A solid keeps its shape due to which of the following factors?
Space between particles
Attractive forces between particles
The type of element in the solid
The container it is placed in
Answer:
c
Explanation:
because whatever elements are used like gas uses water evaporation so yeah
Answer:
(A) Space between particles!
Explanation:
Solids can hold their shape because their molecules are tightly packed together!
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A steam catapult applies a very large force to supplement the engine thrust of a jet fighter as it takes off from an
aircraft carrier. The jet has a mass of 4000 kg and must reach its take off speed of 70 m/s at the end of the 220
m flight deck. What is the combined force of the catapult and engines? (Use the work-energy theorem)
Answer:
The combined force of the catapult and engines is 44545.455 newtons.
Explanation:
Let suppose that platform of the aircraft carrier is horizontal, such that changes in gravitational potential energy can be neglected. In addition, effects of non-conservative forces are neglected. By using Principle of Energy Conservation and Work-Energy Theorem we find that steam catapult-jet aircraft is represented by:
[tex]K_{1}+W_{1\rightarrow 2} = K_{2}[/tex] (1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energies of the jet fighter, measured in joules.
[tex]W_{1\rightarrow 2}[/tex] - Work done on the jet fighter due to the combined force of the catapult and engines, measured in joules.
By applying definitions of translational kinetic energy and work, we expand and simplify the equation above as follows:
[tex]F\cdot \Delta s = \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})[/tex]
[tex]F = \frac{m\cdot (v_{2}^{2}-v_{1}^{2})}{2\cdot \Delta s}[/tex] (2)
Where:
[tex]F[/tex] - Combined force of the catapult and engines, measured in newtons.
[tex]m[/tex] - Mass of the jet fighter, measured in kilograms.
[tex]\Delta s[/tex] - Length of the catapult, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the jet aircraft, measured in meters per second.
If we know that [tex]m = 4000\,kg[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]v_{2} = 70\,\frac{m}{s}[/tex] and [tex]\Delta s = 220\,m[/tex], then the combined force of the catapult and engines is:
[tex]F = \frac{(4000\,kg)\cdot \left[\left(70\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (220\,m)}[/tex]
[tex]F = 44545.455\,N[/tex]
The combined force of the catapult and engines is 44545.455 newtons.
The combined force of the catapult and engines is 44,545.5 N.
The given parameters;
mass of the jet, = 4000 kgspeed of the jet, v = 70 m/slength of the deck, d = 220 mThe combined force of the catapult and engines is determined by applying work energy theorem as shown below;
[tex]W = \Delta E\\\\Fd = \frac{1}{2}m(v^2 -u^2) \\\\220F = \frac{1}{2}(4000)(70^2 - 0^2) \ \\\\220 F = 9,800,000\\\\F = \frac{9,800,000}{220} \\\\F = 44,545.5 \ N[/tex]
Thus, the combined force of the catapult and engines is 44,545.5 N.
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Which describes an image that a concave mirror can make?
A. The image is always virtual.
B. The image can be either virtual or real.
C. The image is never upside down.
D. The image can only be the same size as the object.
Answer:
A
Explanation:
The image formed by the concave mirror is always virtual. An enlarged image is caused when the concave mirror is positioned too near to the object.
What is a concave mirror?When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.
A concave mirror is a name for this sort of mirror. An enlarged and fake image is caused when the concave mirror is positioned too near to the object.
When the distance between the item and the mirror is increased, however, the size of the image decreases, and a true picture is generated.
The picture created by the concave mirror may be manipulated. The image formed by the concave mirror is always virtual.
Hence, option A is correct.
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You tie the loose end of a 0.1 kg yo-yo string to your finger and then release the yo-yo so that it spins down toward the ground (the yo-yo is released from rest and the end of the string tied to your finger remains motionless). After the yo-yo falls a distance of 0.9 m, it has a translational speed of 4 m/s and an angular speed of 180 rad/s. What is the moment of inertia of the yo-yo
Answer:
The answer is "[tex]5.06 \times 10^{-6} \ kg \ m^2[/tex]"
Explanation:
[tex]\to E_1=0..............(i)\\\\\to E_2= \frac{mV^2}{2} +\frac{Iw^2}{2} - mgh.............(ii)\\\\ \Delta E=0\\\\\to mgh= \frac{mV^2}{2} +\frac{Iw^2}{2} \\\\ \to 2 \ mgh= mV^2 +Iw^2\\\\ \to 2 \ mgh- mV^2 =Iw^2\\\\ \to m(2gh- V^2) =Iw^2\\\\ \to I= \frac{m(2gh- V^2)}{w^2}[/tex]
[tex]= 5.06 \times 10^{-6} \ kg \ m^2[/tex]
What is the density of the paint if the mass of a tin containing 5000 cm3 paint is 7 kg. If the mass of the empty tin, including the lid is 0.5 kg.
