Answer:
The answer is below
Explanation:
a) Using the formula:
[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=15\ rev/s,w_o=initial\ velocity=4.4\ rev/s, \\\theta=distance=60\ rev\\\\Substituting:\\\\15^2=4.4^2+2(60)\alpha\\\\2(60)\alpha=15^2-4.4^2\\\\2(60)\alpha=205.64\\\\\alpha=1.71\ rev/s^2[/tex]
b) The disk is initially at rest. Using the formula:
[tex]\theta=\omega_ot+\frac{1}{2}\alpha t^2 \\\\but\ \omega_o=0(rest), \theta=60\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\60=0+\frac{1}{2}(1.71) t^2\\\\t=8.4\ s[/tex]
c)
[tex]\omega=\omega_o+\alpha t \\\\but\ \omega_o=0(rest), \omega=4.4 \ rev/s\ rev,\alpha=1.71\ rev/s^2\\\\Subsituting:\\\\4.4=1.71t\\\\t=2.6\ s[/tex]
d)
[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ velocity=4.4\ rev/s,w_o=initial\ velocity=0\ rev/s, \\\alpha=1.71\ rev/s^2\\\\Substituting:\\\\4.4^2=0+2(1.71)\theta\\\\19.36=3.42\theta\\\\\theta=5.66\ rev[/tex]
A conveyor belt dumps 2500 yd3 of gravel to form a cone-shaped pile. How high could this pile of gravel be, and what could the circumference of the pile of gravel be at ground level
Answer:
Explanation:
Volume of the cone = 1/3πr²h
r is the radius
h is the height
2500 = 1/3πr²h
Criss multiply
r²h = 2500 × 3π
Rewrite the expression
r²h = 50² × 3π
Compare values
r² = 50²
r = √50²
r = 50yd
The radius is 50yards
Also;
h = 3π yards
h = 3(3.14)
h = 9.42yards
Hence the pile is 9.42yards high
Circumference of the pile = 2πr
Given r = 50
Substitute
Circumference = 2π(50)
Circumference = 100π
Circumference = 100(3.14)
Circumference = 314yards
Hence the circumference of the pile at the ground level is 314yards
PLS HELP PLS 10. This graph represents the velocity of an object over time. What is the average acceleration of the
object during the first 4 seconds?
Answer:
8 M/S
Explanation:
This is because the points match from 4 seconds (X) and it matches up to 8 (Y), therefore it went up 8 M/S in 4 seconds.
find the current of 20 c of charge pass a piticular point in a circuit in 10 seconds
A pencil is dropped from rest from a height of 1.8 meters above the ground. How much time does it take the pencil to hit the ground?
Answer:
0.60s
Explanation:
Using the motion equation as follows:
S = ut + 1/2gt²
Where;
S = distance travelled (m)
u = initial velocity (m/s)
t = time (s)
g = acceleration due to gravity (m/s²)
Based on the information provided, S = 1.8m, u = 0m/s, t = ?
S = ut + 1/2gt²
1.8 = (0×t) + 1/2 (9.8t²)
1.8 = 0 + 4.9t²
1.8 = 4.9t²
t² = 1.8/4.9
t² = 0.3673
t = √0.3673
t = 0.6060
t = 0.60s
A bus accelerates from rest at 2.25 m/s2 for 12.3 seconds.
Answer:
6.15m for 12.3 seconds
Explanation:
What is the weight, on Earth, of a book with a mass of 1.5 kg?
0 1,5 N
0 6,5 N
O 11.3N
0 14 7 N
Answer:
[tex]\boxed {\tt 14.7 \ Newtons}[/tex]
Explanation:
Weight is force due to gravity.
The formula for weight is:
[tex]W=m*g[/tex]
where m is the mass and g is the acceleration of gravity.
On Earth, the acceleration due to gravity is 9.8 meters per seconds squared.
The mass of the book is 1.5 kilograms. Therefore:
[tex]m= 1.5 \ kg \\g=9.8 \ m/s^2[/tex]
Substitute the values into the formula.
[tex]W= 1.5 \ kg *9.8 \ m/s^2[/tex]
Multiply.
