Catching a wave, a 77 kg surfer starts with a speed of 1.3 m/s, drops through a height of 1.65 m, and ends with a speed of 8.2 m/s. How much non-conservative work was done on the surfer?
Answer:
Explanation:
The total work done by the wave is expressed as;
Workdone = Potential energy + Kinetic energy
Workdone = mgh + 1/2mv²
m is the mass = 77kg
g is the acceleration due to gravity = 9.8m/s²
v is the velocity = 8.2m/s
h is the height = 1.65m
Substitute into the formula;
Workdone = 77(9.8)(1.65) + 1/2(77)8.2²
Workdone = 1245.09 + 2588.74
Workdone = 3833.83Joules
Hence the amount of non conservative work done on the sofa is 3833.83Joules
Given:
Velocity, v = 8.2 m/sHeight, h = 1.65 mMass, m = 77 kgWe know,
→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]
or,
[tex]= mgh +\frac{1}{2} mv^2[/tex]
By putting the values,
[tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]
[tex]= 1245.09+2588.74[/tex]
[tex]= 3833.83 \ Joules[/tex]
Thus the above approach is right.
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If the velocity of Homer the astronaut (mass =200 kg) is 5 m/s and he runs into and grabs his stationary pal Larry (mass = 150 kg), what is the new velocity of the astronauts after the collision?
We are given:
Homer the Astronaut:
Mass of Homer the astronaut(m1) = 200 kg
initial velocity of Homer the astronaut(u1) = 5 m/s
Larry the Pal:
Mass of Larry the Pal (m2) = 150 kg
initial velocity of Larry the Pal (u2) = 0 m/s
Since they will move together after the collision, they will have the same velocity:
v1 = v2 = V
Solving for the Final velocity:
from the law of conservation of momentum:
m1u1 + m2u2 = m1v1 + m2v2
since v1 = v2 = V:
m1u1 + m2u2 = V(m1 + m2)
replacing the variables with the given values
200 * 5 + 150 * 0 = V(200 + 150)
1000 = 350V
V = 1000 / 350
V = 2.86 m/s
* WILL GIVE BRAINLIEST TO CORRECT ANSWER *
Name a product that people commonly purchase by mass and not by weight.
I would say food but like they take weight too
In football we see unbalanced forces. When 1 player exerts an unbalanced force on another player and causes a player to
Answer:
Fall
Explanation:
Can someone help plzzzzzzz I need it ASAP thank you
Answer:
Explanation:
Frequency is oscillations per second.
So we have to find the Number of seconds she paced.
2 minutes = 2 X 60
= 120 seconds
Therefore,
Her frequency = 10 / 120
= 1/12 Hertz
Help me guys please with this question
Answer:
[tex]\mid \vec C\mid=31.9[/tex]
Explanation:
Consider the vectors:
[tex]\vec A=9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}}[/tex]
[tex]\vec B=-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}}[/tex]
Calculate the magnitude of
[tex]\vec C=-2\vec B-\vec A[/tex]
Substitute the values of the vectors:
[tex]\vec C=-2(-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}})-(9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}})[/tex]
Operate and remove parentheses:
[tex]\vec C=19\mathbf{\hat{i}}+26.8\mathbf{\hat{j}}-9.4\mathbf{\hat{i}}+3.6\mathbf {\hat{j}}[/tex]
Operating both components separately:
[tex]\vec C=9.6\mathbf{\hat{i}}+30.4\mathbf{\hat{j}}[/tex]
Now find the magnitude of C:
[tex]\mid \vec C\mid=\sqrt{9.6^2+30.4^2}[/tex]
[tex]\mid \vec C\mid=\sqrt{1016.32}[/tex]
[tex]\mathbf{\mid \vec C\mid=31.9}[/tex]
The CERN particle accelerator is circular with a circumference of 7.0 km.
Required:
a. What is the acceleration of the protons (m=1.67×10^−27kg) that move around the accelerator at of the speed of light? (The speed of light is v=3.00×10^8m/s.)
b. What is the force on the protons?