We are given:
Mass of the Paint bucket (with paint) = 7000 grams
Mass of the paint bucket (without paint) = 500 grams
Volume of Paint in the Bucket = 5000 cm³
Mass of Paint in the Bucket:
To get the mass of the paint in the bucket, we will subtract the mass of the bucket from the mass of the paint bucket (with paint)
Mass of Paint = Mass of Paint bucket (with paint) - Mass of the paint Bucket (without paint)
Mass of Paint = 7000 - 500
Mass of Paint = 6500 grams
Density of the Paint:
We know that density = Mass / Volume
Density of Paint = Mass of Paint / Volume occupied by Paint
Density of Paint = 6500/5000
Density of Paint = 1.3 grams / cm³
Explain the "war of the currents"
Answer:
the war of currents is
Explanation:
The war of currents was a series of events surrounding the introduction of competing electric power transmission systems in the late 1880s and early 1890s.
Answer:
a series of events surrounding the introduction of competing electric power transmission systems in the late 1880s and early 1890s
Explanation:
a. A remote-control car with a constant velocity drives off the top of a wall that is 10.0 m high and lands 4.60 m from the base of the wall. Draw a diagram of the problem. Label the known and unknown quantities. b. What is the car's speed before it drives off the top of the wall? c. What is the car's vertical speed just before impact? d. How far from the base of the wall would the remote-control car have landed if it had been subject to the gravitational field on the Moon (1.7 N/kg) ?
Answer:
a. Please see the attached graph and drawing combined created with Microsoft Excel
b. Approximately 3.2217 m/s
c. Approximately 14.0 m/s
d. Approximately 11.05 meters
Explanation:
a. The given parameters are;
The height of the wall, h = 10.0 m
The distance from the base of the wall the car lands = 4.60 m
The time, t, it takes the car to land is given by the equation for free fall as follows;
h = 1/2·g·t²
Where;
g = The acceleration due to gravity = 9.81 m/s²
From the equation for free fall, we have;
h/(1/2·g) = t²
∴ t = √(h/(1/2·g) ) = √(10/(1/2·9.81) ) ≈ 1.4278
The time it takes the car to land, t ≈ 1.4278 seconds
b. The horizontal speed of the car = Horizontal distance/(Time) = 4.6/1.4278 ≈ 3.2217 m/s
The horizontal speed of the car before it drives off the wall ≈ 3.2217 m/s
c. The car's vertical speed just before impact is given by the following equation;
v = u + gt
Where;
u = The initial vertical speed = 0
t = The time it takes before impact ≈ 1.4278 seconds
∴ v = 9.81 × 1.4278 ≈ 14.0 m/s
The car's vertical speed just before impact, v ≈ 14.0 m/s
d. Whereby the car is subject to the gravitational field of the moon, we have;
Gravitational force per kilogram = 1.7 N/kg
∴ Gravitational acceleration = 1.7 m/s².
The time it takes the car to land whilst subject to the gravitational field of the moon is therefore;
[tex]t_{Moon}[/tex] = √(h/(1/2·g)) = √(10/(1/2 × 1.7)) ≈ 3.43
[tex]t_{Moon}[/tex] ≈ 3.43 seconds
The horizontal distance covered, at the car's horizontal speed in the time of free fall ≈ 3.2217 m/s × 3.43 seconds ≈ 11.05 meters
The horizontal distance covered, at the car's horizontal speed in the time of free fall = The distance the car will land from the base of the wall ≈ 11.05 meters
The distance the car will land from the base of the wall ≈ 11.05 meters.
HELP PLEASE I need to finish this asap
Answer:
I'm not 100% sure, but I think the answer would be the first one because there's a force pushing the object in every direction, so they would cancel eachother out and make the object stay in the same place.
Explanation:
pls vote brainliest
A car starts from rest and reaches a velocity of 30 m/s in 10 s. If the car
has a mass of 1500 kg, what net force is exerted upon the car?
Answer:
f= ma
Explanation:
v-u/ t
30-0/10
3m/s
maas 1500
1500*3
4500
A current of 15 Amps flows through a circuit with a resistance of 0.10 Ohms. What is the voltage?
A. 0.006 Volts
OB. 1.5 Volts
O C. 15 Volts
D. 150 Volts
Voltage = (current) x (resistance) (Ohm's law)
Voltage = (15 A) x (0.10 ohm)
Voltage = 1.5 volts (B)
What type of numbers in a measurement are always significant?
Answer:
Non-zero digits
Explanation:
umm (ꏿ﹏ꏿ;)
Answer:
There's three rules on determining how many important quantities are in a number: Non-zero digits are always predominant. Any zeros between two significant digits are always major. The final zero or dangling zeros in the decimal segment are really only important.
