The pail’s minimum speed at the top of the vertical circle is 4.72 m/s.
The given parameters:
Mass of the water, m = 2000 g = 2 kgRadius of the circle, r = 1 mTension on the string, T = 25 NThe net force on the pail of water at the top of the vertical circle is calculated as follows;
[tex]T = ma - mg\\\\T = \frac{mv^2}{r} - mg\\\\T + mg = \frac{mv^2}{r} \\\\mv^2 = r(T + mg)\\\\v^2 = \frac{r(T + mg)}{m} \\\\v= \sqrt{\frac{r(T + mg)}{m}} \\\\v = \sqrt{\frac{1 (25\ + \ 2 \times 9.8)}{2}} \\\\v = 4.72 \ m/s[/tex]
Thus, the pail’s minimum speed at the top of the vertical circle is 4.72 m/s.
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The pail’s minimum speed at the top of the circle if no water is to spill out is; v = 4.722 m/s
We are given;
Mass of the pail of water; m = 2000 g = 2kg
Radius of the circle; r = 1 m
Tension exerted by string; T = 25 N
To calculate the net force on the pail of water at the top of the vertical circle, let us take equilibrium of forces to give;
W - T = ma
Where;
W is weight = mg
T is tension on the string
Now, in circular motion, we know that;
Acceleration is; a = v²/r
Thus;
mg - T = m(v²/r)
Making v the subject of the formula gives us;
v = √[(r(T + mg)/m)
Plugging in the relevant values gives;
v = √[(1(25 + (2 * 9.8))/2)
v = 4.722 m/s
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A large atomic bomb produces around 8.00x1015J of energy. How much matter does that bomb turn into energy?
A)2.67x10^79kg
B)2.67x10^7kg
C)7.20x10^32kg
D)0.0889kg
Answer:
D)0.0889kg
Explanation:
If a large atomic bomb produces around 8.00x10¹⁵J of energy, then from Einstein's energy law i.e E=mc² we can calculate that the amount of matter is 0.0889kg. The correct option is D.
What is Einstein's energy law?
The Einstein Energy Law, often referred to as Einstein's famous equation relates the energy of a particle (E) to its mass (m) and the speed of light (c). The equation is as follows:
E = mc²
This equation shows that energy (E) and mass (m) are equivalent and interchangeable and that a small amount of mass can be converted into a large amount of energy. This is the basis for understanding nuclear reactions and the energy released in processes like nuclear fusion and fission.
Here in the question,
To calculate the mass that is converted into energy in an 8.00x10¹⁵J nuclear reaction, we can rearrange the equation to solve for m:
m = E / c²
where c = 3 x 10⁸ m/s is the speed of light in a vacuum.
Substituting the values, we get:
m = 8.00x10¹⁵ J / (3 x 10⁸ m/s)²
m = 0.0889kg
Therefore, a large atomic bomb that produces around 8.00x10¹⁵ J of energy converts approximately 0.0889kg of matter into energy. This is a very small amount of matter, but because c² is such a large number, a huge amount of energy can be released from a small amount of matter in a nuclear reaction.
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personal and unbiased errors belong to
personal to systematic errors
unbiased to random errors
a car moves at a speed of 40km/h. it is stopped by applying brake which produce a uniform acceleration of-0.5m/s^2. how much distance will it move before coming to stop ?
Answer:
Explanation:
We first need to convert the 40 km/h to m/s. Going by the fact that 40 has only 1 significant figure in it, 40 km/h = 10 m/s. The rest of the values are in their proper labels. We will use the equation:
[tex]v^2=v_0^2+2a[/tex]Δx where the final velocity is 0 because the car is coming to a stop at the end; the initial velocity is 10 m/s, the acceleration (or, rather, deceleration) is -.5 m/s/s, and our unknown which is displacement. Filling in:
[tex]0=(10)^2+2(-.5)[/tex]Δx and solving for Δx:
Δx = [tex]\frac{-100}{2(-.5)}[/tex] which ends up being simply that
Δx = 100 m
A body having uniform velocity has zero acceleration? give reason
Answer:
A body having uniform velocity has zero acceleration because
there is not change in velocity.
