Ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.6 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.

Required:
a. How far from the wall is she?
b. What is the period of her up-and-down motion?

Answer:

a. 1 Newton = 100000 Dyne

b. a represents acceleration.

Explanation:

Newton is the standard unit (S.I) of measurement of force. Converting 1 Newton to dyne we have;

1 Newton = 10⁵ Dyne

1 Newton = 100000 Dyne

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

Force = mass * acceleration

[tex] F = ma[/tex]

Hence, we can deduce that a represents the acceleration of an object and it's measured in meters per seconds square.

Answer: hello some data related to your question is missing attached below is the missing data and diagram related to the solution

answer:

P = 141.21 N

Explanation:

Given data:

Mass of crate = 50 kg

coefficient  of static friction ( μ ) = 0.25

Calculate minimum horizontal force ( P ) that holds the crate from sliding

∑fx = 0

     = P + Fcos θ - N*sinθ = 0

     = P + 0.25N cos 30° - Nsin30°  = 0

∴ P = 0.2835 N = 0

P - 0.2853 N = 0 ------- ( 1 )

∑fy = 0

     - 50g + Ncosθ + Fsinθ

     - 50*9.81 + Ncos30° + 0.25Nsin30°

∴ N = 494.942 N ----- ( 2 )

input 2 into 1

P - 0.2853 ( 494.942 ) = 0

P = 141.21 N

Answer:

The right solution is "50200 days".

Explanation:

Given:

Calories intake,

= 6000 kcal,

or,

= [tex]2.52\times 10^7 \ J[/tex]

Force,

= 500 N

As we know,

⇒ [tex]Work \ done = Force\times distance[/tex]

Or,

⇒ [tex]distance = \frac{Work \ done}{Force}[/tex]

By putting the values, we get

                  [tex]=\frac{2.52\times 10^7}{500}[/tex]

                  [tex]=0.502\times 10^5[/tex]

                  [tex]=50200 \ m[/tex]

hence,

The number of days will be:

= [tex]\frac{50200}{1}[/tex]

= [tex]50200 \ days[/tex]

Answer:

[tex]P=217600Pa[/tex]

Explanation:

From the question we are told that:

Density [tex]\rho=1000kg/m^3[/tex]

Depth of Water [tex]d=12.0m[/tex]

Generally the equation for Pressure is mathematically given by

 [tex]P=\rho gh[/tex]

 [tex]P=1000*9.8*12[/tex]

 [tex]P=117600N/m^2[/tex]

Therefore

Absolute Pressure=P+P'

Where

P=Pressure under water

P'=Atmospheric Pressure

Therefore

 [tex]P_A=P+P'[/tex]

 [tex]P_A=117,600+10^5[/tex]

 [tex]P=217600Pa[/tex]

Explanation:

time=Distance/speed

t=32/8

t=4 seconds

Answer:

30

Explanation:

since  [tex]\frac{30}{5}[/tex]=6

         [tex]\frac{30}{6}[/tex]=5

then both 5 and 6 are factors of 30

Have a nice day

Answer:

Explanation:

From the given information:

the car's momentum = momentum of the truck

∴

(a) 816 kg × v = 2650 kg × 16.0 km/h

v = (2650 kg × 16.0 km/h) /  816 kg

v = 51.96 km/hr

(b) 816 kg × v = 9080 kg × 16.0 km/h

v = (9080 kg × 16.0 km/h) /  816 kg

v = 178.04 km/hr

Two connection methods are used for brushless DC motors. One method is to connect the coils in a loop as we compared it with the rotor winding of DC motors in Fig. 2.27. This method is called a Δ (delta) connection.

CORRECT ME IF IM WRONG!!

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Answer:

Answer:

A). The paper airplane’s motion is due to horizontal inertia and the vertical pull of gravity.

Explanation:

Edge.

Answer:

The motion of the paper airplane  is best explained by horizontal inertia and vertical pull of gravity.

Explanation:

Inertia is the property by which the body wants to remain in its position unless any external for is applied. Here horizontal inertia is inertia of motion which is acting horizontally .

While vertical pull is due to the earth .

Hence horizontal inertia and vertical pull best explain the projectile motion of paper airplane.

Also read it;

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Complete question is;

Use a variation model to solve for the unknown value.

The stopping distance of a car is directly proportional to the square of the speed of the car.

a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.

b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?

Answer:

A) d = 333.2 ft

B) 60 mph

Explanation:

Let the stopping distance be d

Let the speed of the car be v

We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;

d ∝ v²

Therefore, d = kv²

Where k is constant of variation.

