From the diagram, the angular speed will increase clockwise, the sign of the angular acceleration will be negative and the direction of rotation will be clockwise direction and the sign of the angular acceleration is negative. The correct answer is option B
Given that two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Counterclockwise is considered the positive rotational direction.
If m1 is 24 kilograms, m2 is 12 kilograms, and mbar is 10 kilograms, The moment of object m1 will be equal to the moment of object m2 without the Mbar
Let assume that the length L of the seesaw is 9 cm.
Anticlockwise moment = 24 x 9/3 = 72Nm
Clockwise moment = 12 x 2(9/3) = 72 Nm
With the consideration of mass of the bar Mbar, this will add to clockwise moment of the seesaw.
Therefore, the direction of rotation will be clockwise direction.
Angular acceleration is positive when object is speeding up and negative when slowing down. Also, angular acceleration is positive when speed increases in an anticlockwise direction and negative when speed increases in the clockwise direction.
From the diagram, since the angular speed increase clockwise, the sign of the angular acceleration will be negative.
We can conclude that the direction of rotation will be clockwise direction and the sign of the angular acceleration is negative.
The correct answer is option B
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describe the motion of objects that are viewed from your reference frame both inside and outside while you travel inside a moving vehicle
Answer:
The objects outside the reference frame aren't moving. It appears this way since the vehicle you are inside is moving, but unless the objects are people, animals, or other vehicles, the objects aren't moving.
The symbol delta x (x) is used to find what value?
Answer:
Explanation:
Δx means a change in the magnitude of the x variable, often used in reference to a number line on the horizontal axis of a plot.
A race car traveling at 100 m/s enters an unbanked turn of 400 m radius. The coefficient of (static) friction between the tires and the track is 1.1. The track has both an inner and an outer wall. Which statement is correct
Answer:
The race car will crash into the outer wall
Explanation:
max fr = μsN = 1.1 mg = 11 m
mv2/R = m(100)2/(400) = 25 m > fr
how does the structure of compounds determines the properties of the compounds?
Answer:
The chemical structure of the molecule is responsible for each of these characteristics. The chemical structure is comprised of the bonding angle, the kind of bonds, the size of the molecule, and the interactions that occur among the molecules. Even little changes in the chemical structure of a molecule may have a significant impact on the characteristics of the substance.
Explanation:
Hope it helps:)
what is the acceleration of the cart at t=8 seconds?
a) 0 m/s^2
b) 10 m/s^2
c) 20 m/s^3
d) -20m/s^2
What is the acceleration of the cart at t=8 seconds?
a) 0 m/s^2b) 10 m/s^2c) 20 m/s^3d) -20m/s^2Hence the answer us letter a) 0 m/s^2.
That's all I know, Hope it help :)
A farm tractor starting from the rest attains a speed of 36000 m/s after covering a distance of 2000 m. Work out the magnitude of the net force the tractor weighs 5000 kg.
Answer:
the answer is 3,888.7
Explanation:
Hope this answer helped!:)
I need help been struggling on this question
Answer:
440 m
Explanation:
S=(u+v) t / 2
S = (11+33) × 20/2
S= 44× 20/2
S=440 m
numerical problems:
a.) convert 300K into the celsius scale.
b.) convert 220 centigrade scale into kelvin scale.
c.) convert 20 ventigrade scale into Fahrenheit scale.
d.) convert 260 Fahrenheit into centigrade. pls help me to solve this problems
How does a balanced chemical equation demonstrate the Law of Conservation of Mass? it shows that only physical changes follow the Law of Conservation of Mass it shows that only physical changes follow the Law of Conservation of Mass it shows that the properties of the elements stay the same after the reaction it shows that the properties of the elements stay the same after the reaction it shows that all compounds remain bonded after the reaction it shows that all compounds remain bonded after the reaction it shows that no atoms have been gained or lost during the reaction
Answer:
it shows that the properties of the elements stay the same after the reaction
it shows that the properties of the elements stay the same after the reaction
it shows that all compounds remain bonded after the reaction
it shows that all compounds remain bonded after the reaction
it shows that only physical changes follow the Law of Conservation of Mass
it shows that only physical changes follow the Law of Conservation of Mass
it shows that no atoms have been gained or lost during the reaction
it shows that no atoms have been gained or lost during the reaction
A 2457 kg car moves with initial speed of 18 ms-l. It is stopped in 62 m by its brakes.
What is the force applied by the brakes?
Answer:
Explanation:
The work of the brakes will equal the initial kinetic energy of the car
Fd = ½mv²
F = mv²/2d
F = 2457(18²) / (2(62))
F = 6,419.903...