I hope this answered your question, have a nice day!
Based on the measured force between objects that are 10 meters apart, how can you find the force between objects that are any distance apart ?
Answer:
F' = 100 F/r²
Explanation:
The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:
F = Gm₁m₂/r²
where,
F = Force between objects
G = Universal Gravitational Constant
m₁ = mass of first object
m₂ = mass of second object
r = distance between objects = 10 m
Therefore,
F = Gm₁m₂/10²
Gm₁m₂ = 100F --------------------- equation (1)
Now, we consider these objects at any distance r apart. So, the force becomes:
F' = Gm₁m₂/r²
using equation (1), we get:
F' = 100 F/r²
So, if the force (F) between objects 10 m apart is known, we can find it at any distance from the above formula.
The force between objects that are any distance apart is expressed as [tex]P'=\frac{100P}{r^2}[/tex]
According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,
[tex]P=\frac{GMm}{r^2}[/tex]
M and m are the masses
r is the distance between the masses
If the force between objects that are 10 meters apart, hence;
[tex]P=\frac{GMm}{10^2}\\P=\frac{GMm}{100}\\GMm = 100P[/tex]
To find the force between objects that are any distance apart, we will use the same formula above to have;
[tex]P'=\frac{GMm}{r^2}\\[/tex]
Substitute the result above into the expression to have:
[tex]P'=\frac{100P}{r^2}[/tex]
Hence the force between objects that are any distance apart is expressed as [tex]P'=\frac{100P}{r^2}[/tex]
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Calculate the velocity of an object moved around a circle with a radius of 1.65 m and an
acceleration of 3.5 m/s2.
Answer:
2.4m
Explanation:
Given parameters:
Radius of circle = 1.65m
Acceleration = 3.5m/s²
Unknown:
Velocity of the object = ?
Solution:
To solve this problem, we have to apply the formula of centripetal acceleration;
Centripetal acceleration = [tex]\frac{v^{2} }{r}[/tex]
v is the velocity
r is the radius
v² = centripetal acceleration x radius
Insert parameters and solve;
v² = 3.5 x 1.65
v = √5.775 = 2.4m
The velocity of the object as it moves around a circle with the given radius and acceleration is 2.4m/s.
Given the data in the question;
Radius; [tex]r = 1.65m[/tex]Centripetal acceleration; [tex]a_c = 3.5m/s^2[/tex]Velocity; [tex]v = \ ?[/tex]
From the centripetal force equation, we use the expression for centripetal acceleration:
[tex]a_c = \frac{v^2}{r}[/tex]
Where [tex]a_c[/tex] is the centripetal acceleration, v is the velocity and r is radius.
We substitute our given values into the equation
[tex]3.5m/s^2 = \frac{v^2}{1.65m} \\\\v^2 = 3.5m/s^2\ *\ 1.65m\\\\v^2 = 5.775m^2/s^2\\\\v = \sqrt{5.775m^2/s^2} \\\\v = 2.4m/s[/tex]
Therefore, the velocity of the object as it moves around a circle with the given radius and acceleration is 2.4m/s.
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pls answer this
quick as possible
Answer:
The answer is the 4th one
Explanation:
Good luck man
carter found the rock shown in the picture below near a river
Answer:
there's no picture
Explanation:
and what would the question be anyways?
Answer:
what is the question
Explanation:
Which of the following is NOT a correct statement?*
As the volume of a contained gas increases, the pressure in the container will
decrease.
As the temperature decreases, the kinetic energy of the particles decreases.
As pressure increases with a constant temperature, the volume decreases.
O As temperature increases, the kinetic energy of the particles decreases.
AnswerAmontons's law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. ... If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure 1).:
Explanation:
What holds the metal ions together in a lattice?
A. covalent bonds
B. metallic bonds
C. hydrogen bonds
D. ionic bonds
Answer:
B. Metallic bonds.
If the brakes are applied and the speed of the car is reduced to 13 m/s in 14 s, determine the constant deceleration of the car.Express your answer to three significant figures and include the appropriate units. Assume ais negative if it acts in the opposite sense to the positive velocity.
Answer:
Explanation:
v = u + at
v is final velocity , u is initial velocity , a is acceleration and t is time .
here v = 0 , u = 13 m /s , t = 14 s
putting the values
0 = 13 + 14 a
a = 13 / 14 m /s²
= - .9286
= - 0.929 m /s²
1500 POINTS TO WHOEVER ANSWERS THIS FIRST!!!!!!!!!!
1. You drop a bowling ball 15m from the top of the science building. What is its
velocity as it impacts the ground?
2. A cheetah accelerates from 0 to 13m/s in 3 seconds. How far does it travel
during that time?
3. Your foolish friend shoots a bullet straight in the air at a muzzle velocity of
380 m/s. How high does the bullet go (neglecting air resistance)?
