Answer:
ghgivjgifigo ra together aigig disgust u hoodie
Raghu studies in grade 6th. He wants a cricket bat to be made by a carpenter. He tells the
that the length of the bat should be 7 hand spans. The tall carpenter tells Raghu that it
will be ready by tomorrow. When Raghu went to collect the bat the next day, he was very
disappointed. Why? Was the bat longer or shorter than what Raghu expected? Give reason.
carpenter
Based on the information given, it can be noted that the bat was either shorter or longer than what he expected.
From the information given, it was stated that Raghu wants a cricket bat to be made by a carpenter and he tells the carpenter that the length of the bat should be 7 hand spans.
Since he got disappointed when he collected the bat, the reason for this will be because the bat was either shorter or longer than what he requested.
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Hello I am absolutely stumped on these six physics problems. Please help me on them.
Answer:
Explanation:
20° from the normal = 110° from parallel
1a) τ = (200sin90)[6] + (75sin110)[3] - (100sin110)[3]
τ = 1,129.52305... = 1100 N•m CCW
1b) τ = 200(6)(sin90) + 75(-3)(sin(360-110)) + 100(3)(sin(270 + 20)
τ = 1,129.52305... = 1100 N•m CCW
1c) My directions agree, both are positive z by right hand rule.
1d) Moment of inertia for a thin rod about an axis perpendicular to its center is
I = (1/12)mL²
τ = Iα
α = τ/I = 1129.523 / ((1/12)(200)(12²)) = 0.4706 rad/s²
θ = ½αt²
θ = ½(0.4706)(2•60)² = 3,388.56916... radians
θ = 3400 radians
at which time is is spinning about 9 revolutions per second
Two rods of mass m, length L are stuck together to form an X shape and spun around the center. What is the rotational inertia
9. P waves move faster than S waves A. True B. False
a uniform thin rod of length l and mass m is allowed to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position. a.) What is the speed of the center of gravity when the rod reaches its lowest position? b.) What is the tangential speed of the lowest point of the rod when the rod reaches its lowest position?
Answer:
Explanation:
Potential energy gets converted to rotational kinetic energy
a) ½Iω² = mgh
½(mL²/3)ω² = mgL/2
(L/3)ω² = g
ω = [tex]\sqrt{3g/L}[/tex]
v(CG) = (L/2) [tex]\sqrt{3g/L}[/tex]
Not sure if you wanted angular speed or tangential speed of the CG so I gave both.
b) v = Rω = L [tex]\sqrt{3g/L}[/tex]
Sophie applies a 50 n force to push a box 2 meter across the floor calculate the smount of work done in the box
PLEASE HELP ASP WILL GIVE 50 POINT AND BRAINLIEST!!!!!!!!!!!!!
A. In the Bohr model of the hydrogen atom, the speed of the electron is approximately
2.16 × 10⁶ m/s. Find the central force acting on the electron as it revolves in a circular orbit of radius 5.17 × 10⁻¹¹ m.
Answer in units of N.
B. Find the centripetal acceleration of the electron
Explanation:
A. The centripetal force experienced by an electron as it goes around a hydrogen nucleus is given by
[tex]F_c = m_e\dfrac{v^2}{r}[/tex]
where [tex]m_e = \text{electron\:mass} = 9.11×10^{-31}\:\text{kg}[/tex]
[tex]r = 5.17×10^{-11}\:\text{m}[/tex] = orbital radius
[tex]v = 2.16×10^6\;\text{m/s}[/tex] = orbital velocity
so the centripetal force is
[tex]F_c = (9.11×10^{-31}\:\text{kg})\dfrac{(2.16×10^6\;\text{m/s})^2}{5.17×10^{-11}\:\text{m}}[/tex]
[tex]\;\;\;=8.22×10^{-8}\:\text{N}[/tex]
B. The electron's centripetal acceleration is given by
[tex]a_c = \dfrac{v^2}{r}[/tex]
Using the values from (A), we get
[tex]a_c = \dfrac{(2.16×10^6\;\text{m/s})^2}{5.17×10^{-11}\:\text{m}}[/tex]
[tex]\;\;\;=9.02×10^{22}\:\text{m/s}^2[/tex]
True or False. Abraham and Sarah were in agreement regarding God's command to sacrifice Isaac.
