A boat is cruising in a straight line at a constant speed of 2.5 m/s when it is shifted into neutral. After coasting 14 mm the engine is engaged again, and the boat resumes cruising at the reduced constant speed of 1.5 m/s. Assuming constant acceleration while coasting,

a. How much time did it take for the boat to coast the 12 m?
b. What was the boat’s acceleration while it was coasting?
c. What was the speed of the boat when it had coasted for 6.0 m? Explain.

Answer:

100 or 95

Explanation:

GPE=M*GFS*change in height

GPE=4.0*10*2.5

=100

For the gravitational field strength you can use 10 or 9.5 but if it says specifically to use a certain value, than use that

Answer:

form 1 question??????????

Answer:

The strong person should carry the ladder at the  front end and the weak person should carry it at the back end.

Explanation:

this is because in such a case the strong person has to pull the ladder whereas the weak person at the back end have to push the ladder. In such case it is easier to push because the weak person can use the force of gravity of his own body for pushing the ladde.

However in case of pulling the ladder one has to overcome his own gravity to pull the heavy object

Answer:

  weak person: near the end of the ladder

  strong person: nearer the center of the ladder

Explanation:

The closer the strong person is to the center of mass of the ladder, the greater the fraction of the ladder's weight that person will carry. The weak person will feel less weight if the strong person is nearer to the center of mass.

For example, if the strong person is 3/4 of the ladder's length from the end where the weak person is, then the strong person will carry twice the weight the weak person is having to carry. (Assuming the mass is uniformly distributed.)

Answer:

A rider.

Explanation:

A rider designates a clause or group of clauses added to an initial contract, being an annex to it.  These added clauses are attached to the initial contract, integrating to it and losing the character of an independent document. Therefore, it would then be a piece of writing that is to be treated as part of the main contract. Usually, the purpose of the riders is to include particular clauses for the insured, that is, clauses that are specific to them outside of the company's generic contract.

Answer:

A rider, for plato users

Explanation:

Answer:

It's true I took the test on Edge.

Explanation:

Answer:

True

Explanation:

Got it right on edg

Corrected, it's 2) 200

Answer:

30.63 m

Explanation:

Using y = ut + 1/2gt² where u = initial speed of block = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time of fall = 2.5 s and y = height of fall.

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

y = ut + 1/2gt²

y = 0 m/s × 2.5 s + 1/2 × 9.8 m/s² × (2.5 s)²

y = 0 m + 4.9 m/s² × 6.25 s²

y = 0 m + 30.625 m

y = 30.625 m

y ≅ 30.63 m

So, the brick fell 30.63 m

Answer:

When the mass increases or when distance between the bodies reduces

Explanation:

According to Newton's law of universal gravitation, the gravitational attraction between two bodies always increase if the mass increases and the distance between the bodies reduces.

The law of universal gravitation states that "the gravitational force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".

Mathematically;

      Fg  = [tex]\frac{G m1 m2}{r^{2} }[/tex]  

G is the universal gravitation constant

m is the mass

r is the distance

Explanation:

the gas will be hit by high energy particles as the radioactive decay happens and the decaying nuclei are breaking apart. these particles energize the gas and cause electrons to move around the orbitals. the flow of electrons is detected by the sensor.

Answer:

277.5 N.

Explanation:

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

Mass (m) of astronaut = 75 kg

Acceleration due to gravity on Mar (gₘ) = 3.7 m/s²

Weight of astronaut on Mar (Wₘ) =.?

Weight of an object is simply defined by the following equation:

Weight (W) = mass × acceleration due to gravity (g)

W = m × g

With the above formula, we can obtain the weight of the astronaut on Mar as follow:

Mass (m) of astronaut = 75 kg

Acceleration due to gravity on Mar (gₘ) = 3.7 m/s²

Weight of astronaut on Mar (Wₘ) =.?

Wₘ = m × gₘ

Wₘ = 75 × 3.7

Wₘ = 277.5 N

Thus, the weight of the astronaut on Mar is 277.5 N

Answer:

the average net force on the truck is 3744 N.