A) Speed is 50 mph and stopping distance of 170 ft.

v = 50 mph

d = 170 ft = 0.032197 miles

Thus,from d = kv², we have;

0.032197 = k(50²)

0.032197 = 2500k

k = 0.032197/2500

k = 0.0000128788

If the car is now travelling at 70 mph, then;

d = 0.0000128788 × 70²

d = 0.06310612 miles

Converting to ft gives;

d = 333.2 ft

B) stopping distance is now 244.8 ft

Converting to miles = 0.046363636 miles

Thus from d = kv², we have;

0.046363636 = 0.0000128788(v²)

v² = 0.046363636/0.0000128788

v² = 3599.99658

v = √3599.99658

v ≈ 60 mph

Explanation:

Energy input = F×d = (150 N)(5.5 m) = 825 J

Energy output = mgh = (66 kg)(9.8 m/s^2)(1.10 m) = 711 J

efficiency = [tex]\dfrac{\text{output}}{\text{input}}[/tex]×100% = 86.2%

Answer:

a) [tex]T=0.01s[/tex]

b) [tex]T=0.001s[/tex]

c) [tex]T=0.00001s[/tex]

Explanation:

From the question we are told that:

Given Frequencies

a. 100 Hz,

b. 1 kHz,

c. 100 kHz.

Generally the equation for Waveform Period is mathematically given by

[tex]T=\frac{1}{f}[/tex]

Therefore

a)

For

[tex]T=100 Hz[/tex]

[tex]T=\frac{1}{100}[/tex]

[tex]T=0.01s[/tex]

b)

For

[tex]F=1kHz[/tex]

[tex]T=\frac{1}{1000}[/tex]

[tex]T=0.001s[/tex]

c)

For

[tex]F=100kHz[/tex]

[tex]T=\frac{1}{100*100}[/tex]

[tex]T=0.00001s[/tex]

Answer:

So the answer is B. A is wrong because negative answer = deceleration

We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...

Answer: The weight of the object is 29.4 N

Explanation:

To calculate the weight of the object, we use the equation:

[tex]W=m\times g[/tex]

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = [tex]9.8m/s^2[/tex]

Putting values in above equation, we get:

[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]

Hence, the weight of the object is 29.4 N

Answer:

Explanation:

Here's what we know and in which dimension:

y dimension:

[tex]v_0=30[/tex] m/s

v = 0 (I'll get to that injust a second)

a = -9.8 m/s/s

The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.

x dimension:

Δx = 70 m

v = ??

Velocity is our unknown.

Solving for the time in the y dimension:

[tex]v=v_0+at[/tex] and filling in:

0 = 30 + (-9.8)t and

-30 = -9.8t so

t = 3.1 seconds

We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.

In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.

Δx = vt and

70 = v(6.2) so

v = 11.3 m/s

Answer:

B

Explanation:

The seasons are measured in how far or close the earth is to the sun.

Answer:

By measure the effective length and the time period of the pendulum.

Explanation:

Let the student take the oscillating pendulum at the planet.

He measure the time period of the pendulum  by using the stop watch or the ordinary watch.

Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.

Now, use the formula of the time period of the pendulum,

[tex]T =2\pi\sqrt\frac{L}{g}[/tex]

Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.  

By rearranging the terms, we get

[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]

Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.

Answer:

less than the weight of the block.

Explanation:

From the free body diagram, we get.

The normal force is N = Mg cosθ

The tension in the string is T = Mg sinθ

Wight of the block when the block is static, W = Mg

Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,

therefore, the normal force is less than the weight of the static block.

Answer:

A

Explanation:

I guess not that much confidential!

Answer:

[tex]\mu=0.185[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=17kg[/tex]

Force [tex]F=33N[/tex]

Velocity [tex]v=1.6m/s[/tex]

Distance [tex]d= 9.8m[/tex]

Generally the equation for Work done is mathematically given by

 [tex]W=\triangle K.E+\triangle P.E[/tex]

Where

 [tex]\triangle K.E=(F-F_f)*2[/tex]

 [tex]F_f=F+\frac{\triangle K.E}{d}[/tex]

 [tex]F_f=33+\frac{0.5*17*1.6^2}{9.8}[/tex]

 [tex]F_f=30.8N[/tex]

Since

 [tex]f = \mu*m*g[/tex]

 [tex]\mu= 30.8/(m*g)[/tex]

 [tex]\mu= 30.8/(17*9.81)[/tex]

 [tex]\mu=0.185[/tex]

Answer:

B.

Explanation:

sana makatulong sayo

Answer:

The speed of the combined mass after the collision is 2.1 m/s.