Answer:
Explanation:
a) centripetal acceleration is the acceleration of a body in a circular path. It is expressed as;
a = mv²/r
m is the mass of proton = 1.67×10^−27kg
v is the velocity = 3.00×10^8m/s
r is the radius
Since C = 2πr
7000m = 2πr
r = 7000/2π
r = 1114.08m
Substitute
a = 1.67×10^−27 (3.00×10^8)²/1114.08
a = 1.67×10^−27 * 9×10^16/1114.08
a = 15.03*10^-11/1114.08
a = 0.001346*10^-11
a = 1.346*10^-14m/s²
b) Force on the proton = mass * acceleration
Force = 1.67×10^−27kg * 1.346*10^-14
Force = 2.246*10^-41N
hence the force on the proton is 2.246*10^-41N
A 2.0 cm thick brass plate (k_r = 105 W/K-m) is sealed to a glass sheet (kg = 0.80 W/K m), and both have the same area. The exposed face of the brass plate is at 80°C, while the exposed face of the glass is at 20 °C. How thick is the glass if the glass brass interface is at 65 C? Ans. 0.46 mm
Which statement is part of Dalton's atomic theory?
Matter is composed of small particles called atoms.
Atoms can be divided into their subatomic particles.
Atoms are able to be seen with proper spectroscopy equipment.
Chemical reactions can change atoms from one type to another.
Answer:
Dalton’s atomic theory was a scientific theory on the nature of matter put forward by the English physicist and chemist John Dalton in the year 1808. It stated that all matter was made up of small, indivisible particles known as ‘atoms’.
Explanation:
Atoms can be divided into their subatomic particles.
The statement that is part of Dalton's atomic theory is as follows: Matter is composed of small particles called atoms.
What is Dalton's atomic theory?John Dalton is a scientist that first stated the theory of chemical combination in 1803.
The components of these theory are as follows:
Elements consist of indivisible small particles called atoms.All atoms of the same element are identical i.e. different elements have different types of atom. Atoms can neither be created nor destroyedTherefore, according to this question, the statement that is part of Dalton's atomic theory is as follows: Matter is composed of small particles called atoms.
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Assume you are in the car and the car is moving at a certain speed to
school. Are you at rest or in motion with respect to the school? With
respect to the car?
Hmmm... Who can answer this question?
How did life begin?
Just a practice
Answer:
The earliest known life-forms are putative fossilized microorganisms, found in hydrothermal vent precipitates, that may have lived as early as 4.28 Gya (billion years ago), relatively soon after the oceans formed 4.41 Gya, and not long after the formation of the Earth 4.54 Gya.
Explanation:
Is this what your looking for?
How much thermal energy (in kcal) is required to change a 43 g ice cube from a solid at - 16.5 oC to steam at 11.5 oC above boiling
Answer:
The total thermal energy required is 8.93 Kcal
Explanation:
Given;
mass of the ice cube, m = 43 g
specific heat capacity of water, Cp = 4.18 J/gc
specific latent heat of fusion of ice, Cf = 334 J/g
First step, determine the heat needed to raise the temperature of the ice from -16.5 °C to 0° C
Q₁ = mCp[0 - (-16.5)]
Q₁ = 43 x 4.18(16.5)
Q₁ = 2965.71 J
Second step, determine the latent heat of fusion of ice at 0°C
Q₂ = mCf
Q₂ = 43 x 334
Q₂ = 14362 J
Third step, determine the quantity of heat required to raise the temperature of the water initially at 0°C to above 11.5 °C of boiling point of water.
The final temperature of the water = 11.5 °C + 100° C = 111.5 °C
Q₃ = mCp Δθ
Q₃ = 43 x 4.18 (111.5 - 0)
Q₃ = 20041 J
Total thermal energy required = Q₁ + Q₂ + Q₃
Total thermal energy required = 2965.71 J + 14362 J + 20041 J
Total thermal energy required = 37,368.71 J
Total thermal energy required = 8.93 Kcal
In ultimate the disc may be passed in any direction.
True
False
I think is True: Because...
when you throw a disc you can throw it in any direction, many people call it "Flying Saucer" I'll give you an example ... When you throw something, for example a paper, you want to throw it at your classmate. You already know what address you want to send it to, then I say it is: True ...