Answer:
Explanation:
Not in Genesis. The proposed sacrifice of Isaac was all Abraham's doing. Sarah was not present and could not, therefore offer any input. It might have been covered in other Jewish writings, but it is not recorded in Genesis.
Since the book most widely used by Everyone is the Bible, I would answer false.
Which ball moved at the same speed as Ball 3?
Answer:
you forgot to attach the image
What is the half-life of the imaginary element Lokium? Show your work
Answer:What is the half life of the element Lokium?
The half-live of the element Lokium is 4.
Explanation:
sorry if im wrong, have good day
Answer:
The half-life of the element Lokium is 4
What is half-life?
Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life. The half-life of a radioactive isotope is the amount of time it takes for one half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.
#SPJ2
In the following free body diagram, what is the net force on the object?
A.10n
B.5 N to the right
C.20 N to the right
D.7 N to the right
Answer:
B
Explanation:
Simply take all forces pointing to the right of the box as positive and all of the forces pointing to the left of the box as negative and add all values.
ΣF = 7 + 18 + (-20) = 5N to the right
a spring with a spring constant of 65 n/m is displaced -94cm. what is the magnitude of the force exerted by the spring
Answer:
Explanation:
F = kx = 65 N/m(0.94 m) = 61.1 = 61 N
How can whales descend quickly and face no problems
Answer:
Whales face an increasing number of threats including: For hundreds of years people hunted whales for their oil to fuel lamps and candles, to lubricate machinery and to make margarine, lipsticks and other products. They also used baleen whales to make tennis racquets and corsets!
Explanation:
HELP ME the mean free path λ and the mean collision time τ of the molecules of a diatomic gas of molecular mass 6.00 × 10⁻²⁵ kg and radius r = 1.0 x 10⁻¹⁰ m are measured. From these microscopic data can we obtain macroscopic properties such as temperature T and pressure P? If so, consider λ = 4.32 x 10⁻⁸ m and τ = 3.00 x 10⁻¹⁰ s and calculate T and P.
The temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.
The given parameters;
Mass of the gas molecules, m = 6 x 10⁻²⁵ kgRadius of the gas, r = 1 x 10⁻¹⁰ mMean free path, [tex]\lambda_{rms}[/tex] = 4.32 x 10⁻⁸ mMean collision time, [tex]\tau = 3 \times 10^{-10} \ s[/tex]The mean velocity of the gas molecules is calculated as follows;
[tex]\tau = \frac{\lambda _{rms}}{V_{rms}} \\\\V_{rms} = \frac{\lambda _{rms}}{\tau} \\\\V_{rms} = \frac{4.32 \times 10^{-8} }{3 \times 10^{-10}} \\\\V_{rms} = 144 \ m/s[/tex]
The temperature of the gas molecules is calculated as follows;
[tex]V_{rms} = \sqrt{\frac{3kT}{M} } \\\\V_{rms}^2 = \frac{3kT}{M} \\\\T = \frac{V_{rms} ^2 M}{3k}[/tex]
where;
k is Boltzmann constant[tex]T = \frac{V_{rms} ^2 M}{3k} \\\\T = \frac{(144)^2 \times (6.0 \times 10^{-25})}{3 \times 1.38 \times 10^{-23}} \\\\T = 300.5 \ K[/tex]
The number of gas molecules per unit volume is calculated as follows;
[tex]\lambda = \frac{1}{4\pi \sqrt{2} \ r^2 n} \\\\n = \frac{1}{\lambda 4\pi \sqrt{2} \ r^2} \\\\n = \frac{1}{(4.32 \times 10^{-8}) \times 4 \pi \times \sqrt{2} \ \times (1\times 10^{-10})^2} \\\\n = 1.303 \times 10^{26} \ molcules/m^3[/tex]
The pressure of the gas molecule is calculated as follows;
[tex]n = \frac{P}{kT} \\\\P = nkT\\\\P = (1.303 \times 10^{26} ) \times (1.38 \times 10^{-23}) \times (300.5)\\\\P = 540,341.07 \ Pa\\\\P = 5.33 \ atm[/tex]
Thus, the temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.