F = 6.4 kN
Saturn's mass is 5.68 x 1024 kg and its radius is 6.03 x 107 m. A. Calculate the gravitational field strength at Saturn's surface. (2 marks) B. Calculate the force of gravity at Saturn's surface on an object with a mass of 50 kg.
Hi there!
A.
We can calculate the gravitational field strength using the following equation:
[tex]g = \frac{Gm_p}{r^2}[/tex]
G = Gravitational Constant
mp = mass of planet (kg)
r = radius (m)
Plug in the given values:
[tex]g = \frac{(6.67*10^{-11})*(5.68*10^{24})}{(6.03*10^7)^2} = \boxed{0.104 N/kg}[/tex]
B.
The force can be calculated using:
[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]
Plug in the values:
[tex]F_g = \frac{(6.67*10^{-11})(5.68*10^{24})(50)}{(6.04*10^7)^2} = \boxed{5.209N}[/tex]
Answer:
[tex]\boxed {\boxed {\sf g=0.104 \ N/kg \ and \ F_g= 5.2 \ N }}[/tex]
Explanation:
A. Gravitational Field Strength
The gravitational field strength can be calculated using the following formula:
[tex]g= \frac{Gm}{r^2}[/tex]
G, or the universal gravitational constant, is 6.67 × 10⁻¹¹ N*m²/kg². The mass of Saturn is 5.68 × 10²⁴ kilograms. The radius of Saturn is 6.03×10⁷ meters.
Substitute these values into the formula.
[tex]g= \frac{ (6.67 \times 10^{-11} \ N*m^2/kg^2) (5.68 \times 10^{24} \ kg)}{(6.03 \times 10^{7} \ m )^2}[/tex]
Multiply the numerator and square the denominator.
[tex]g= \frac{ 3.78856 \times 10^{14} \ N *m^2/kg }{3.63609 \times 10^{15} \ m^2}[/tex]
Divide.
[tex]g= 0.1041932405 \ N/kg[/tex]
The original measurements of mass and radius have 3 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 1 in the ten-thousandth place tells us to leave the 4 in the thousandth place.
[tex]\boxed {g \approx 0.104 \ N/kg}[/tex]
B. Force of Gravity
The force of gravity is calculated using the following formula:
[tex]F_g= mg[/tex]
The mass of the object is 50 kilograms. We just calculated the gravitational field strength, which is 0.104 Newtons per kilogram. Substitute these values into the formula.
[tex]F_g= (50 \ kg)(0.104 \ N/kg)[/tex]
Multiply. The units of kilograms cancel.
[tex]\boxed {F_g=5.20 \ N}[/tex]
What type of equilibrium maintains body position during sudden motion?
dynamic
rotational
static
balanced
I think static is the correct answer
Standing at a crosswalk, you hear a frequency of 550 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 475 Hz. Determine the ambulance's speed from these observations. (Take the speed of sound to be 343 m/s.)
a box has a mass of 4 kg and surface area 4 metre square calculate the pressure exerted by the box on the floor
- BRAINLIEST answerer ❤️✌
Answer:
9.8
Explanation:
we know ,
f=m.g(g=9.8m/s^2)
now,
p=force/area
p=m.g/a
p=4×9.8/4
p=4 pascal.
if u put g as 10 then ypu will get 10 pascal
Does it appear that true average HAZ depth is larger for the high current condition than for the nonhigh current condition
Answer:
The data suggest that the true mean HAZ depth is larger when the current setting is higher.
define heterotrophic
Answer:
Heterotrophic requiring complex organic compounds of nitrogen and carbon (such as that obtained from plant or animal matter) for metabolic synthesis.
Answer -:
⟹ It is a mode of nutrition in which organism are unable to synthesize organic substance by themselves and obtain part of whole of organic substance from external environment. the organism that obtain their food by this method are called heterotphs.
tha organism which lack green pigments chlorophyll are included in this group.all animal ,fungi and most of bacteria belong to this group.a large number of higher plant also Lack chlorophyll . they are also unable to synthesize their organic substance food .On the basis of types of food and feeding habit's nutrition classified into following group -:
(1) Holozoic Nutrition -:
nutrition and animal consumes a plant or an animal us whole or a part of it in solid or liquid form most of free living a cellular protist and all animal show Holozoic Nutrition
(2) Herbivores -:
animal eating grass or other plant material are called Herbivores.
Example - Grazers- horse , cow , goat etc .(3) Carnivorous
flesh eating animal are called carnivorous.
Example -: Lion , tiger , Wolf etc(4) Omnivorous
animal eating food or plant us well us animal origin are called omnivorous.
(5) Insectivorous
insect eating animal are calledinsectivorous .
(6) Frugivores
animal which mainly depend on fruit are called frugivores.
a convex mirror and a plane mirror both give virtual and erect images still a convex mirror is used in vehicles. why?