Explanation:

Given;

mass of the truck, m = 1200 kg

initial velocity of the truck, u = 25 m/s

final velocity of the truck, v = 53 m/s

distance traveled by the truck, d = 350 m

The acceleration of the truck is calculated as;

v² = u² + 2ad

53² = 25² + (2 x 350)a

700a = 53² - 25²

700a = 2184

a = 2184 / 700

a = 3.12 m/s²

The average net force on the truck is calculated as;

F = ma

F = 1200 x 3.12

F = 3744 N

Therefore, the average net force on the truck is 3744 N.

Answer:

The Carnot engine has zero power

Explanation:

Although theoretically the Carnot engine has more efficiency than the real engine. In practice however they tend to have zero power.

This is because all its processes are reversible (that is isothermic and adiabatic).

So the system equilibrates with its surroundings at every point in time. This makes work done very slow and the power generated is zero.

Carnot cycles requires attaining isothermal heat transfers which is quite difficult and take a long time. Also a pump that can handle liquid-vapour phase mixture will be required.

This is not practical.

Answer:

Explanation:

The tank is half full so height of water level is at 1 m . Centre of gravity of water will be at height of .5 m . Pressure will act at this point .

Pressure = h d g where h is height of centre of mass of water column , d is density of water and g is acceleration due to gravity .

Pressure of water column  = .5 x 10³ x 9.8 = 4.9 k Pa .

Air is pressurized to 5 kPa so

resultant pressure  on one side of the tank due to the air and water

= 4.9 + 5 kPa = 9.9 kPa .

Total force on one face = pressure x area of one face under water

= 9.9 x 10³ x .5 x 2²

= 19.8 kN .

Answer:

a) A:170572.5 J

   C: 55794.9J

b) 170572.5 J

c) 41.4413265306m

d) 2.7455874717m/s

Explanation:

a) Kinetic energy = 0.5*m*v²

KE at A = 0.5*420*28.5² = 170572.5 J

KE at C = 0.5*420*16.3² = 55794.9 J

b) Mechanical energy is the total kinetic energy plus potential energy at a point on the system. There is no potential energy at A.

ANSWER: 170572.5 J

c) v²=u²+2as

28.5²=2(9.8)s

812.25/19.6=s

s=41.4413265306m

d) h=height from part c, r=radius of loop

v²=u²+2as

v²=gr or a=v²/r

Ei=Ef

mgh=0.5mv²+mg(2r)

gh=0.5v²+2gr

h=0.5r+2r

h=5/2r

r=2/5h=(2/5)(41.4413265306)=16.5765306122

F=ma

mg=m(v²/r)

g=v²/r

v²=(9.8)(16.5765306122)

v=√162.45

=12.7455874717m/s

Answer:

6.86 m/s

Explanation:

The minimal velocity needed is when we have only vertical motion, then i will think in the problem only in one axis.

I suppose that the only force, in this case, is the gravitational force acting on the fish.

Then the gravitational equation of the fish will be:

a(t) = -9.8m/s^2

For the velocity equation we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial velocity of the fish and is what we want to find.

For the position equation we need to integrate over time again to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + v0*t + p0

p0 is the initial position of the fish, and because he starts one the water, the initial position is p0 = 0 m

Then the equation is:

p(t) = (1/2)*(-9.8 m /s^2)*t^2 + v0*t

p(t) = (-4.9 m/s^2)*t^2 + v0*t

We know that the maximum height is 2.4m

The value of time at which the fish gets his maximum height is when the velocity of the fish is equal to zero, then we first need to solve:

v(t) = (-9.8m/s^2)*t + v0 = 0

      t = v0/9.8m/s^2

Now we replace this in the position equation to get the maxmimum height, which is equal to 2.4m

2.4m = p( v0/9.8m/s^2) =  (1/2)*(-9.8 m /s^2)*(v0/9.8m/s^2)^2 + v0*(v0/9.8m/s^2)

2.4m = (1/2)(-v0)^2(-9.8 m /s^2) + v0^2/(9.8m/s^2))

2.4m = (1 - 1/2)*v0^2/(9.8m/s^2)

2.4m = 0.5*v0^2/(9.8m/s^2)

2.4m/0.5 = v0^2/(9.8m/s^2)

4.8m*(9.8m/s^2) = v0^2

√(4.8m*(9.8m/s^2)) = v0 = 6.86 m/s

Answer:

Ps eso me salio

Physical education, sports education or sports education are terms that refer to the teaching and learning of physical exercises whose main objective is education and health. This has been the decisive reason for the introduction of physical exercises in elementary school in the 19th century.

