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
On the periodic table , the vertical columns that extend down the periodic table are called ?
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. :)
If there is "waste" energy, does the Law of Conservation of Energy still apply? please don't type something random if so i'll just report it.
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.
If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.True or False. The larger a waves wavelength, the more energy it carries. (1 Point) True O False O Maybe
Explanation:
False, it is oppisite the shorter the wavelength the more energy it carries.
ts
An electric light bulb mixer is used for 19.2 seconds. In that time, it transfers 1
536 J of energy.
Calculate the power output of the cake mixer.
A
I DONT KNOW
A spring has a constant of 100 N/m. What Force does the spring exert on you if you stretch it a distance of 0.5 m?
Answer:
F = - K x force is opposed to direction of extension
F = -100 N / m * .5 m = -50 N
A school is creating a small parking lot next to the school. They will need
998,140 grams of rock for the parking lot. How many pounds of rock does
the school need? (1 kg = 2.205 lb)
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
Your family is moving, and you are asked
to help move some boxes. One box is so
heavy that you must push it across the
room rather than lift it. What are some
ways you could reduce friction to make
moving the box easier?
Answer:
Explanation:
I would use a hand truck dollies
Choose all the answers that apply.
Which of the following will cause an increase in the acceleration of an object?
increase force
decrease force
increase mass
decrease mass
Whats the answer!?
This cat is A tired B lazy or C Dreaming
the answer is obviously c because cats cant have nightmares
A 40 kg boy standing on a skateboard throws a 2 kg ball 20 m/s to the left.
a. What is the ball's momentum?
O 10 kg m/s
O 20 kg mis
O 40 kg m/s
O 1 kg m/s
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
g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament
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
What type of energy is stored in a battery?
pls give all the possible answer and pls if you can give an explanation
thx I'm really stuck
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:
Curtis, a student in our class, makes the following statement: The puck reached a slightly higher location on the ramp than I predicted. This is because I used the wrong mass for the puck when I did all my calculations. I accidentally used the mass of the smaller puck rather than the mass of the larger puck in my video." Is this a plausible explanation? Would the using the wrong mass for the puck during the calculations mean the puck would reach a greater height? Explain your reasoning.
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.
Name and explain the
various types of friction.
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:
A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 4.00 kg particle such that the center of mass of the three-particle system has the coordinates (-0.500 m, -0.700 m)
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.
A mass m is gently placed on the end of a freely hanging spring. The mass then falls 33 cm before it stops and begins to rise. What is the frequency of the oscillation
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 .
An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is a constant 1.6 m/s2, how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed).
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.
greyhound 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. Compare the speed of the
hound and the hare,
need full solution:-
[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..!
(5/3)a × 525/3a metresNow the distance travelled by the hare in it's 6 leaps..!
6a metresNow let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!
25/3a = 6a25/3 = 625:18What is a work out time setting? (Gym)
Answer:
It determines how long you do a certain workout.
A car has a mass of 1500 kg. If the driver applies the brakes while on a gravel road, the maximum friction force that the tires can provide without skidding is about 7000 N. how long are the skid marks
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
a lamp with flux 1400 lm, has an intensity of
a.) 1400 cd
b.) 100 cd
c.) 1000 cd
d.) 111 cd
One component of a metal sculpture consists of a solid cube with an edge of length 24.5 cm . The alloy used to make the cube has a density of 8050 kg/m3 . Find the cube's mass.
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
2.19 The drag characteristics of a blimp traveling at 4 m/s are to be studied by experiments in a water tunnel. The prototype is 20 m in diameter and 110 m long. The model is one-twentieth scale. What velocity must the model have for dynamic similarity
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]
Energy can be transferred from one place to another through?
A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 8.11 m below its highest point?
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
A constant force FA is applied to an object of mass M, initially at rest. The object moves in the horizontal x-direction, and the force is applied in the same direction. After the force has been applied, the object has a speed of vf. Which mathematical routines can be used to determine the time in which the force is applied to the object of mas
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
A truck travelling down the street suddenly brakes, applying a 14 N force over 3.5 seconds. What was the impulse over the given time.
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
Two point charges, A and B, are separated by a distance of 19.0 cm . The magnitude of the charge on A is twice that of the charge on B. If each charge exerts a force of magnitude 45.0 N on the other, find the magnitudes of the charges.
Answer:
QA = 19μC
QB = 9.5 μC
Explanation:
The force that each charge exerts on the other must obey Coulomb's Law, as follows:[tex]F_{AB} = \frac{k*Q_{A} * Q_{B}}{r_{AB}^{2}} (1)[/tex]
We know that the value of the magnitude of FAB is 45.0 N, the distance between QA and QB is 0.19 m, and that QA = 2*QB.Replacing in (1), we can solve for QB, as follows:[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]
Since QA = 2*QB⇒ QA = 2* 9.5μC = 19.0 μC⇒ QB = 9.5μCIt would be impossible to build a microscope that could use visible light to see the molecular structure of a crystal because. It would be impossible to build a microscope that could use visible light to see the molecular structure of a crystal because. lenses cannot be ground with fine enough precision. diffraction limits the resolving power to about the size of the wavelength of the light used. lenses with enough magnification cannot be made. lenses cannot be placed in the correct place with enough precision. More than one of the above is correct.
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
The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grains are there in the ball
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]