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A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) what is the mass of the block office? (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?
Answer:
(a) 91 kg (2 s.f.) (b) 22 m
Explanation:
Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.
(a)
[tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}[/tex]
Subsequently,
[tex]F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})[/tex]
*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.
(b) To find the final velocity of the ice block at the end of the first 5 seconds,
[tex]v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}[/tex]
According to Newton's First Law which states objects will remain at rest
or in uniform motion (moving at constant velocity) unless acted upon by
an external force. Hence, the block of ice by the end of the first 5
seconds, experiences no acceleration (a = 0) but travels with a constant
velocity of 4.4 [tex]m \ s^{-1}[/tex].
[tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}[/tex]
Therefore, the ice block traveled 22 m in the next 5 seconds after the
worker stops pushing it.
Can someone help me solve this problems please? It's a physics problem.
Answer:
i cant see
Explanation:
but im smart
14. Earthworms are crucial for forming soil. As they search for food by digging tunnels,
they expose rocks and minerals to the effects of weathering. Over time, this process
creates new soil. Worms are not the only living things that help to create soil. Plants
also play a part in the weathering process. As the roots of plants grow and seek out
water and nutrients, they help to break large rock fragments into smaller ones. Have
you ever seen a plant growing in a sidewalk? As the plants grows, its roots spread into
tiny cracks in the sidewalk. These roots apply pressure to the cracks, and, over time,
the cracks become larger, ice wedging can occur more readily. As the cracks expand,
more water can flow into them. When the water freezes, it expands and presses against
the walls of the cracks, which makes the cracks larger. Over time, the weathering
caused by water, plants, and worms help to form soil. QUESTION: Ice wedging, as
described in the passage, is an example of which of the following?
A mechanical weathering
B. Oxidation
C. chemical weathering
D. Hydrolysis
Ice wedging, as described in the passage, is an example of mechanical
weathering.
Mechanical weathering is also known as physical weathering and it
involves the breaking of rock into smaller particles without causing changes
in the chemical properties.Mechanical weathering is usually carried through
physical processes such as freezing and thawing etc.
In this scenario, we were told that water freezes, expands and presses
against the walls of the crack thereby breaking into smaller parts which is a
physical process hence mechanical weathering being present.
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a 2000kg car initially traveling at a speed of 15 m/s is accelerated by a constant force of 10000 n for 3 seconds. the new speed of the car is
Answer:
The new speed of the car is 30 m/s.
Hope you could get an idea from here.
Doubt clarification - use comment section.
A 2.2 kg model rocket is shot straight up in the air from the ground, with an initial velocity of 36.4 m/s. The rocket reaches its maximum height, and falls back to the ground. What is the maximum height of the rocket? Round your answer to 2 decimal places.
Answer:
Explanation:
Ignoring friction, the initial kinetic energy will convert to maximum potential energy at its highest point.
PE = KE
mgh = ½mv²
h = v²/2g
h = 36.4²/ (2(9.81))
h = 67.53109...
h = 67.53 m
A moving car initially has kinetic energy K. The car then moves in the opposite direction with four times its initial speed. What is the kinetic energy now
If initial speed was v, Kinetic energy was K = 1/2mv^2
When speed will be 4v, KE will be:
= 1/2 m (4v)^2
=1/2m 16 v^2
=16K
Now kinetic energy will become 16 times of initial Kinetic Energy
Predict changes in state according to change in particle motion. Know the vocabulary used to describe changes of state.
The change in the state of matter causes change in the motion of the particles of the matter. The gaseous state of matter has the greatest speed while the solid state has the least speed.
The change in state of every matter is accompanied by lost or gained of energy.
Example is water.
The solid state of water is ice. The motion of particles of the water is relatively zero because the molecules are held at a fixed position.
The liquid state of water occurs when the temperature of the ice is increased above zero degree Celsius. The speed of the particles of water in liquid state is greater than solid state.
The gaseous state of water occurs when the temperature of the liquid water is increased beyond 100 degree Celsius. The speed of water in gaseous state is greater than liquid state.
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PLEASE ANSWER QUICK GIVING BRAINLIEST TO THE ONE WHO ANSWERS
Describe what happens when iron and oxygen combine. Can the change be reversed?
