A level test track has a coefficient of road adhesion of 0.80, and a car being tested has a coefficient of rolling friction that is approximated as 0.018 for all speeds. The vehicle is tested unloaded and achieves the theoretical minimum stop in 180 ft (from brake application). The initial speed was 60 mi/h. Ignoring aerodynamic resistance, what is the unloaded braking efficiency

Answer:

Displacement

Explanation:

The quantity 45m north is a typical example of displacement.

Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.

Answer:

Explanation:

a

Answer:

It determines how long you do a certain workout.

Answer:

The answer is -30.95.

Explanation:

Use the lens equation: 1/focal length = 1/object distance + 1/image distance. The answer comes out to -30.95. This is correct on Acellus.

When it is held 5.20 cm in front of an object, the image distance will be "-30.95 cm". To understand the calcultaion, check below.

According to the question,

Object distance, u = -5.20 cm

Focal length, f = 6.25 cm

By using the Lens formula, we get

→ [tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u}[/tex]

or,

→ [tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]

By substituting the values, we get

     [tex]= \frac{1}{6.25} - \frac{1}{5.20}[/tex]

  [tex]\frac{1}{v} = -\frac{21}{650}[/tex]

By applying cross-multiplication,

   v = -30.95 cm

Thus the above answer is correct.  

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Without friction, NO.

The speed at D depends only on the difference in height between A and D. Whatever happens between them doesn't matter.

The speed of the coaster car at point D will be affected if  the height of point C is changed.

Potencial Energy:

It is the enrgy in a body due to the position of differnt part of the object or system.

As we increase the the hight of the car the potetial enrgy increase, the gravitational acceleration on car will be more due to the high of the point C.

Therefore, the speed of the coaster car at point D will be affected if  the height of point C is changed.

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Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

so

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

v₂ = √( 3457.44 - 325.164 )  

v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

Answer:

49 Ns

Explanation:

Given data

Force= 14N

time = 3.5seconds

Applying the expression for impulse

P= Ft

substitute

P=14*3.5

P=49 Ns

Hence the impulse is 49 Ns

Answer:

a delta

Explanation:

The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.

A delta is a depositional landform where a smaller body of water enters into a larger one.

The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.

So, this feature is a delta.  

Answer:

c

Explanation:

Answer:

c

Explanation:

Answer:

 t = [tex]\frac{ v \ F}{ m}[/tex]

Explanation:

The question is a bit strange, for this exercise we must use the mathematical relationship of Newton's second law to find the acceleration of the body

          F = m a

          a = F / m        (1)

with this acceleration the mathematical relations of kinematics of accelerated motion must be used

          v = v₀ + a t

with the body starting from rest its initial velocity is zero

          v = a t

          t = v / a          (2)

if we substitute the equation 1 in 2

          t = [tex]\frac{ v \ F}{ m}[/tex]

this is the final mathematical expression that allows to find the time based on the data of the problem

Answer:

the total distance the bird travels before the trains collide is 69.54 km

Explanation:

Given the data in the question,

dA = 22km/h × t

dB = 51km - 22km/h × t

now we find the time, when the two trains collide

dA = Db

22km/h × t =  51km - 22km/h × t

44km/h × t =  51km

t = 51km / 44km/h

t = 1.159 hrs

so the bird can fly back and forth for 1.159 hrs  before the train collide.

hence, distance travelled by the bird in total will be;

d = v × t

we substitute

d = 60 km/h × 1.159 h

d = 69.54 km

Therefore, the total distance the bird travels before the trains collide is 69.54 km

Answer:

40 kg m/s

Explanation:

Given the following data;

Mass of boy = 40kg

Mass of ball = 2kg

Velocity = 20m/s

To find the momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum =mass * velocity

Substituting into the equation, we have

Momentum = 2 * 20

Momentum = 40 kg m/s

Answer:

T = 10010 N

Explanation:

To solve this problem we must use the translational equilibrium relation, let's set a reference frame

X axis

       Fₓ-Fₓ = 0

       Fₓ = Fₓ

whereby the horizontal components of the tension in the cable cancel

Y Axis  

        [tex]F_{y} + F_{y} - W =0[/tex]

        2[tex]F_{y}[/tex] = W

let's use trigonometry to find the angles

        tan θ = y / x

        θ = tan⁻¹ (0.30 / 0.50 L)

        θ = tan⁻¹ (0.30 / 0.50 15)

        θ = 2.29º

the components of stress are

         F_{y} = T sin θ

we substitute

       2 T sin θ = W

       T = W / 2sin θ

        T = [tex]\frac{ 800}{ 2sin 2.29}[/tex]

        T = 10010 N

Answer:

70 kg

Explanation:

divide it by 2

Hope this helped!