Sorry if it's wrong :(
significance of practicals in the discipline of geography
The second-order dark fringe in a single-slit diffraction pattern is 1.40 mm from the center of the central maximum. Assuming the screen is 89.0 cm from a slit of width 0.710 mm and assuming monochromatic incident light, calculate the wavelength of the incident light.
We know, for single slit :
[tex]y =\dfrac{ n\lambda L}{a}\\\\\lambda = \dfrac{ya}{nL}[/tex] ...1)
[tex]y = 1.4\ mm = 1.4 \times 10^{-3}\ m[/tex]
n = 2
L = 89 cm = 0.89 m
[tex]a=7.1\times 10^{-4}\ m[/tex]
Putting all these in equation 1), we get :
[tex]\lambda = \dfrac{ya}{nL}\\\\\lambda = \dfrac{1.4\times 10^{-3}\times 7.1\times 10^{-4}}{2\times 0.89 }\\\\\lambda = 5.584 \times 10^{-7}\ m[/tex]
Therefore, wavelength of the incident light is [tex]5.584 \times 10^{-7}\ m[/tex] or 558.4 nm.
Hence, this is the required solution.
A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it looks like this group varied the amount of mass sitting on the block with each trial - this is not recommended). Nonetheless, what is their average coefficient of static friction?
Trial Mass of block (g) Hanging mass (kg)
1 105 0.053
2 165 0.081
3 220 0.118
4 280 0.149
5 315 0.180
6 385 0.198
Answer:
0.130
Explanation:
From the given data, the coefficient of static friction for each trial are:
1. 0.053
2. 0.081
3. 0.118
4. 0.149
5. 0.180
6. 0.198
The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198
= 0.779
So that;
the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]
= [tex]\frac{0.779}{6}[/tex]
= 0.12983
The average coefficient of static friction is 0.130
The average coefficient of static friction is 0.13.
The coefficient of static friction is obtained using the formula; μ = F/R
Where;
F = force acting on the body
R = reaction
μ = coefficient of static friction
The average of measurements is given as; ∑summation of measurements/number of measurements
We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;
0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198/6
= 0.13
The average coefficient of static friction is 0.13
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we feel cold in winter when we come out from the quilt but the same room becomes warmer after coming back from outside the room
Answer:
Yes
Explanation:
I think this is because when you go out of the room and going to a hotter room you then get the heat from that room. It then becomes warmer in the room you are coming from because your body got the heat from the outside the room. I think it is because of body temperature.
HOPE THIS HELPED
If you jump upward with a speed of 1.70 m/s how high will you be when you stop rising?
Answer:
How long do you jump (sec) ?
Explanation:
An oil refinery uses a Venturi tube to measure the flow rate of gasoline. The density of the gasoline is
ρ = 7.40 ✕ 102 kg/m3,
the inlet and outlet tubes, respectively, have a radius of 3.74 cm and 1.87 cm, and the difference in input and output pressure is
P1 − P2 = 1.20 kPa.
a) find the speed of the gasoline as it leaves the hose
b) find the fluid flow rate in cubic meters per second
Answer:
(a) V₂ = 1.86 m/s
(b) Q = 5.1 x 10⁻⁴ m³/s
Explanation:
(a)
The formula derived for Venturi tube is as follows:
P₁ - P₂ = (ρ/2)(V₂² - V₁²)
where,
P₁ - P₂ = Difference in Pressure of Inlet and Outlet = 1.2 KPa = 1200 Pa
ρ = Density of Gasoline = 7.4 x 10² kg/m³
V₂ = Exit Velocity = ?