Answer:
iron combines with oxygen to produce rust, which is the compound named iron oxide.
Explanation:
A wire that is 1.0 m long with a mass of 90 g is under a tension of 710 N. When a transverse wave travels on the wire, its wavelength is 0.10 m. What is the frequency of this wave?
Answer:
890 HzI hope you liked my answer. Thank You!
Suppose a bulldozer is being hauled at 50 km/h to a construction site on the back of a flatbed truck. From which reference point would the bulldozer not appear to be moving?
Answer:
from someone watching the bulldozer from the cab of the truck or from elsewhere on the flatbed.
Explanation:
How large is the tension in a rope that is being used to accelerate a 100 kg box upward at 2m/s2?
[tex]\\ \sf\Rrightarrow T=F[/tex]
[tex]\\ \sf\Rrightarrow T=ma[/tex]
[tex]\\ \sf\Rrightarrow T=100(2)[/tex]
[tex]\\ \sf\Rrightarrow T=200N[/tex]
Which of the following is NOT a function of the lens in the eye?
It can perform minor adjustments for distance.
It flattens when light rays from distant objects are to be focused.
It is a light receptor that generates nerve signals that are sent to the brain.
It maintains its spherical shape to view nearby objects.
Answer:
it is a light receptor that generates nerve signals that are sent to the brain
Explanation:
the lens are like the glasses, this means that is used to see things better. You just put them in your eye and that's all it's not connected to the brain
Answer:
c
Explanation:
Select the best reason for studying the past and its effect on us today based on "The Terror of the Middle Ages." A. to learn what people did on a daily basis B. to enjoy stories about where people used to live O C. to study the causes of diseases and learn to prevent them D. to learn about earlier cultures and lifestyles
it is D for sure im good with history
Answer: D
Explanation: Its just s that guy ca get brainlist :D
2. Define Lightning conductor. How does it work?
In Building, there are a host of protective devices that are installed to protect lives and properties, one of them is the Lightning Conductor that a metal rod mounted on a structure and intended to protect the structure from a lightning strike. It works on the principle of induction
Principle of Operation of Lightning ConductorThe lightning conductor works on the principle of induction:
When a charged cloud passes by the building hosting the Lightning conductor, it gets a charge opposite to that of the cloud through the process of induction. They are Typically made from copper material.Most Lightening conductors are made from copper materials.
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As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constants. The angular speed changes from 3.70 rad/s at t = 0 to 2.00 rad/s at t = 8.60 s.
a. Use this information to determine σ and ω0.
σ = _______s−1
ωo = ______rad/s
b. Determine the magnitude of the angular acceleration at t = 3.00 s.
______rad/s2
c. Determine the number of revolutions the wheel makes in the first 2.50 s
_______rev
d. Determine the number of revolutions it makes before coming to rest.
_______rev
Hi there!
a.
We can use the initial conditions to solve for w₀.
It is given that:
[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]
We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:
[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]
Now, we can solve for sigma using the other given condition:
[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]
b.
The angular acceleration is the DERIVATIVE of the angular velocity function, so:
[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]
c.
The angular displacement is the INTEGRAL of the angular velocity function.
[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]
[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]
[tex]\theta = 8.471 rad[/tex]
Convert this to rev:
[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]
d.
We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.
[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]
Evaluate the improper integral:
[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]
Convert to rev:
[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]
Use the equation of motion to answer the question. Use the equation of motion to answer the question.
x=x0+v0t+12at2
An object has a starting position of x = 2 m, a starting velocity of 4.5 m/s, and no acceleration. Which option shows the final position of the object after 2 s?
The final position of the object after 2 s is 11 m.
Motion: This can be defined as the change in position of a body.
⇒ Formula:
x = x₀+v₀t+1/2(at²)........................ Equation 1⇒ Where:
x = Final position of the objectx₀ = Starting positionv₀ = Starting velocityt = timea = accelerationFrom the question,
⇒ Given:
x₀ = 4.5 m/st = 2 sx₀ = 2ma = 0 m/s²⇒ Substitute these values into equation 1
x = 2+(4.5×2)+1/2(0²×2)x = 2+9+0x = 11 mHence, The final position of the object after 2 s is 11 m
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