Answer:

63.502932 Kilograms

Explanation:

Answer:

[tex]a_2=24.5\ \text{m/s}^2[/tex] towards right

[tex]a_1=2.1875\ \text{m/s}^2[/tex] towards left

Explanation:

[tex]m_1[/tex] = Mass of skater = 56 kg

[tex]m_2[/tex] = Mass of ball = 5 kg

v = Final velocity of ball = 7 m/s

u = Initial velocity of ball = 0

s = Distance the ball moved in the hand of the skater = 1 m

Moving left is considered and moving right is considered positive.

From kinematic equations of motion we have

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{7^2-0^2}{2\times 1}\\\Rightarrow a=24.5\ \text{m/s}^2[/tex]

So, the ball will move towards right with a magnitude of acceleration [tex]a_2=24.5\ \text{m/s}^2[/tex].

The force on the ball will be

[tex]F_2=m_2a_2\\\Rightarrow F_2=5\times 24.5\\\Rightarrow F_2=122.5\ \text{N}[/tex]

The force on the ball is [tex]122.5\ \text{N}[/tex]

The reaction force on the skater will be equal to the force on the ball but will have opposite direction.

[tex]-F_1=F_2\\\Rightarrow -m_1a_1=F_2\\\Rightarrow a_1=-\dfrac{122.5}{56}\\\Rightarrow a_1=-2.1875\ \text{m/s}^2[/tex]

So, the skater will move towards left with a magnitude of acceleration [tex]2.1875\ \text{m/s}^2[/tex]

After the ball is thrown, the skater will accelerate to the left at 2.19 m/s² while the ball will accelerate to the right at 24.5 m/s².

The given parameters

The acceleration of the ball as it leaves the skaters hand is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\ v^2 = 0 + 2as\\\\ a = \frac{v^2}{2s} \\\\ a = \frac{7^2}{2(1)} \\\\ a = 24.5 \ m/s^2[/tex]

The force experienced by the ball is calculated as follows;

[tex]F = ma\\\\ F_b = 5 \times 24.5\\\\ F_b = 122.5 \ N[/tex]

The force experienced by the skater is equal in magnitude to that of the ball but opposite in direction.

[tex]F_s = -F_b\\\\ F_s = -122.5 \ N[/tex]

The acceleration of the skater after the ball was thrown is calculated as follows;

[tex]F = ma \\\\ a = \frac{F}{m} \\\\ a = \frac{-122.5 }{56} \\\\ a =-2.19 \ m/s^2[/tex]

Thus, after the ball is thrown, the skater will accelerate to the left at 2.19 m/s² while the ball will accelerate to the right at 24.5 m/s².

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Answer:

Explanation:

The same as its weight before adding the duck

Answer and Explanation: No, the explanation is not plausible. The puck sliding on the ice is an example of the Principle of Conservation of Energy, which can be enunciated as "total energy of a system is constant. It can be changed or transferred but the total is always the same".

When a player hit the pluck, it starts to move, gaining kinetic energy (K). As it goes up a ramp, kinetic energy decreases and potential energy (P) increases until it reaches its maximum. When potential energy is maximum, kinetic energy is zero and vice-versa.

So, at the beginning of the movement the puck only has kinetic energy. At the end, it gains potential energy until its maximum.

The representation is as followed:

[tex]K_{i}+P_{i}=K_{f}+P_{f}[/tex]

[tex]K_{i}+0=0+P_{f}[/tex]

[tex]\frac{1}{2}mv^{2} = mgh[/tex]

As we noticed, mass of the object can be cancelled from the equation, making height be:

[tex]h=\frac{v^{2}}{2g}[/tex]

So, the height the puck reaches depends on velocity and acceleration due to gravity, not mass of the puck.

Answer:

Explanation:

Its 165.5m

Answer:

2577 K

Explanation:

Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.

So, T = ⁴√(P/σεA)

Since P = 60 W, we substitute the vales of the variables into T. So,

T = ⁴√(P/σεA)

= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)

= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)

= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)

= ⁴√(0.00441 × 10¹⁶K⁴)

= 0.2577 × 10⁴ K

= 2577 K

Explanation:

False, it is oppisite the shorter the wavelength the more energy it carries.

[tex]{\large{\bold{\rm{\underline{Given \; that}}}}}[/tex]

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

[tex]{\large{\bold{\rm{\underline{To\; find}}}}}[/tex]

★ The speed of the hound and the hare

[tex]{\large{\bold{\rm{\underline{Solution}}}}}[/tex]

★ The speed of the hound and the hare = 25:18

[tex]{\large{\bold{\rm{\underline{Full \; Solution}}}}}[/tex]

[tex]\dashrightarrow[/tex]  As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

 So firstly let us assume a metres as the distance covered by the hare in one leap.

Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.

 But 3 leaps of the hound are equal to 5 leaps of the hare.

Henceforth, (5/3)a meters is the distance that is covered by the hound.

 Now according to the question,

Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)

Now the distance travelled by the hound in it's 5 leaps..!

 Now the distance travelled by the hare in it's 6 leaps..!

 Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!

