On a particular day the ocean swell has a period of 10 s, a wave velocity of 5.0 m/s, and a wave amplitude of 0.520 m. An anchored boat is bobbing up and down as the wave passes by. What is the peak vertical velocity of the boat

Answer:

Explanation:

I would use a hand truck dollies

Answer:

Batteries use chemistry, in the form of chemical potential, to store energy, just like many other everyday energy sources. For example, logs store energy in their chemical bonds until burning converts the energy to heat.

Explanation:

Answer:

118kg

Explanation:

answered

Given

density of the cube= 8050 kg/m3 .

length of the sizes of the cube=24.5 cm

We can convert the length to cm for unit consistency.

It's length =24.5 cm =0.245m

✓ the length of sizes of the cube is the same, then the volume can be calculated as

Volume= L^3

= (0.245m)^3

=0.01470625 m^3

✓ but we know that

Density = mass/ volume

Then,

Mass= (Volume × density)

= (0.01470625)(8050)

= 118 kg

Hence, the mass of the cube is 118 kg

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 unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Explanation:

Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:

Unmarked police car

[tex]s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}[/tex] (1)

Speeder

[tex]s = s_{o} + v_{o,S}\cdot t[/tex] (2)

Where:

[tex]s_{o}[/tex] - Initial position, measured in meters.

[tex]s[/tex] - Final position, measured in meters.

[tex]v_{o,P}[/tex], [tex]v_{o,S}[/tex] - Initial velocities of the unmarked police car and the speeder, measured in meters per second.

[tex]a[/tex] - Acceleration of the unmarked police car, measured in meters per square second.

[tex]t[/tex] - Time, measured in seconds.

[tex]t'[/tex] - Initial instant for the unmarked police car, measured in seconds.

By equalizing (1) and (2), we expand and simplify the resulting expression:

[tex]v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t[/tex]

[tex]v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t[/tex]

[tex]\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0[/tex]

If we know that [tex]a = 1.6\,\frac{m}{s^{2}}[/tex], [tex]v_{o,P} = 0\,\frac{m}{s}[/tex], [tex]v_{o,S} = 53.4\,\frac{m}{s}[/tex] and [tex]t' = 2.2\,s[/tex], then we solve the resulting second order polynomial:

[tex]0.8\cdot t^{2}-56.92\cdot t +3.872 = 0[/tex] (3)

[tex]t_{1} \approx 71.082\,s[/tex], [tex]t_{2}\approx 0.068[/tex]

Please notice that second root is due to error margin for approximations in coefficients. The required solution is the first root.

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Answer:

It determines how long you do a certain workout.

Answer:

F = - K x        force is opposed to direction of extension

F = -100 N / m * .5 m = -50 N

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

Answer:

Explanation:

The mass falls by .33 m before it begins to rise . At that point loss of potential energy is equal to gain of elastic energy .

1/2 k x² = mgx

.5 x k x .33² = m x 9.8 x .33

k / m = 59.4

frequency of oscillation =  [tex]\frac{1}{2\pi} \times\sqrt{\frac{k}{m} }[/tex]

= [tex]\frac{1}{2\pi} \times\sqrt{59.4}[/tex]

= 1.22 per second .

Explanation:

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

Answer:

There are four types of friction: static, sliding, rolling, and fluid friction. Static, sliding, and rolling friction occur between solid surfaces. Static friction is strongest, followed by sliding friction, and then rolling friction, which is weakest. Fluid friction occurs in fluids, which are liquids or gases.

Explanation:

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:

groups

Explanation:

Answer: Groups

Explanation: They are in the same group! Like the Alkaline Metals are all the group. They all lose an electron. :)

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]

Answer:

QA = 19μC

QB = 9.5 μC

Explanation:

       [tex]F_{AB} = \frac{k*Q_{A} * Q_{B}}{r_{AB}^{2}} (1)[/tex]

      [tex]Q_{B} = \sqrt{\frac{F_{AB}*r_{AB} ^{2}}{2*k} } = \sqrt{\frac{45.0N*(0.19m) ^{2}}{2*9e9N*m2/C2} } = 9.5e-6 C (2)[/tex]

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.

Read more about diffraction here https://brainly.com/question/16749356

[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..!

Explanation:

Yes, the law of conservation of energy still applies even if there is waste energy.

The waste energy are the transformation products of energy from one form to another.

According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".

But of then times, energy is lost as heat or sound within a system.

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:

2200.9lb

Explanation:

This is a conversion problem.

 We have been given that:

            Mass of rock the school needed  = 998140g

Unknown:

       Pound of rocks the park needed  = ?

To solve this problem, we have to convert from:

     grams to kilograms and then to pounds

            1000g of the rock will weigh 1kg

  So;       998140g of the rock will weight 998.14kg

Therefore:

               1kg of a substance weighs 2.205lb

           998.14g will weight2.205 x 998.14  = 2200.9lb

Explanation:

Given data:

mass of the car = 1500 kg

maximum friction force = 7000 N

initial velocity v_i = 20 m/s ( it is not given in the question just an assumption)

final velocity v_f = 0 m/s

[tex]\begin{array}{l}

\sum F_{y}=M g-F_{n}=0 \\

\sum F_{x}=-F_{s}=m a_{x} \\

-F_{s}=m a_{x}

\end{array}[/tex]  

[tex]a_{x}=\frac{-F_{s}}{m}=\frac{-7000}{1500}[/tex]

[tex]a_{x}=-4.7 \mathrm{~m} / \mathrm{s}^{2}[/tex]

Now we can find the distance from this formula:

[tex]v_{f x}^{2}=v_{i x}^{2}+2 a_{x}(\Delta x)[/tex]

[tex]0=20^{2}+(2 \times-4.7 \times \Delta x)[/tex]

[tex]20^{2}=9.4 \Delta x[/tex]

[tex]\Delta x=\frac{20^{2}}{9.4}=42.55 \mathrm{~m}[/tex]

So, the shortest distance in which the car can stop safely without kidding

=42.55 m

Given :

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12.

To Find :

How many grains are there in the ball?

Solution :

Volume of ball of the ballpoint is :

[tex]V = \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\times 3.14 \times 0.5^3}{3}\ mm^3\\\\V = 0.523\ mm^3[/tex]

Now, grain size of 12 has about 520000 grains/mm³.

Therefore, number of grains are :

[tex]n = 520000\times 0.523\ grains\\\\n = 271960\ grains[/tex]

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.

the answer is obviously c because cats cant have nightmares

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:

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