The energy required to change the state of a substance was determined to be E(H).
If the mass of the substance was DOUBLED, the value of E(H) will
(A) be halved
(B) be doubled
(C) be quadrupled
(D) remain constant
Answer:
ccccccccc or bbbbbbb or aaaaaaaa or ddddddd
Una rueda tiene 4 cm de diámetro y realiza
200 vueltas cada 5 minutos. Calcular:
a. la frecuencia
b. el período
c. la velocidad lineal o tangencial
d. La velocidad angular
e. La aceleración centrípeta.
Answer:
Una rueda tiene 4 cm de diametro y realiza 200 vueltas cada 4 minutos calcular a frecuencia b el periodo c la velocidad lineal l tangencial
The direction of applied force has to be .............. to the distance in order to say work is done
Answer:
Parallel
Explanation:
The work done is defined as the force applied on an object and the displacement in the position of the object in the direction of force.
W = F s cos A
where, F is the force, s is the displacement and A is the angle between force and displacement.
When the angle between the force and the displacement is 90 degree, the work done is zero.
To get the maximum work the angle between the force and the displacement is 0 degree.
So, to get the work done by the force the angle between the force and displacement is 0 degree that means the force and displacement is parallel to each other.
What's an area of the earth that suggest that " the Earth is like one giant jigsaw" idea could be true.
Which choice shows the correct sequence of features formed by continued wave erosion?
Answer:
Wave-cut cliff, sea arch, sea stacks
Explanation:
The effect of a wave erosion is made obvious by the structures formed by the wave action.
The high areas of land adjacent to the incoming wave develop the early features or the formation of wave action, which includes the wave-cut cliff
The continuous undercutting of the cliff by the wave results in the formation of the wave cut platform
The effect of the wave further on a cliff, results in the formation of a sea arch and finally a sea stack
Therefore, the correct sequence is the wave-cut cliff, sea arch, sea stacks
a. A man of 100 kg mass is standing on the surface of the earth of mass 6 × 1024 kg and
radius 6400km. Calculate the gravitational force between them. [Ans: 977.05 N]
Answer:
m1=100kg
m2=6×10^24kg
r=6400km
G=6.6×10^-11
F=?
as we know that
F=G m1.m2/r²
F=6.6×10^-11×100×6×10^24/(6400)²
F=6.6×10^-11×6×10^26/4.096×10^7
F=6.6×10^-11×1.465×10^19
F=9.67×10^8N
If we aim a radio telescope at a distant spiral arm of the Milky Way Galaxy, we will probably observe a 21-cm line. If we point a large optical telescope at this same region, we will probably not be able to detect the neutral hydrogen that gives rise to the 21-cm radio signal. Why not
Answer:
Because of the interstellar dust and interior location of the solar system.
Explanation:
We will probably not be able to detect the neutral hydrogen that gives rise to the 21-cm radio signal if we point a large optical telescope to the region because, the interstellar dust obscures the location of the spiral arm of the Milky way galaxy and this makes neutral hydrogen that gives rise to the 21-cm radio signal difficult to detect.
Also, the interior location of the solar system also makes the neutral hydrogen that gives rise to the 21-cm radio signal difficult to detect.
So, the interstellar dust and the interior location of the solar system makes it difficult to detect the neutral hydrogen that gives rise to the 21-cm radio signal with a large optical telescope.
20. A semiconductor is a
A) crystalline solid that conducts current under any condition
B) metallic solid that conducts current under any condition
C) metallic solid that conducts current under certain conditions
D) crystalline solid that conducts current under certain conditions
Answer:
D) crystalline solid that conducts current under certain conditions
Explanation:
A semiconductor is any crystalline solid material whose electrical conductivity is intermediary between that of insulators and conductors. That is, a semiconductor does not conduct like a conductor but does more than an insulator.