[tex]W=14.7 \ kg*m/s^2[/tex]
1 kg meter per second squared is equal to 1 Newton, so our answer of 14.7 kg*m/s² is equal to 14.7 Newtons.
[tex]W=14.7 \ N[/tex]
The weight of a 1.5 kilogram book on Earth is 14.7 Newtons.
Answer:
Answer:
Explanation:
Weight is force due to gravity.
The formula for weight is:
where m is the mass and g is the acceleration of gravity.
On Earth, the acceleration due to gravity is 9.8 meters per seconds squared.
The mass of the book is 1.5 kilograms. Therefore:
Substitute the values into the formula.
Multiply.
1 kg meter per second squared is equal to 1 Newton, so our answer of 14.7 kg*m/s² is equal to 14.7 Newtons.
The weight of a 1.5 kilogram book on Earth is 14.7 Newtons.
Explanation:
All of the following would motivate someone toward attaining a career goal except
A expectation of money
B. low self-esteem
C. belief in success
D. financial stability
All would motivate someone toward attaining a career goal except low self-esteem. The correct option is B.
What is self-esteem?We tend to feel better about ourselves and about life in general when we have healthy self-esteem.
It improves our ability to deal with life's ups and downs. When we have low self-esteem, we tend to view ourselves and our lives in a more negative and critical light.
People's choices and decisions are heavily influenced by their self-esteem. In other words, self-esteem motivates people by making it more or less likely that they will take care of themselves and reach their full potential.
Except for low self-esteem, everyone would not motivate somebody to pursue a career goal.
Thus, the correct option is B as low self esteem cannot motivate anyone for career goal.
For more details regarding self-esteem, visit:
https://brainly.com/question/28768312
#SPJ5
I need help with this science work.
Answer:
1. Radiation
2. Conduction
3. Conduction
4. Convection
5. Convection
6. Convection
7. Radiation
the whale shark is the largest of all fish and can have the mass of 3 adult elephants. suppose that a crane is lifting a whale shark into a tank for delivery to an aquarium. The crane must exert a force of 25000 N to lift the shark from rest. if the sharks acceleration equals 1.25m/s^2, what is the sharks mass ?
Answer: The whale shark is the largest type of fish in the world. Its mass can be as large as 2.00 x 10^4kg, which is the equivalent mass of three average adult elephants. Suppose a crane lifts a net with a 2.00 x 10^4kg whale shark off the ground. The net is steadily accelerated from rest over an interval of 2.5s until the net reaches a speed of 1.0 m/s.
Explanation:
9. A 0.25-kg mass is attached to a string and swung in a vertical circle whose radius is 0.75 m. At the bottom of the circle, the mass is observed to have a speed of 10 m/s. What is the magnitude of the tension in the string at that point
Answer:
T = 33.34 N
Explanation:
Given that,
Mass attached to the spring, m = 0.25 kg
Radius of the vertical circle, r = 0.75 m
Speed of the mass, v = 10 m/s
We need to find the magnitude of the tension in the string at that point. We know that the tension in the string is equal to the centripetal force acting on it. It can be given by
[tex]T=\dfrac{mv^2}{r}\\\\T=\dfrac{0.25\times (10)^2}{0.75}\\\\T=33.34\ N[/tex]
So, the tension in the string is 33.34 N.
If the first stage provides a thrust of 5.25 mega-newtons [MN] and the space shuttle has a mass of 4,470,000 pound-mass [lbm], what is the acceleration of the spacecraft in miles per hour squared [mi/h2]
Answer:
20,861.65 mi/h²
Explanation:
We convert 4,470,000 pound-mass [lbm], to kg. Since 2.205 lbm = 1 kg, then 4,470,000 lbm = 4,470,000 lbm × 1 kg/2.205 = 2,027,210.88 kg
Since Force , F = ma where m = mass and a = acceleration, and our force of thrust , F = 5.25 MN = 5,250,000 N and or mass = mass of spacecraft = 2,027,210.88 kg, we then find the acceleration, a.
a = F/m = 5,250,000 N/2,027,210.88 kg = 2.59 m/s².