V₁ = Inlet Velocity
Therefore,
1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)
V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)
V₂² - V₁² = 3.24 m²/s² ------------------- equation (1)
Now, we will use continuity equation:
A₁V₁ = A₂V₂
where,
A₁ = Inlet Area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²
A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²
Therefore,
(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂
V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)
V₁ = 0.25 V₂
using this value in equation (1):
V₂² - (0.25 V₂)² = 3.24 m²/s²
0.9375 V₂² = 3.24 m²/s²
V₂² = (3.24 m²/s²)/0.9375
V₂ = √(3.456 m²/s²)
V₂ = 1.86 m/s
(b)
For fluid flow rate we use the following equation:
Flow Rate = Q = A₂V₂ = (2.746 x 10⁻⁴ m²)(1.86 m/s)
Q = 5.1 x 10⁻⁴ m³/s
The formula for finding variables in a Venturi tube is shown below:
The speed of the gasolineP₁ - P₂ = (ρ/2)(V₂² - V₁²)
where, P₁ - P₂ is difference in pressure of Inlet and outlet, ρ = density, V₂ = exit velocity and V₁ is inlet velocity
P₁ - P₂ = 1.2 KPa = 1200 Pa
ρ = 7.4 x 10² kg/m³
V₂ = Exit Velocity = ?
V₁ = Inlet Velocity
We then substitute the variables into this equation.
P₁ - P₂ = (ρ/2)(V₂² - V₁²)
1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)
V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)
V₂² - V₁² = 3.24 m²/s² ------ equation (1)
The continuity equation A₁V₁ = A₂V₂ is then used
where,A₁ = Inlet area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²
A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²
(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂
V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)
V₁ = 0.25 V₂
We then substitute the value into equation 1
V₂² - (0.25 V₂)² = 3.24 m²/s²
0.9375 V₂² = 3.24 m²/s²
V₂² = (3.24 m²/s²)/0.9375
V₂ = √(3.456 m²/s²)
V₂ = 1.86 m/s
The fluid flow rate we use the following equation:This can be calculated using the formula
Flow Rate = Q = A₂V₂
= (2.746 x 10⁻⁴ m²)(1.86 m/s)
= 5.1 x 10⁻⁴ m³/s
How does the abundance of hydrogen and helium support the Big Bang Theory?
It is the proportion predicted to be present in the early universe.
The hydrogen and helium abundance helps us to model the expansion rate of the early universe.
In the abundance of hydrogen and helium, we can say that they account for nearly all the nuclear matter in today's universe.
In big Bang model, the universe is mostly light or protons.
This abundance of hydrogen and helium is consistent with this big bang model. The process of forming this hydrogen and helium is often called big bang nucleosynthesis.The Schramm's model for relative abundances indicate that helium is about 25% by mass and hydrogen about 73% with all other elements constituting less than 2%.
Several proponents of big Bang theory has proposed similar relative abundance for hydrogen and helium. In all it is clear that hydrogen and helium constitute of more than 98% of the ordinary matter in the universe.
Finally, the hydrogen and helium abundance helps us to model the expansion rate of the early universe.
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2. Heather and Matthew walk with an average velocity of +0.87 m/s eastward.
If it takes them 27 minutes to walk to the store, what is their
displacement? (include direction)
(5 points)
HELP ASAP !!! !!!!!!!
Answer:
they are cooler than the rest if the sun
C4. A 50.0 kg boy runs at 10.0 m/s, jumps on a cart and rolls off at 2.50 m/s. What is the mass of the cart
Answer:
The mass of the cart is 150 kg.
Explanation:
Given that,
Mass of a boy, m₁ = 50 kg
Initial speed of boy, u₁ = 10 m/s
Initial speed of car, u₂ = 0 (at rest)
The speed of the cart with the boy on it is 2.50 m/s, V = 2.5 m/s
Let m₂ is the mass of the cart. Using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\50(10)+m_2(0)=(50+m_2)(2.5)\\\\500=125+2.5m_2\\\\375=2.5m_2\\\\m_2=150\ kg[/tex]
So, the mass of the cart is 150 kg.
When you turn off your bedroom lights, the kitchen lights can stay on. This is because your home is wired using which of the following? (AKS 10b / DOK 1)
A.
Fuses
B.
Series circuits
C.
Batteries
D.
Parallel circuits
When you turn off your bedroom lights, the kitchen lights can stay on. This is because your home is wired by using the series circuit. Hence, option B is correct.
What is a Series circuit?The amount of current in the flow circuits of each element in a series circuit is the same. This is due to the recent flow having a single path. The flow rate (marble speed) at any point in a circuit (tube) at any given moment in time has to be equal because electric current moves through a conductor like marble in a duct.
The requirement that all values (voltage, current, resistance, and power) relate to one another in terms of the identical two places in a circuit is a significant exception to Ohm's law.