Pls answer thiss
Answer:Convex mirrors are used because these mirrors provide a wider viewing angle than a plane mirror. This wide angle will help you getting more information/overview than what is happening at a narrow spot right behind the car if you use a plane mirror.
With a convex mirror you are for example able to detect an overtake (by the car behind you) early, if you for some reason wanted to turn left into another lane at the same moment the overtake took place - so you then can prevent a collision. Convex mirrors are simply covering a much larger area behind the car than plane mirrors do. And in the US, on the mirrors there is a text explaining that the vehicle behind you is closer than it appears - some kind of an idiot explanation in case some driver took the mirror image literally….because in the mirror image of a convex mirror, everything looks smaller and further away than they actually are.
Explanation: mark me as brainliest this is my best answer till now
what memory are you using to remember who the president of the united states is
Answer:
The First 8 Presidents
For this exercise, we're going to use a silly story made of silly sentences. The letters that represent the last names of these presidents are W, A, J, M, M, A, J, V. One silly sentence to help you remember this sequence is: Wilma and John made merry and just vanished
working memory.
sensory memory.
short-term memory.
long-term memory.
A race car, starting from rest, travels around a circular turn of radius 22.5 m. At a certain instant the car is still speeding up, and its angular speed is 0.541 rad/s. At this time, the car’s total acceleration vector (centripetal plus tangential) makes an angle of 39.0 with respect to the car’s velocity. What is the magnitude of the car’s total acceleration
Answer:
Explanation:
The answer:
https://www.chegg.com/homework-help/questions-and-answers/race-car-starting-rest-travels-around-circular-turn-radius-247-m-certain-instant-car-still-q402991
Total acceleration of car is 148.31 m/s².
What is centripetal acceleration?Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.
Given parameters:
Radius of the circular path: r = 22.5 m.
Angular speed: ω = 0.541 rad/s.
So, centripetal acceleration; α = ω²r = (0.541)² × 22.5 m/s² = 6.58 m/s².
Tangential acceleration: [tex]\alpha_t[/tex] = αr = 6.58 × 22.5 = 148.16 m/s².
Hence, total acceleration = √(α² + [tex]\alpha_t[/tex]²) = √(6.58² +148.16²) = 148.31 m/s².
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Just need the answer
Answer:
1. 1, 2, 4 all show some form of refraction as the bending of a light ray when passing from one media to another.
Explanation:
Number 4 is the most accurate as it also shows some light being reflected and the bending of the refracted light ray in the correct direction for going from a medium of low refractive index (air) into a higher refractive index material (crown glass)
a body of mass 15 kg accelerates from rest of the rate of 4.0 ms^-2. determine the distance with the body travel in 25 seconds
The distance traveled by the body in the given time is 1,250 m.
The given parameters;
mass of the body, m = 15 kgacceleration of the body, a = 4 m/s²time of motion, t = 25 sinitial velocity, u = 0The distance traveled by the body in the given time is calculated as follows;
[tex]s =ut + \frac{1}{2} at^2\\\\s = 0 \ + \ \frac{1}{2} (4)(25^2)\\\\s =1,250 \ m[/tex]
Thus, the distance traveled by the body in the given time is 1,250 m.
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How is the wavelength of a sound affected when (a) a sound source moves toward a stationary observer and (b) the observer moves away from a stationary sound source
Answer:
If the observer is stationary but the source moves toward the observer at a speed vs, the observer still intercepts more waves per second and the frequency goes up. This time it is the wavelength of the wave received by the observer that is effectively shifted by the motion, rather than the speed.
It velocity of light scalar or vector equality
Answer:
velocity is a vector quantity
Explanation:
velocity is a vector quantity because it has a mass and a direction
An empty cylindrical barrel is open at one end and rolls without slipping straight down a hill. The barrel has a mass of 15.0 kg, a radius of 0.400 m, and a length of 0.800 m. The mass of the end of the barrel equals a fourth of the mass of its side, and the thickness of the barrel is negligible. The acceleration due to gravity is =9.80 m/s2.
What is the translational speed f of the barrel at the bottom of the hill if released from rest at a height of 33.0 m above the bottom?
Hi there!
We can use work and energy to solve this problem.
We know that:
Ei = Ef
Ei = Potential energy = mgh
Ef = Rotational kinetic + Translational kinetic = 1/2Iω² + 1/2mv²
The barrel is comprised of a hollow cylinder and disk-shaped bottom, so:
I (hollow cylinder) = mr²
I (disk) = 1/2mr²
Calculate the moment of inertias of each.