Answer:

the correct one is: a diffraction limits the resolving power to approximately the size of the wavelength of the light used

Explanation:

To be able to solve two structures with a light source, the Rayleigh criterion must be met that stable the two structures are solved when the first minimum of diffraction at one point is in the code of the first maximum of the other point

Using this criterion we can find an expression for the first minimization of the diffraction spectrum m = 1

     sin θ tea = λ / a

now the structure of the comatose has a separation of around 1 nm and the wavelength of visible light ranges from 400 to 700 nm, when substituting we find

           sin θ = 400/1 10

           sin θ  = 400

           sin θ  = 700/1

           sin θ  = 700

These values ​​are neither impossible since the sin function is bounded between -1 to 1, so we cannot see the diffraction

When reviewing the different statements, the correct one is: a diffraction limits the resolving power to approximately the size of the wavelength of the light used:

It would be impossible to build a microscope that could use visible light to

see the molecular structure of a crystal because diffraction limits the

resolving power to about the size of the wavelength of the light used.

Diffraction occurs when waves encounter an obstacle which leads to it

bending around it. Visible light when diffracted results in the overlapping of

the patterns thereby formation of images as one instead of in different parts.

This thereby leads to the resolving power being limited to the size of the

wavelength of light used and images won't be able to be identified

separately.

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Answer:

Explanation:

a ) distance s = 12 m .

constant speed = 1.5 m/s

time taken to coast 12 m

= distance / speed

= 12 / 1.5 = 8 s

b ) initial velocity u = 2.5 m /s

final velocity v = 1.5 m /s

displacement s = 14 m

acceleration a = ?

v² = u² + 2 as

1.5² = 2.5² + 2 a x 14

a = - .1428 m /s²

= - 14.28 cm / s²

c )

v = ?

u = 2.5 m /s

s = 6 m

a = -   .1428 m /s²

v² = u² - 2 as

= 2.5² - 2 x 6 x .1428

= 6.25 - 1.71

= 4.54

v = 2.13 m /s .

Answer:

[tex]Vm=0.894m/s[/tex]

Explanation:

From the question we are told that

Velocity if travel [tex]v=4m/s[/tex]

Diameter of  prototype [tex]d_1=20m[/tex] and [tex]d_2=110m[/tex]

Scale ratio=[tex]\frac{1}{20}[/tex]

Generally Velocity of of the model using Froud's model is mathematically given as

[tex]Fm=Fp[/tex]

[tex]\frac{Vm}{\sqrt{Lmg}} =\frac{Vp}{\sqrt{Lpg}}[/tex]

[tex]Vm=Vp*\frac{Vp}{\sqrt{Lpg} }[/tex]

[tex]Vm=4*\frac{1}{\sqrt{20}}[/tex]

[tex]Vm=0.894m/s[/tex]

The force exerted by the person is an upward force equal to the weight of the box, and that force is perpendicular to the motion. If there is no motion in the direction of the force, then no work is done by that force. Yet you certainly feel like you are doing work if you carry a heavy box.

For the special case of a constant force, the work may be calculated by multiplying the distance times the component of force which acts in the direction of motion.*picture 2*

credits-goo.gle,hyper physics

A woman walks in a  straight line with the sun to her right at six o'clock in the morning.

The sun rises East of her, so the woman is walking toward the North pole.

A man walks in a straight line with the sun to his right at six o'clock in the  evening.

The sun sets West of him, so the man is walking toward the South pole.

The woman and the man are both walking along lines of constant longitude.

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ([tex]\vec r_{cm}[/tex]), measured in meters, is defined by this weighted average:

[tex]\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}[/tex] (1)

Where:

[tex]m_{i}[/tex] - Mass of the i-th particle, measured in kilograms.

[tex]\vec r_{i}[/tex] - Location of the i-th particle with respect to origin, measured in meters.

If we know that [tex]\vec r_{cm} = (-0.500\,m,-0.700\,m)[/tex], [tex]m_{1} = 1\,kg[/tex], [tex]\vec r_{1} = (-1.20\,m, 0.500\,m)[/tex], [tex]m_{2} = 4.50\,kg[/tex], [tex]\vec r_{2} = (0.600\,m, -0.750\,m)[/tex] and [tex]m_{3} = 4\,kg[/tex], then the coordinates of the third particle are:

[tex](-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}[/tex]

[tex](-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})[/tex]

[tex](4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)[/tex]

[tex](x_{3},y_{3}) = (-1.562\,m,-0.944\,m)[/tex]

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Answer:

Explanation:

Kinetic energy of block will be converted into heat energy by friction .

Heat energy produced = 1/2 m v²

= .5 x 4.8 x 1.2²

= 3.456 J

85% of energy is converted into heat energy , so heat energy produced

= .85 x 3.456 = 2.9376 J .

If Q heat is given to m mass of object having s as specific heat and Δt is increase in temperature

Q = msΔt

specific heat of iron s = 462 J / kg C

Putting the values ,

2.9376 = 4.8 x 462 x Δt

Δt = 13.24 x 10⁻⁴ ⁰C.