Therefore, the wiring of the home is in series combination. So, option B is correct.
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Objects are lighter on the moon than they are on earth. if an object A weighs 25lbs on the Moon and another object B weighs 25 Newtons on earth, which has more mass?
a. Object a
b. Object b
c. Same mass
d. Other
Answer:
a. Object A
Explanation:
The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.
The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.
So that,
For object A on the moon, mass = 25 lbs
For object B on the earth, W = 25 N,
W = m x g
25 = m x 10 (g = 10 m/[tex]s^{2}[/tex])
m = [tex]\frac{25}{10}[/tex]
= 2.5 lbs
Mass of object B is 2.5 lbs.
Therefore, the mass of the object A is more than that of B.
from a flying aeroplane abody should be dropped in advance to hit the target why
From a flying plane a body should dropped in advance to hit the target,Why? ... The body should be dropped in advance as when the body is dropped it has the velocity of the plane. So, in air the body moves forward which we have to take into consideration in order to hit the target.
A spring gun fires a ball horizontally at 15 m/s. It is mounted on a flat car moving in a straight line at 25 m/s. relative to the ground, what is the horizontal speed of the ball when the gun is aimed forward?
Answer:
15 m/s
Explanation:
The Speed Of The Car Does Not Add To The Speed Of The Bullet
Fast and safe heart rate for workouts is called muscular strength? True or false
Answer:
False
Explanation:
Answer:
False
Explanation:
Hope this helped, Have a Wonderful Day/Night!!
A golf ball (m=26.7g) is struck a blow that makes an angle of 33.6 degrees with the horizontal. The drive lands 190m away on a flat fairway. The acceleration of gravity is 9.8 m/s^2 . If the golf club and ball are in contact for 7.13 ms, what is the average force of impact?
Answer:
Th average force impact is [tex]F = 168.298 \ N[/tex]
Explanation:
From the question we are told that
The mass of the golf ball is [tex]m_g = 26.7 \ g = 0.0267 \ kg[/tex]
The angle made is [tex]\theta = 33.6 ^o[/tex]
The range of the golf ball is [tex]R = 190 \ m[/tex]
The duration of contact is [tex]\Delta t = 7.13 \ ms = 7.13 *10^{-3} \ s[/tex]
Generally the range of the golf ball is mathematically represented as
[tex]R = \frac{v^2 sin2(\theta)}{g}[/tex]
Here v is the velocity with which the golf club propelled it with, making v the subject
[tex]v = \sqrt{\frac{R * g}{sin 2 (\theta)} }[/tex]
=> [tex]v = \sqrt{\frac{190 * 9.8}{sin 2 (33.6)} }[/tex]
=> [tex]v = 44.94 \ m/s[/tex]
Generally the change in momentum of the golf ball is mathematically represented as
[tex]\Delta p = m * (v - u )[/tex]
here u is the initial velocity of the ball before being stroked and the value is 0 m/s
[tex]\Delta p = 0.0267 * ( 44.94 - 0 )[/tex]
=> [tex]\Delta p = 1.19996 \ kg \cdot m/s[/tex]
Generally the average force of impact is mathematically represented as
[tex]F = \frac{\Delta p }{\Delta t}[/tex]
=> [tex]F = \frac{1.19996 }{7.13 *10^{-3}}[/tex]
=> [tex]F = 168.298 \ N[/tex]
A ball has a diameter of 3.77 cm and average density of 0.0839 g/cm3. What force is required to hold it completely submerged under water?
magnitude _________ N
The force required to hold it completely submerged under water is 0.252 N
As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.
Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation
F = Buoyant force - weight of sphere
Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m
Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³
Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg
Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N
Volume of water displaced = 4/3 π r³ = 2.805 e-5
Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N
F = 0.275 - 0.023 = 0.252 N
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The force required to hold it completely submerged under water is 0.25 N
The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³
The density of water [tex]\rho_w[/tex] = 1000 kg/m³
Diameter = 3.77 cm = 0.0377 m
radius of ball = 0.0377/2 = 0.01885 m
The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]
Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:
The force is required to hold it completely submerged under water (F) is:
[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]
F = 0.25 N
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