Explanation:

Answer:

[tex]1270.64\ \text{J}[/tex]

Explanation:

m = Mass of object = [tex]\dfrac{mg}{g}[/tex]

mg = Weight of object = 20 N

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

[tex]\theta[/tex] = Angle of slope = [tex]30^{\circ}[/tex]

f = Force of friction

fd = Work done against friction

The force balance of the system is

[tex]\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}[/tex]

The work done against friction is [tex]1270.64\ \text{J}[/tex].

Answer:

the peak vertical velocity of the boat is 0.3267 m/s

Explanation:

Given that;

Time period T = 10 sec

wave velocity = 5.0 m/s

wave amplitude A = 0.520 m

peak vertical velocity of in the wave can be expressed as;

Vp = Aw

where w is angular frequency,  ( w = 2π/T)

so

Vp = A × 2π/T

so we substitute

Vp = 0.520 × 2π/10

Vp = 3.2673 / 10

Vp = 0.3267 m/s

Therefore, the peak vertical velocity of the boat is 0.3267 m/s

Answer:

The magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

Explanation:

Since vector A has magnitude 50 cm and a direction of 30, its x - component is A' = 50cos30 = 43.3 cm and its y - component is A" = 50sin30 = 25.

Also, Since vector B has magnitude 35 cm and a direction of 110, its x - component is A' = 35cos110 = -11.97 cm and its y - component is A" = 35sin110 = 32.89 cm.

So, the vector sum R = A + B

The x-component of the vector sum is R' = A'+ B' = 43.3 cm + (-11.97 cm) = 43.3 cm - 11.97 cm = 31.33 cm

The y-component of the vector sum is R" = A"+ B" = 25 cm + 32.89 cm = 57.89 cm

So, the magnitude of R = √(R'² + R"²)

= √((31.33 cm)² + (57.89 cm)²)

= √(981.5689 cm² + 3,351.2521 cm²)

= √(4,332.821 cm²)

= 65.82 cm

≅ 65.8 cm

The direction of R is Ф = tan⁻¹(R"/R')

= tan⁻¹(57.89 cm/31.33 cm)

= tan⁻¹(1.84775)

= 61.58°

≅ 61.6°

So, the magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

Answer:

3.33m

Explanation:

Given parameters:

Force applied  =500N

Elastic constant  = 150N/m

Unknown:

Amount of stretch or extension  = ?

Solution:

To solve this problem use the expression below:

    F  = k e

F is the force applied

k is the elastic constant

e is the extension

  So;

           500  = 150 x e

             e  = [tex]\frac{500}{150}[/tex]    = 3.33m

Answer:

The force and texture of an object matter a lot.

Explanation:

If you were to try and run up a glass hill, could you? You maybe could, but it would be harder to than up carpet.

Answer:

40 N

Explanation:

According to the scenario, computation of given data are as follows:

Mass (m) = 20 kg

Initially moving (v) = 3

Actual distance (d) = 2.25 m

So, we can calculate friction (f) by using following formula,

f × d = [tex]\frac{1}{2} mv^{2}[/tex]

By putting the value, we get

f × 2.25 = [tex]\frac{1}{2}[/tex] × 20 × [tex]3^{2}[/tex]

f × 2.25 = 10 × 9

f = 90 ÷ 2.25

= 40 N.

Answer:

Unbalanced Forces.

Explanation:

Answer:

Unbalanced

Explanation:

Unbalanced forces are not equal and they always cause the motion of an object to change the speed and or direction of the object.

Balanced forces are equal and will not cause the motion of an object to change.

From Newton's law of inertia "an object will continue in its state of rest or of uniform motion unless acted upon by an external force".

The net force for a body with unbalanced forces is greater than zero.

For a body with balanced forces, the net force is zero

Answer:

a = 5 [m/s²]

Explanation:

We know from Newton's second law that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = total force = 50 [N]

m = mass = 10 [kg]

a = acceleration [m/s²]

Now replacing:

[tex]a = \frac{F}{m} \\a=50/10\\a = 5 [m/s^{2} ][/tex]

Answer:

12J

Explanation:

Given parameters:

Mass  = 1.5kg

Impulse  = 6kgm/s

 let us start first by find the velocity with which this body moves;

       Impulse  = mass x velocity

         Velocity  = Impulse /  mass  = 6/ 1.5  = 4m/s

Initial velocity  = 0m/s

Unknown:

Resulting kinetic energy  = ?