Explanation:

mass of runner, m = 70 kg

speed  of runner, u = 2.7 m/s

mass of shortstop, m' = 85 kg

speed  of shortstop, u' = 1.6 m/s

Let the velocity of combined system is v.

Use conservation of momentum

Momentum before collision = momentum after collision

m u + m' u' = (m + m') v

70 x 2.7 + 85 x 1.6 = (70 + 85) v

189 + 136 = 155 v

v = 2.1 m/s

Answer:

Mercury - 800°F (430°C) during the day, -290°F (-180°C) at night. Venus - 880°F (471°C) Earth - 61°F (16°C) Mars - minus 20°F (-28°C)30-Jan-2018

Answer:

belpw

Explanation:

The distance prior to the sliding friction dispersing her energy would be:

- The distance will remain unaffected by the sliding friction i.e. 354m

As we know, When Sliding friction dissolves her energy, leading her Kinetic Energy to turn 0 on coming to the state of rest. So,

[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]            (∵ Work in -ve denotes it is done opposite to friction)

Given that,

m(mass) [tex]= 50 kg[/tex]

v(velocity) [tex]= 30 km/hr[/tex] or [tex]8.33 m/s[/tex]

The coefficient of Kinetic Friction [tex]= 0.01[/tex]

g(gravitational force) [tex]= 9.8 m/s^2[/tex]

Initial Velocity(u) [tex]= 30[/tex] × [tex]1000/3600 m/s[/tex]

[tex]= 8.33 m/s[/tex]

Now by employing the provided values,

[tex]F =[/tex] μ[tex]mg[/tex]

[tex]= (0.01) (50) (9.8)[/tex]

[tex]= 4.9[/tex]

∵ [tex]F = 4.9 N[/tex]

By using the above expression, we will find the distance;

[tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]

⇒ [tex]1/2 (50) (0)^2 - 1/2 (50) (8.33)^2 = -4.9(S)[/tex]

⇒ [tex]1734.7225 = 4.9S[/tex]

⇒ [tex]S = 1734.7225/4.9[/tex]

∵ [tex]S = 354 m[/tex]

Because [tex]1/2 mv^2 - 1/2 mu^2 = -W[/tex]  [tex]= -[/tex] μmgS

⇒ [tex]S = (u^2 - v^2)[/tex]/2μ[tex]g[/tex]

Thus, the distance will remain unaffected by the sliding friction i.e. 354m

Learn more about "Friction" here:

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Complete question is;

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:

(a) the work done by the force of gravity on the projectile,

(b) the change in kinetic energy of the projectile since it was fired, and

(c) the final kinetic energy of the projectile.

(d) Are any of the answers changed if the initial angle is changed?

Answer:

A) W = mgh

B) ΔKE = mgh

C) K2 = mgh + ½mv_o²

D) No they wouldn't change

Explanation:

We are expressing in terms of m, v0​, h, and g. They are;

m is mass

v0 is initial velocity

h is height of projectile fired

g is acceleration due to gravity

A) Now, the formula for workdone by force of gravity on projectile is;

W = F × h

Now, Force(F) can be expressed as mg since it is force of gravity.

Thus; W = mgh

Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.

Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.

B) Change in kinetic energy is simply;

ΔKE = K2 - K1

Where K2 is final kinetic energy and K1 is initial kinetic energy.

However, from conservation of energy, we now that change in kinetic energy = change in potential energy.

Thus;

ΔKE = ΔPE

ΔPE = U2 - U1

U2 is final potential energy = mgh

U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.

Thus;

ΔKE = ΔPE = mgh

Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.

C) As seen in B above,

ΔKE = ΔPE

Thus;

½mv² - ½mv_o² = mgh

Where final kinetic energy, K2 = ½mv²

And initial kinetic energy = ½mv_o²

Thus;

K2 = mgh + ½mv_o²

Similar to a and B above, this will not change even if initial angle is changed

D) All of the answers wouldn't change because their equations don't depend on the angle.

Answer:

The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of  second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of  fourth coil.

Explanation:

The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.

So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of  second coil.

And

The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of  fourth coil.

Answer:

[tex] = \frac{30 \times {10}^{ - 3} }{1} \\ = 0.03 \: km \: per \: second[/tex]

Answer:

108 km/hr or 0.03 km/s

Explanation:

conversion factor for m/s to km/hr is 5/18

conversion factor for m/s to km/s is 1/1000

Answer:

The correct answer is "4000 J".

Explanation:

Given that,

Force,

= 200 N

Displacement,

= 20 m

Now,

The work done will be:

⇒ [tex]Work=Force\times displacement[/tex]

By putting the values, we get

              [tex]=200\times 20[/tex]

              [tex]=4000 \ J[/tex]