Based on this, a semiconductor is said to only conduct electricity under certain conditions as opposed to conductors that do so under any condition. Example of semiconductors are silicon, carbon, antimony, arsenic etc.
Annie has a soccer ball and a cake ball she kicks each ball with the same Force the soccer ball accelerates at 3 m per second squared and the kickball accelerates at 5 m per second squared use Newton's law to describe why the kickball has a greater acceleration
describe how air resistance would affect a falling object
Answer:
With air resistance, acceleration throughout a fall gets less than gravity (g) because air resistance affects the movement of the falling object by slowing it down. How much it slows the object down depends on the surface area of the object and its speed
Explanation:
PLZ MARK AS THE BRAINLIEST
HAVE A GOOD DAY
: )
diffusion in physics
Answer:
sorry but which class your talking 'bout
Use the equation d=st where d= distance, s= speed, and t= time. If you ride a skateboard at a speed of 8m/s for 12 seconds, how far do you travel
Answer:
96 meters
Explanation:
d = st
d = (8 m/s)(12 s)
d = 96 m
use the particle model to explain why the water expands
Answer:
Take water as an example; the size, shape and chemical composition of the water particles remains the same whether it is solid water (ice) or gaseous water (steam) - but how those particles move and are arranged is different for
Why are houses in hot areas painted in white while those in cold areas painted with dull colours
Answer:
Because white color reflects back the heat coming from the sun while dull colors absorb the heat.
Angie walked a distance of 90 meters east in 70 seconds. What was her
velocity?
A. 0.78 m/s east
B. 1.3 m/s east
O C. 7 m/s east
D. 9 m/s east
According to Moseley, the properties of elements are a
periodic function of their
Answer:
According to Henry Moseley, Atomic number is the fundamental property of the elements because elements are made up of atoms . ... So the physical and chemical properties of elements are periodic functions of their atomic number.
A 550N object has a coefficient of .012 against a smooth surface. What
force is needed to keep the object sliding?(ANSWER to 1 decimal place, no
units)
Vertically, the object is in equilibrium, so that the net force in this direction is
∑ F (vertical) = n - mg = 0
where n is the magnitude of the normal force due to the contact between the object and surface. You're given that the object's weight is mg = 550 N, so n = 550 N as well.
Horizontally, the net force would be
∑ F (horizontal) = p - f = 0
where p is the magnitude of the applied force and f is the magnitude of (kinetic) friction opposing p. Now,
f = 0.012n = 0.012 (550 N) = 6.6 N
so that you need to apply a force of p = 6.6 N to keep the object sliding at a steady pace.
When the system pressures are lower than the pressure in the refrigerant tank, such as when the system has just been evacuated or if the system is out of refrigerant, refrigerant can be added to both the high and low pressure sides of the system. A. True B. False
Answer: True
Explanation:
It should be noted that when refrigerant is added in vapor form to the operating refrigeration system, then the addition of the refrigerant should be to the low-pressure side of the refrigeration system.
Furthermore, when system pressures are lower than the pressure in the refrigerant tank as stated in the question, then the refrigerant can be added to both the high and low pressure sides of the system.
Therefore, the correct option is true.