Since the mass on the base is one-fourth of its side:
x = mass of side
x + x/4 = 15
4x + x = 60
5x = 60
x = 12 kg
end mass = 3 kg
Solve for each moment of inertia:
Side: (12)(0.4²) = 1.92 Kgm²
Bottom: 1/2(3)(0.4²) = 0.24 Kgm²
Side + bottom = 2.16 Kgm²
We can now solve:
mgh = 1/2mv² + 1/2(2.16)v²/r²
(15)(9.8)(33) = 1/2(15)v² + 1/2(13.5)v²
4851 = 14.25v²
v = 18.45 m/s
A wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it (see figure). The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope. The tension in the rope during the descent of the block is 20 N. What is the moment of inertia of the wheel?
The moment of inertia of the wheel is 4.27 kg.m²
The kinematics equation explains the variables associated and related of motion.
From the information given, applying the kinematic equation of motion to determine the acceleration of the block, we have:
[tex]\mathbf{y = ut + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = (0)t + \dfrac{1}{2}at^2}[/tex]
[tex]\mathbf{y = \dfrac{1}{2}at^2}[/tex]
Making acceleration (a) the subject, we have:
[tex]\mathbf{a = \dfrac{2y}{t^2}}[/tex]
where;
y = 1.5 mt = 2.0 s[tex]\mathbf{a = \dfrac{2\times 1.5 }{2.0^2}}[/tex]
a = 0.75 m/s²
The angular acceleration of the wheel can be estimated by the formula:
[tex]\mathbf{\alpha = \dfrac{a}{r}}[/tex]
[tex]\mathbf{\alpha = \dfrac{0.75 \ m/s^2}{0.40 \ m}}[/tex]
[tex]\mathbf{\alpha = 1.875 \ rad/s^2}[/tex]
Finally, the torque acting on the wheel is:
[tex]\mathbf{\tau = I \alpha}[/tex]
[tex]\mathbf{Tr = I \alpha}[/tex]
where;
T = tensionr = radiusI = moment of inertia∝ = angular acceleration∴
[tex]\mathbf{I =\dfrac{T\times r}{\alpha} }[/tex]
[tex]\mathbf{I =\dfrac{20 \ N\times 0.40 \ m}{1.875 \ rad/s^2} }[/tex]
I = 4.27 kg.m²
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The current in a resistor is 2.0 A, and its power is 78 W. What is the voltage?
Answer:
39 volts
Explanation:
Use the equation [tex]P=VI[/tex]
[tex]78=V(2)[/tex]
[tex]V=39[/tex]
This is for Lipor only.
Answer:
im here\
Explanation:
A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during the first 2 minutes of flight, the main engines operate for a total of 8.5 minutes after the launch. Once the SRBs are released, the main engines alone accelerate the rocket from about 1341 m/s to 7600 m/s.
What is the acceleration of the SRB and main engine during the first 2.0 minutes of flight?
A. 52 m/s2
B. 13 m/s2
C. 9.8 m/s2
D. 11 m/s2
The acceleration of the SRB and main engine during the first 2.0 minutes of flight is 52.16 m/s².
The given parameters;
initial velocity of the engine, u = 1341 m/sfinal velocity of the engine, v = 7600 m/stime of motion, t = 2 minutes = 2 x 60 s = 120 sThe acceleration of the SRB and main engine is calculated as follows;
[tex]a = \frac{\Delta v}{\Delta t } \\\\a = \frac{7600 - 1341}{2 \times 60 s} \\\\a = 52.16 \ m/s^2[/tex]
Thus, the acceleration of the SRB and main engine during the first 2.0 minutes of flight is 52.16 m/s².
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Four small 0.600-kg spheres, each of which you can regard as a point mass, are arranged in a square 0.400 m on a side and connected by light rods. Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane at point O.
Answer:
.192 kg x m^2
Explanation:
I= mass of a times radius of a squared + mass of b times radius of b squared +...
I= .6 kg x .4m^2 + .6 kg x .4m^2
= .192 kg x m^2
Hope this helps :)
The center of mass of a 1600 kg car is midway between the wheels and 0.7 m above the ground. The wheels are 2.6 m apart. (a) What is the minimum acceleration A of the car so that the front wheels just begin to lift off the ground
Answer:
Explanation:
I guess we are ASSUMING that this is a rear wheel drive car as a front wheel drive car will never get the front wheel normal force to zero
If we consider it as a statics problem and choose our moment center carefully...say 0.7 m above the rear wheel to ground contact point.
Call the traction force at the rear wheels F
The normal force on the front wheels will be zero, so no moment generated by the front wheels.
Summing moments about our chosen point to zero
1600(9.8)[2.6 / 2] - F[0.7] = 0
F = 291,200
this force will create an acceleration of
a = F/m
a = 291200/1600
a = 182 m/s²
which is about 18.6 times gravity acceleration