed of the body and the time taken to travel the distance.

Given information-

Car A has a head start and is a distance DA beyond the starting line at,

[tex]t=0[/tex]

Car A travels at a constant speed [tex]v_A[/tex].

Car B travels at a constant speed [tex]v_B[/tex].

The distance is the product of the speed of the body and the time taken to travel the distance.

The position equation from the motion for car A can be given as,

[tex]x_A=v_At+D_A[/tex]

The position equation from the motion for car B can be given as,

[tex]x_B=v_Bt[/tex]

The distance traveled by the car A and car B  should be equal to the as they meet at the same position. Thus,

[tex]x_A=x_B[/tex]

Put the values,

[tex]v_At+D_A=v_Bt\\v_At-v_Bt=-D_A\\t(v_B-v_A)=D_A\\t=\dfrac{D_A}{v_B-v_A}[/tex]

Hence the time car B will catch the car A after is,

[tex]\dfrac{D_A}{v_B-v_A}[/tex]

Learn more about the speed of the object here;

https://brainly.com/question/4931057

The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. ... Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces.

Answer:

In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force

Explanation:

option a is right

Answer:

land becomes unfit for food production

Answer:

6.52 m/s

1.72 m/s

5.38 m/s

Explanation:

this question requires us to find the average velocity.

1. velocity in straight down direction:

velocity = distance/time

= 15.0/2.30

= 6.52 m/s

2. velocity in straight backward direction:

velocity = distance/time

= 3.00 /1.74

= 1.72m/s

3. velocity in straight forward direction

velocity = distance/time

= 28.0/5.20

= 5.38 m/s

these are the his velocities for each if the intervals.

thank you!

Answer:

i think c

Explanation:

Answer:

Explanation:

FBEDAC

Answer: d

Explanation:

Answer:

so easy add the subtract then multiplay the add

Explanation:

Answer:

16 meters

Explanation:

When the rope is suddenly cut the object moving tangent at the circle. In that moment the gravity act in the object making it falls.

First we need to find how much time de object take to reach at the ground.

g=acceleration of gravity=10m/s²

v= vertical velocity =0m/s

h=vertical altitude =20m

We will find t such that h(t)=0

[tex]0=20-5t^{2} \\\\5t^{2} =20\\\\t^{2} =4\\\\t=2s[/tex]

v=horizontal velocity

D(t=2) is the horizontal distancetravelled by the object:

[tex]D(2)=8*2\\\\D(2)=16m[/tex]

Answer:

I believe you put how you think you'd feel it's that simple

Answer:

utilizing positive values

Explanation:

Answer:

d

Explanation:

Answer:

Explanation:

Total time taken = 0.85 s .

Time taken by sound to travel 5.43 m + time taken by coin to fall by 5.43 m = .85

5.43 / 343 + time taken by coin to fall by 5.43 m = .85

time taken by coin to fall by 5.43 m = .85 - 5.43/343 = .834 s

Let the initial velocity of throw of coin = u

displacement of coin s  = 5.43 m

time take to fall t =  .834 s

s = ut + 1/2 gt²

5.43 = u x .834 + .5 x 9.8 x .834²

5.43 = u x .834 + 3.41

u x .834 = 2.02

u = 2.42 m /s .

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

[tex]V_{tot} = V_{q1} + V_{q2}[/tex]

[tex]V_q = \dfrac{k \cdot q}{r}[/tex]

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

[tex]V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0[/tex]

The electric field at the point exactly midway between the plates, [tex]V_{tot}[/tex] = 0

3) The electric field, 'E', between plates is given as follows;

[tex]E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C[/tex]

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, [tex]F_e[/tex] = E × e

∴ [tex]F_e[/tex] = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, [tex]F_e[/tex] ≈ 1.807 × 10⁻⁷ N

Answer:

Explanation:

1.00 km = 1000 m .