We now convert this acceleration into miles per hour. Since 1 mile = 1609 meters and 60 × 60 s = 1 hour ⇒ 3600 s = 1 hour, Our conversion factor for meter to mile is 1 mile/1609 m and that for second to hour is 3600 s/1 hour. We square the conversion factor for the time so we have (3600 s/1 hour)².
Multiplying both conversion factors with our acceleration, we have
a = 2.59 m/s²
= 2.59 m/s² × 1 mile/1609 m × (3600 s/1 hour)²
= 33566440/1609 miles/hour²
= 20,861.65 mi/h²
= 20,861.65 miles per hour squared
what is the answer to life and everything in the universe?
In theory, if the airspeed of an airplane is doubled while in level flight, parasite drag will become
Answer:
Four times greater
Explanation:
Drag is comprised of parasite drag and induced drag. The Parasite drag is the drag that acts on an object when the said object happens to be moving through a fluid. Parasite drag is also a combination of two forms of drag. These two forms of drag are called form drag and skin friction drag.
As the speed of an airplane increases, the parasite drag also increases while the induced drag of the airplane decreases. The parasite drag of an airplane in theory becomes four times greater, when it's airspeed is doubled.
It takes an airplane nearly 3/4 of a mile to stop. Which law is being used?
A) Newton's first law
B) Newton's second law
C) Newton's third law
D) Law of conservation of momentum
E) Law of universal gravitation
Answer:
I think it is D
Explanation:
might be wrong
Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about 5200 K.
Answer: Root mean square speed, Vrms=5,694m/s
Explanation:
Root mean square speed is defined as the the square root of the average of the square of the velocity and given as [tex]\sqrt 3RT/M[/tex]
Where R =8.3145J/mol.k
M molar mass of the molecule in kg/mol
Helium= He=4g/mol = 0.004kg/mol
T= Temperature = 5200k
Vrms= [tex]\sqrt 3RT/M[/tex]
=[tex]\sqrt 3 x 8.3145 x 5200/0.004[/tex]
=[tex]\sqrt32426550[/tex]
Vrms =5,694m/s
A biker pushes on the pedal and accelerates her 50 kg bike from 0 m/s to 10 m/s. What is the the impulse?
Answer:
3.7
Explanation:
hello what is momentum?
Answer:
Momentum is vector quantity and it is defined as the product of mass of the object and velocity of the object.
As it is a vector quantity it has both direction and magnitude, and momentum is written as
p = m × v
p = momentum
m = mass of the object
v = velocity
Two kinds of momentum are there and they are:
Linear MomentumAngular MomentumA 80-kg person stands at rest on a scale while pulling vertically downwards on a rope that is above them. Use g = 9.80 m/s2.
What is the critical magnitude of tension that the rope is pulling on the person when the person just begins to lift off the scale?
Answer:
784 N
Explanation:
Given that a 80-kg person stands at rest on a scale while pulling vertically downwards on a rope that is above them. Use g = 9.80 m/s2.
The weight of the person = mg
The weight = 80 × 9.8
The weight = 784 N
The critical magnitude of tension that the rope is pulling on the person when the person just begins to lift off the scale will be equal to the weight of the person which is 784 N
I WILL MARK YOU AS BRAINLIEST IF RIGHT
What is the magnitude of the net force acting on this object? And what direction?
Answer:
20N to the North
Explanation:
the west and east forces cancel out, that leaves us with 50 to the north, and 30 to the south, the difference would be 20 N to the North
Neil travelled 36km at a speed of 8km/h Grant travelled 48km at a speed of 10km/h whose journey was quickest
Answer:
[tex]\boxed {\tt Neil's \ journey}[/tex]
Explanation:
We want to find who travelled faster, so we must find the time for each person's journey. Time can be found by dividing distance by speed.
Neil
[tex]t=\frac{d}{s}[/tex]
Neil travelled 36 kilometers at a speed of 8 kilometers per hour.
[tex]d= 36 \ km \\s= 8 \ km/hr[/tex]
[tex]d=\frac{36 \ km}{8 \ km/hr}[/tex]
Divide (the kilometers or "km" will cancel each other out).