A CROW BAR WITH LENGTH 200 CM IS USED TO LIFT A LOAD OF 600N . IF THE DISTANCE BETWEEN FULCRUM AND LOAD IS 0.75. CALCULATE ; a, effort b, MA c, VR
Answer:
a. Effort = 960 Newton
b. Mechanical advantage (M.A) = 0.625
c. Velocity ratio (V.R) = 1.67
Explanation:
Given the following data;
Load = 600 NLength of crowbar = 200 cmLength of load arm = 0.75 mConversion:
100 cm = 1 m
X cm = 0.75 m
Cross-multiplying, we have;
X = 0.75 * 100 = 75 cm
First of all, we would find the effort arm;
Effort arm = length of crow bar - length of load arm
Effort arm = 200 - 75
Effort arm = 125 cm
Next, we would determine the mechanical advantage (M.A) of the crow bar;
[tex] M.A = \frac {Effort \; arm}{Load \; arm} [/tex]
Substituting the values into the formula, we have;
[tex] M.A = \frac {125}{200} [/tex]
M.A = 0.625
To find the effort of the crow bar;
[tex] M.A = \frac {Load}{Effort} [/tex]
Making "effort" the subject of formula, we have;
[tex] Effort = \frac {Load}{M.A} [/tex]
[tex] Effort = \frac {600}{0.625} [/tex]
Effort = 960 Newton
Lastly, we would determine the velocity ratio (V.R);
[tex] V.R = \frac {length \; of \; effort \; arm}{length \; of \; load \; arm} [/tex]
[tex] V.R = \frac {125}{75} [/tex]
V.R = 1.67
2. Speed limits on curves help to reduce the effect of ........force on a vehicle
A. enertia
b. centrifugal
c. gravity
Answer:
enertia
Explanation:
enertia
The correct answer is (B) Centrifugal Force
The centrifugal force on a vehicle on the curves is outwards or can say away from the curve and pulls the out of the track.
the centrifugal force is given by [tex]F=\frac{mv^{2} }{r}[/tex], which is directly proportional to speed v.
Speed limits on the curves help in reducing the centrifugal force by reducing the speed.
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mention various medium level and higher level human resources related to engineering.
Answer:
Higher-level human resources related to engineering are those who:
Assist other engineers with solving complex problems. Work with management to keep the project on track. Mentor younger engineersMedium Level engineers will normally comprise the following:
They are responsible for developing engineering plans and performing calculations with respect to the same. Project workflow management Engineering Support facilitation Project Reporting Coordination of Team MembersCheers
power of a lens whose focal length is -40 d.
Answer:
as we know
p=1/f
p=1/-40m
p= -0.025d
you have written the unit of focal length wrong it is the unit of power
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A jet airplane is in level flight. The mass of the airplane is m=8950 kg. The airplane travels at a constant speed around a circular path of radius R=9.33 mi and makes one revolution every T=0.123 h. Given that the lift force acts perpendicularly upward from the plane defined by the wings, what is the magnitude of the lift force acting on the airplane?
Answer:
The net force is 91780.8 N.
Explanation:
mass, m = 8950 kg
Radius, R = 9.33 miles = 15015.2 m
Time, T = 0.123 h = 442.8 s
There are two forces acting on the plane.
Horizontal force is the centripetal force and the vertical force is the weight.
[tex]Fx =m R w^2\\\\Fx = m R \frac{4\pi^2}{T^2}\\\\Fx = 8950\times 15015.2\times \frac{4\times 3.14\times 3.14}{442.8\times 442.8}\\\\Fx = 27030.8 N \\\\Fy = m g \\\\ Fy = 8950\times 9.8 \\\\Fy = 87710 N[/tex]
The net force is
[tex]F = \sqrt{Fx^2 + Fy^2}\\\\F = \sqrt {27030.8^2 + 87710^2}\\\\F = 91780.8 N[/tex]
steel wire 8m long and 4mm in diameter is fixed to two rigid b. [1] supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steel, a 12x10 %/K, Young's modulus for steel, Y = 2x10¹¹ N/m².
Answer:
301.6 N
Explanation:
The length of the wire L₀ = 8 m and its diameter, d = 4 mm = 4 × 10⁻³ m. Since its temperature drops by 10°C, it will have a change in length ΔL = L₀αΔθ where α = linear expansivity of steel, a 12 × 10⁻⁶ /K, and Δθ = temperature change = -10°C = -10 K(negative since it is a drop)
So, the strain, ε = ΔL/L₀ = αΔθ = 12 × 10⁻⁶ /K × 10 K = 12 × 10⁻⁵
Now the Young's modulus of steel, Y = σ/ε where σ = stress = T/A where T = increase in tension in steel wire and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 4 × 10⁻³ m and ε = strain = 12 × 10⁻⁵
So, σ = Yε
Since Y = 2 × 10¹¹ N/m².