Blythe's run : -------

Time taken to run 600 m speed

= distance / speed

T₁= 600 / 4.10 = 146.34 s

Next time T₂ = 1 min = 60 s

acceleration of Blythe

a = (7.30 - 4.10) / 60 = .053 m /s²

displacement during acceleration

= ut + 1/2 at²

= 4.10 x 60 + .5 x .053 x 60²

= 246 + 95.4

= 341.4 m

Rest of the distance to be covered = 1000 - ( 600 + 341.4 )

= 58.9 m

Time taken to cover this distance

T₃= 58.9 / 7.3 = 8.06 s

Total time = T₁ + T₂ + T₃ = 214.4 s

Geoff s run : ---------

initial acceleration during first 3 min

= (8.3 - 0 ) / (3 x 60 )

= .046 m /s²

displacement

s = ut + 1/2 a t²

= 0 + .5 x .046 x ( 3 x 60 )²

= 745.2 m

Rest of the distance of race

= 1000 - 745.2 = 254.8 m

This distance is covered at speed of 8.3 m/s

time taken to cover this distance

T₂ = 254.8 / 8.3

= 30.7 s

Total time taken to complete the race

= 180 + 30.7

= 210.7 s .

Answer:

19.5 m/s

87.8 m

Explanation:

The acceleration of block one is:

∑F = ma

-m₁gμ = m₁a

a = -gμ

a = -(9.8 m/s²) (0.22)

a = -2.16 m/s²

The velocity of block one just before the collision is:

v² = v₀² + 2aΔx

v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)

v = 7.63 m/s

Momentum is conserved, so the velocity of block two just after the collision is:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁ = m₂v₂

(18.5 kg) (7.63 m/s) = (7.25 kg) v

v = 19.5 m/s

The acceleration of block two is also -2.16 m/s², so the distance is:

v² = v₀² + 2aΔx

(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx

Δx = 87.8 m

Explanation:

By using conservation of linear momentum and also by equating work done to kinetic energy,  [tex]V_{2}[/tex]  = 15.36 m/s and [tex]d_{2}[/tex] = 4.32 meters

Parameters given are :

[tex]m_{1}[/tex] = 19.5 kg

friction, μk = 0.35

distance d = 2.6 m

mass [tex]m_{2}[/tex] = 8.25 kg.

initial velocity of [tex]U_{1}[/tex] = 6.5 m/s.

a.) Since we assumed that the block one stops after it collides with block two, the final velocity for block one will be zero. That is, [tex]V_{1}[/tex] = 0 so its final momentum = 0

Let us also assume that block two was initially at rest. Therefore, it initial velocity and its momentum will be equal to zero.

The formula to use will be :

[tex]m_{1}U_{1} = m_{2}V_{2}[/tex]

Substitute all the parameters into the formula above

19.5 x 6.5 = 8.25[tex]V_{2}[/tex]

Make [tex]V_{2}[/tex] the subject of formula

[tex]V_{2}[/tex] = 126.75/8.25

[tex]V_{2}[/tex] = 15.36 m/s

b.) Let us first calculate the work done in by block one.

The K.E = [tex]1/2mU^{2}[/tex]

substitute its mass and velocity into the formula

K.E = 1/2 x 19.5 x [tex]6.5^{2}[/tex]

K.E = 411.94 Joule

The work done = Kinetic energy

But the resultant Force F = force f - friction

where Frictional force = 0.35 x 19.5 x 9.8

Frictional force = 66.89N

Work done will be the product of resultant Force F and the distance travelled

(F - 66.89) x 2.6 = 411.94

F - 66.89 = 411.94/2.6

F - 66.89 = 158.44

F = 225.3 N

The second block will experience the same force which is equal to 225.3N

Find the kinetic energy of the second block.

K.E =  [tex]1/2mV^{2}[/tex]

K.E = 0.5 x 8.25 x 15.36^2

K.E = 973.2

Using The work done = Kinetic energy

225.3[tex]d_{2}[/tex] = 973.2

[tex]d_{2}[/tex] = 973.2/225.3

[tex]d_{2}[/tex] = 4.32 meters

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