Solution:

To solve this problem use the formula below:

      K.E  = [tex]\frac{1}{2}[/tex]  m (v  - u)²  

m is the mass

v is the final velocity

u is the initial velocity

  So;

        K.E  =  [tex]\frac{1}{2}[/tex]  x 1.5 x (4 - 0)²  

  K.E  = 1.5 x 8  = 12J

Answer:

P.E = 98 Joules

Explanation:

Given the following data;

Mass = 5kg

Speed = 4m/s

Height = 2m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

[tex] P.E = mgh[/tex]

Where, P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the equation, we have;

[tex] P.E = 5*9.8*2[/tex]

P.E = 98 Joules

Answer:

Explanation:

In the whole process , potential energy of the cart is converted into kinetic energy . At the top of the vertical loop , the whole of potential energy is regained and kinetic energy becomes zero if we release the cart from a height of 2R because difference of height between lowest and highest point of motion  is 2R .  In that case kinetic energy at top = 0 , velocity v = 0

At the top , weight mg is acting which is providing centripetal force . So cart must have some velocity at the top . If it be v

mv²/R = mg

v = √ gR .

For that purpose , the cart must be released from a height greater than 2R .

The extra height beyond 2R will make the velocity at the top non-zero.

Answer:

the magnitude of Vpg = 493.711 km/h

Explanation:

given data

speed Vpg = 560 km/h

speed Vwg = 80 km/h

solution

we get here magnitude of the plane velocity w.r.t. ground is

we know that the Vpg = Vpw + Vwg       .....................1

writing the component of the velocity that is

Vpw = (0 km/h î - 560 km/h j )

Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)

adding these

Vpg = (0+80 cos 45 km/h ) î  + ( -560 + 80 sin 45 km/h j)i

Vpg = (42.025 )  î  (-491.92 km/h)j

now we take magnitude

the magnitude of Vpg = [tex]\sqrt{(42.025^2+(-491.92)^2)} km/h[/tex]

the magnitude of Vpg = 493.711 km/h

Answer:

I think it might be C.

Explanation:

Answer:

c. The large particles in a suspension eventually settle to the bottom of the container.

Explanation:

The correct statement from the given choices is that the large particles in a suspension eventually settle to the bottom of the container.

A suspension is a mixture made up of small insoluble particles of a solid in a liquid or gas.

They have the following properties:

This question is incomplete. the complete question is;

a) Suppose that the estimate of the maximum height that a player can throw a baseball is 25 ft. For how long after the ball leaves the player's hand does it return to the players hand?

b) Suppose that the estimate of the distance that the ball moves while a player is throwing it is 2.5 feet -- that is, the distance from where the ball is when the player starts their throw until it leaves their hand. Calculate the magnitude of the average acceleration in m/s2m/s2 that the ball has while it is being thrown, as it moves from rest to the point where it leaves the player's hand.

Answer:

a) Total time of Flight is 2t is 2.49 seconds

b) The magnitude of the average acceleration is 98.03 m/s²

Explanation:

Given the data in the question;

a)

Maximum height [tex]h_{max}[/tex] = 25 ft = (25 /3.281) = 7.62 m

For how long after the ball leaves the player's hand does it return to the players hand;

[tex]h_{max}[/tex]  = [tex]\frac{1}{2}[/tex]gt²

we know that g = 9.81 m/s²

so we substitute

7.62 m =  [tex]\frac{1}{2}[/tex] × 9.81 m/s² × t²

7.62 m = 4.905m/s² × t²

t² = 7.62 m / 4.905

t² = 1.5535

t = √1.5535

t = 1.246 s

So total time of Flight is 2t = 2 × 1.246 s = 2.49 seconds

b)

given that

distance = 2.5 ft = ( 2.5 / 3.281) = 0.762 m

V = ?

from the First Equation of Motion

v = u + at

0 = u - 9.81 × 1.246 s

u = 12.2232 m/s

so from the Third Equation of Motion : v² = u² + 2as

(12.2232 m/s)² = 0² + 2 × a × 0.762 m

149.4066  = 1.524a

a = 149.4066 / 1.524

a = 98.03 m/s²

Therefore, the magnitude of the average acceleration is 98.03 m/s²