Substituting the values of the variables into the equation, we have
σ = Yε
σ = 2 × 10¹¹ N/m² × 12 × 10⁻⁵
σ = 24 × 10⁶ N/m²
Since σ = T/A
T = σA
T = σπd²/4
Substituting the values of the variables into the equation, we have
T = σπd²/4
T = 24 × 10⁶ N/m² × π × (4 × 10⁻³ m)²/4
T = 24 × 10⁶ N/m² × π × 16 × 10⁻⁶ m²/4
T = 24 × 10⁶ N/m² × π × 4 × 10⁻⁶ m²
T = 96 N × π
T = 301.59 N
T ≅ 301.6 N
So, the increase in tension in the steel wire is 301.6 N
Luis rushes 40m toward his truck. He notices he forgot his shoes and returns back to his house. His total travel time is 60s.
What is Luis’s average speed over the 60s?
What is Luis average velocity over the 60s?
1. The average speed of Luis over the 60 s is 1.33 m/s
2. The average velocity of Luis over the 60 s is zero.
1. The average speed can be calculated with the following equation:
[tex] S = \frac{d}{t} [/tex]
Where:
d: is the total distance traveled
t: is the total time = 60 s
The total distance can be calculated by the sum of the distance traveled to the truck (40 m) and the distance traveled to his home (the same 40 m), as follows:
[tex] d = 40 m + 40 m = 80 m [/tex]
So, the average speed is:
[tex] S = \frac{d}{t} = \frac{80 m}{60 s} = 1.33 m/s [/tex]
Hence, Luis's average speed over the 60 s is 1.33 m/s.
2. Now, the average velocity is a vector given by:
[tex]v = \frac{\Delta x}{\Delta t} = \frac{x_{f} - x_{i}}{t}[/tex]
Where:
[tex]x_{f}[/tex]: is the final distance traveled (the way to his home) = 40 m
[tex]x_{i}[/tex]: is the initial distance traveled (the way to his truck) = 40 m
Δt: is the total time = t = 60 s
Given that Luis in his journey, returns to the same starting point (his house), the total displacement is zero, and so the average velocity is also zero:
[tex]v = \frac{x_{f} - x_{i}}{t} = \frac{40 m - 40 m}{60 s} = 0[/tex]
Therefore, Luis's average velocity over the 60 s is zero.
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As shown in the figure below, Luis rushes 40 m toward his truck. He notices he forgot his shoes and returns back to his house. His total travel time is 60 s.
Answer:
What is Luis’s average velocity over the 60 s period?
Answer: 0 m/s
What is Luis’s average speed over the 60 s period?
Answer: 1.33 m/s
a car of mass 500 kg travelling with a speed of 54km/hr is brought to rest over a distance of 40 m. find the acceleration and force applied on brakes
Answer:
Explanation:
[tex]a=\frac{v_f-v_0}{t}[/tex] We don't yet need the mass of the car, as we can see by this equation. What we do need is the velocity of the car, the initial velocity, in meters per second, and right now it's in km/hr. Not good. We need to convert. The conversion is as follows:
[tex]54\frac{km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex] Ok, that's good. Now we have everything we need but the time element. If the car traveled a distance of 40 meters at 15 m/s, then we can use the d = rt equation to solve for t, and when we find t we plug it into the acceleration equation:
40 = 15t and
t = 2.7 seconds. The car traveled for 2.7 seconds to go that 40 meters. That's the only reason we were given the displacement. We need it for nothing else but that.
Filling in the acceleration equation now:
[tex]a=\frac{0-15}{2.7}=-5.6\frac{m}{s^2}[/tex] and the negative indicates we are in fact slowing down. That's the answer for the acceleration portion of the problem; now we need the force, F, applied to the brakes.
F = ma where m is mass (we get to use that value now!) and a is -5.6 m/s/s.
F = 500(-5.6) and
F = -2800 N and the negative here means that the force of the brakes is acting against the motion of the car: the brakes are pulling the car "backwards" to stop while the car's motion is forward. The negative indicates the direction the force is being applied.