[tex]d=4.5 \ hr[/tex]
Grant
[tex]t=\frac{d}{s}[/tex]
Grant travelled 48 kilometers at a speed of 10 kilometers per hour.
[tex]d= 48 \ km \\s= 10 \ km/hr[/tex]
[tex]d=\frac{48 \ km} { 10 \ km/hr}[/tex]
Divide (the kilometers or "km" will cancel each other out).
[tex]d=4.8 \ hr[/tex]
Neil's journey took 4.5 hours and Grant's took 4.8 hours. We want to find the journey that was quicker and 4.5 is less than 4.8. So Neil's journey was quicker.
How do mass and the type of material affect thermal energy transfer?
Answer:
Warmer objects have faster particles and higher temperatures. If two objects have the same mass, the object with the higher temperature has greater thermal energy. ... Heat is the transfer of thermal energy between objects that have different temperatures.
An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.54 m/s, and its maximum acceleration is 8.13 m/s2. How much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum
Answer:
Explanation:
The spring will move under SHM .
Maximum speed is at the middle equilibrium position . Maximum acceleration is when the particle is at one of the extreme position . So time gap between the two position is one fourth of time period of oscillation .
= T / 4
If A be the amplitude and ω be angular velocity
ω A = maximum velocity = 1.54
ω² A = maximum acceleration = 8.13
dividing
ω = 5.28
2π / T = 5.28
T = 1.189
T / 4 = 0.3 s.
A student is standing on top of the school building and has a height of 44.1m. They kick a rock off the edge of the building with velocity of 6 m/s . how far away from the base of the building does the rock land?
A man accelerates at 2.52 m/s2 for 3.73s. If his initial velocity was 1.75 m/s, what is his final velocity?
Answer:
Final Velocity = 11.15 m/s
Explanation:
By using the equation v = u + at we can find the answer.
v = final velocity
u = initial velocity
a = acceleration
t = time
Plug the numbers in and you get 11.15 m/s, hope this helps!
QUESTION 1 During the time of Pangaea most of the dry land on Earth was joined into one huge landmass that covered nearly a third of the planet's surface. The giant ocean that surrounded the continent is known as what?
QUESTION 2Pangaea existed during what geological time periods, which were times of great change?
QUESTION 3 Most distributions of rocks within Earth's crust, including minerals, fossil fuels, and energy resources, are a direct result of the history of plate motions and collisions and the corresponding changes in the configurations of the continents and ocean basins. Research and explain how plate tectonics could account for these resources exactly.
PLS ANSWeR
Answer:
Pangea existed between about 299 million years ago (at the start of the Permian Period of geological time) to about 180 million years ago (during the Jurassic Period). It remained in its fully assembled state for some 100 million years before it began to break up.
Explanation:
At room temperature, most metals are
Select one:
a. gas
O b. liquid
O c. an alloy
O d. solid
how much tension (fm) must be supplied at the triceps to stabalize the arm against an external force (fe) of 200 N given dm
The missing part of the question is :
given dm-2 cm and de 25 cm
Answer:
The correct answer is = 2500 N
Explanation:
Given:
External force (fe) = 200 N
dm- 2 cm = 0.02 m
de = 25 cm = 0.25 m
Balancing torque about elbow
Fm*dm = Fe*de
By putting value
fm*0.02 = 200*0.25
fm= 200*0.25/0.02
fm = 50/0.02
fm = 2500 N
Thus, the correct answer is = 2500 N
If George Washington had become "King of America," our government might have become?
Answer:
Monachry
Explanation:
Hope this helped!!!
A 5.00 gg bullet is fired horizontally into a 1.20 kgkg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.390 mm along the surface before stopping.
