Answer:
janakajnakanabjjiaj yav good day for you to everyone else to do with my family is it just doesn't want me and I am
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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In football we see unbalanced forces. When 1 player exerts an unbalanced force on another player and causes a player to
Answer:
Fall
Explanation:
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!!
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
A shell traveling with speed, v0 , exactly horizontally and due north explodes into two equal mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed v0 . What is the velocity of the other fragment? Hint: Velocity has both magnitude and direction.
Answer:
yeah
Explanation:
yeah yeah yeah yeah
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
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
* 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
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]
what is Force and Preasure ??
Answer:
Definition: Pressure:- Force per unit area surface is called pressure. P = F/ A Pressure = Force /Area Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed. Gauge pressure is the pressure relative to the ambient pressure.
Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force.
Pressure is defined to be the amount of force exerted per area.
Formula of pressure = P = F÷A
Where, P = Pressure,
F = Force applied and,
A = Unit area.
Examples of Pressure is Air pressure.
A horizontal force of 30N is applied to a mass of 10 kg causing it to accelerate. If the coefficient of friction is 0.20 what is the frictional force?
If work is shown that would be great
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The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.
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In this question ,
surface vertical force = Weight of the object
Thus ;
svf = ( mass ) × ( gravity acceleration )
_________________________________
If gravity acceleration is 10 :
svf = 10 × 10 = 100 N
So ;
frictional force = 100 × 0.20
frictional force = 20 N
##############################
If gravity acceleration is 9.8 :
svf = 10 × 9.8 = 98 N
So ;
frictional force = 98 × 0.20
frictional force = 19.6 N
_________________________________
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a ball of mass 0.5 kg is at point with initial speed 4 m/s at height 10. what is the total energy
Answer:
The total energy is 53 Joule
Explanation:
Mechanical Energy
The mechanical energy of an object of mass m, speed v, and at a height h is:
[tex]\displaystyle E = m.g.h+\frac{mv^2}{2}[/tex]
The ball has a mass of m=0.5 Kg, a speed v = 4 m/s, and at a height of h=10 m. Thus the total energy is:
[tex]\displaystyle E = 0.5\cdot 9.8\cdot 10+\frac{0.5\cdot 4^2}{2}[/tex]
E = 49 J + 4 J = 53 J
The total energy is 53 Joule
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.
Romeo is a 68 kg astronaut. Juliet is a beautiful cosmonaut who is standing on the balcony of a 4.58 x 10^5kg space station that is at rest and out of gas. Romeo is floating 25 meters away from the space station’s center of mass, how strong is the force between Romeo and Juliet?
Answer:
F = 3.32 x 10⁻⁶ N
Explanation:
The force of attraction between two masses is given by Newton's Law of Gravitation, as follows:
F = Gm₁m₂/r²
where,
F = Force between Romeo and Juliet = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = mass of Romeo = 68 kg
m₂ = mass of space station = 4.58 x 10⁵ kg
r = distance = 25 m
Therefore,
F = (6.67 x 10⁻¹¹ N.m²/kg²)(68 kg)(4.58 x 10⁵ kg)/(25 m)²
F = 3.32 x 10⁻⁶ N
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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HELP ASAP !!! !!!!!!!
Answer:
they are cooler than the rest if the sun
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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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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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?
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]
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.
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:
The diagram shows the four primary steps in the production of work in a four-stroke heat engine.
What is the correct order in which these steps occur?
A. Y, W, X, z
B. X, Z, W. Y
C. W, X, Y, Z
D. Z, X, Y, W
Answer:
D- Z, X, Y, W
Explanation:
Correct on edge
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
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 :(
What happens to the mechanical advantage of a machine if the output force is less than the input force? What must happen to output distance? Give an example of a machine that does this?
significance of practicals in the discipline of geography
A car travels at a speed of 55 km/hr and slows down to 10 km/hr in 20 seconds. What is the acceleration?
Answer:
a = 0.62 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics. But first, we must convert speeds from kilometers per hour to meters per second.
[tex]v_{f} =v_{o} -a*t[/tex]
[tex]55[\frac{km}{hr}]*\frac{1hr}{3600s}*\frac{1000m}{1km} =15.27[\frac{m}{s} ]\\10[\frac{km}{hr} ]*\frac{1hr}{3600s}*\frac{1000m}{1km} = 2.77[\frac{m}{s} ][/tex]
where:
Vf = final velocity = 2,77 [m/s]
Vo = initial velocity = 15.27 [m/s]
t = time = 20 [s]
a = acceleration [m/s²]
Now replacing:
[tex]2.77=15.27-a*(20)\\20*a=12.49\\a = 0.62[m/s^{2}][/tex]
Answer:
It would be 10.00
Explanation:
Hope this helps its different for everyone what was it for u it was D for me
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)