Answer:
V= 295.2m/s
Explanation:
Given that mass of bullet = 5.0g = 0.005kg
Mass of wooden block = 1.20kg
The coefficient of kinetic friction between block and surface = 0.20
The force of friction of the block
Fk = (μ) mg
Fk = 0.2×1.2×9.8m/s
F = 2.352N
= Force × distance
The work done = 2.352× 0.390
= 0.91728J
The initial velocity of the block can be calculated
1/2mv^2
0.5×1.2×v^2 = 0.917J
0.6V^2 = 0.917J
V^2 = 0.917J/0.6
V^2 = 1.52
V = √1.52
V = 1.23m/s
Now we shall use conservation of momentum to find the velocity of the bullet
0.005kg×v = 1.2kg × 1.23m/s
0.005v = 1.476
V= 1.476/0.005
V= 295.2m/s
Protons can be accelerated to speeds near the speed of light in particle accelerators. Calculate the wavelength of a proton moving at 2.90 x 108 m/s if the proton was a mass of 1.673 x 10-24 g. 2. Give the values for the quantum numbers associated with the following orbitals a) 2p b) 3s c) 5d
Answer:
1. λ = 1.4x10⁻¹⁵ m
2. a) n=2, l=1, [tex]m_{l}[/tex]= -1, 0, +1, [tex]m_{s}[/tex] = +/- (1/2)
b) n=3, l=0, [tex]m_{l}[/tex]= 0, [tex]m_{s}[/tex] = +/- (1/2)
c) n=5, l=2, [tex]m_{l}[/tex]= -2, -1, 0, +1, +2, [tex]m_{s}[/tex] = +/- (1/2)
Explanation:
1. The proton's wavelength can be found using the Broglie equation:
[tex] \lambda = \frac{h}{mv} [/tex]
Where:
h: is the Planck's constant = 6.62x10⁻³⁴ J.s
m: is the proton's mass = 1.673x10⁻²⁴ g = 1.673x10⁻²⁷ kg
v: is the speed of the proton = 2.90x10⁸ m/s
The wavelength is:
[tex] \lambda = \frac{h}{mv} = \frac{6.62 \cdot 10^{-34} J.s}{1.673 \cdot 10 ^{-27} kg*2.90 \cdot 10^{8} m/s} = 1.4 \cdot 10^{-15} m [/tex]
2. a) 2p
We have:
n: principal quantum number = 2
l: angular momentum quantum number = 1 (since is "p")
[tex]m_{l}[/tex]: magnetic quantum number = {-l,... 0 ... +l}
Since l = 1 → [tex]m_{l} = -1, 0, +1[/tex]
[tex]m_{s}[/tex]: is the spin quantum number = +/- (1/2)
b) 3s:
n = 3
l = 0 (since is "s")
[tex]m_{l}[/tex] = 0
[tex]m_{s}[/tex] = +/- (1/2)
c) 5d:
n = 5
l = 2 (since is "d")
[tex]m_{l}[/tex] = -2, -1, 0, +1, +2
[tex]m_{s}[/tex] = +/- (1/2)
I hope it helps you!
Answer:
(1) The wavelength of the proton is 1.366 x 10⁻¹⁵ m
(2) 2p( l = 1, ml = -1,0,+1)
3s( n = 3, l = 0, ml = 0)
5d ( l = 2, ml = -2,-1,0,+1,+2)
Explanation:
Given;
mass of the proton; m = 1.673 x 10⁻²⁴ g = 1.673 x 10⁻²⁷ kg
velocity of the proton, v = 2.9 x 10⁸ m/s
The wavelength of the proton is calculated by applying De Broglie's equation;
[tex]\lambda = \frac{h}{mv}[/tex]
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
Substitute the given values and solve for wavelength of the proton;
[tex]\lambda = \frac{h}{mv}\\\\ \lambda = \frac{(6.626*10^{-34})}{(1.673*10^{-27})(2.9*10^8)}\\\\\lambda = 1.366 *10^{-15} \ m[/tex]
(2) the values for the quantum numbers associated with the following orbitals is given by;
n, which represents Principal Quantum number
[tex]l,[/tex] which represents Azimuthal Quantum number
[tex]m_l,[/tex] which represents Magnetic Quantum number
(a) 2p (number of orbital = 3):
[tex]l= 1\\m_l = -1,0,+1[/tex]
(b) 3s (number of orbital = 1):
[tex]n= 3\\l=0\\m_l= 0[/tex]
(c) 5d (number of orbital = 5)
[tex]l=2\\m_l = 2, -1, 0, +1, +2[/tex]