4. Your foolish friend shoots a bullet straight in the air at a muzzle velocity of
380 m/s. What is the bullets velocity when it falls back down and hits your
friend?
Answer: Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.
Explanation:
Full moon is located______
a. farthest from the Sun
b. at a 90 degree angle with the Sun and Earth
c. closest to the Sun
Answer:
Im going to take an educated guess and say its B.
Explanation:
A Full moon is located farthest from the Sun. Thus, the correct option for this question is A.
What is a Full moon?A full moon may be defined as one of the types of lunar phases when the Moon significantly appears fully illuminated from Earth's perspective. This event typically occurred when Earth is located between the Sun and the Moon.
When the earth is located between the sum and the moon, it is the time when the moon is located farthest from the sun. And it directs to an event of a full moon. The average time duration of the full moon and the repetition of the same phase normally requires 29.53 days.
Therefore, the Full moon is located farthest from the Sun. Thus, the correct option for this question is A.
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During the slowing down process, there is a time when the angular speed is 0.5 rev/s. The fan arm is 0.8 m long. What is the speed of a point on the end of the fan arm at this time
Answer:
2.51 m/s
Explanation:
Given that,
Angular speed of a fan, [tex]\omega=0.5\ rev/s = 3.14\ rad/s[/tex]
Length of the fan arm, r = 0.8 m
We need to find the speed of a point on the end of the fan arm at this time. Let v is the speed. It is given in terms of angular speed is given by :
[tex]v=r\omega\\\\v=0.8\times 3.14\\\\v=2.51\ m/s[/tex]
So, the required speed is 2.51 m/s.
Light travels fastest through which of the following?
A.
air
B.
glass
C.
vacuum
D.
water
things that affects the demand for electricity
A water molecule is made up of one oxygen and two hydrogen atoms. Why is water considered a pure substance?
Answer:
A mixture is when two or more substances combine physically together.
However, in water, two hydrogen atoms combine with one oxygen atom chemically, forming a new substance that has properties different from hydrogen alone or oxygen alone.
Explanation:
How much tension must a rope withstand if it is used to accelerate a 1200-kg car vertically upwards at 0.8m/s2?
Answer:
T = 12732 [N]
Explanation:
To solve this problem we must use newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
By means of the attached free body diagram, we can better understand the solution to this problem.
As we see in the diagram the tension Force T lifts the car upwards so the movement is upwards. Therefore we have two forces one upward (positive) of the force T and the other negative downward due to the weight of the vehicle.
As the movement is up the acceleration is also up (with a positive sign).
ΣF = m*a
where:
F = forces [N] (units of Newtons)
m = mass = 1200 [kg]
a = acceleration = 0.8 [m/s²]
T - 1200*9.81 = 1200*(0.8)
T = 12732 [N]
Answer:
12,720
Explanation:
What is the momentum of a 8850 kg medium truck that is traveling with a velocity of 55 m/s west on the highway
Answer:
486,750 kg*m/s
Explanation:
Momentum is mass*velocity
M = m*v
M = 8850kg*55m/s
M = 486,750 kg*m/s
The momentum of an 8850 kg medium truck that is traveling with a velocity of 55 m/s west on the highway is 486750 kg m / s.
What is momentum?Momentum is the result of a particle's mass and velocity. Being a vector quantity, momentum possesses both magnitude and direction. According to Isaac Newton's second equation of motion, the force acting on the particle equals the time rate of change of momentum.
According to Newton's second law, if a particle is subjected to a constant force for a specific amount of time, the result of the force and time (referred to as the impulse) is equal to the change in momentum.
Given:
The mass of the truck is, m = 8850 kg,
The velocity of the truck is, v = 55 m/s,
Calculate the momentum of the truck as shown below,
Momentum = m × v
Momentum = 8850 × 55
Momentum = 486750 kg m / s
Thus, the Momentum of the truck is 486750 kg m / s.
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asheed and Sofia are riding a merry-go-round that is spinning steadily (uniform circular motion). Sofia is twice as far from the axis as is Rasheed. Sophia's acceleration is _________ that of Rasheed. *
Answer:
The same as
Explanation:
First of all, they are in a merry go round, which is a circular one. And thus, we would be dealing with circular motions.
The angular velocity of a merry go round, in this instance, is given by
w = Δ/Δt
The formula has nothing whatsoever to do with distance, and as such, both Rasheed and Sofia would have the same angular velocity.
Transferring this further, the angular acceleration is given as
α = Δw/Δt
Remember I said the velocity has nothing to do with distance, well, so does the acceleration as we can see from the formula stated. And therefore, both Rasheed and Sofia would have the same angular acceleration
A 16-V battery powers a series circuit that contains two lightbulbs The
voltage across the first lightbulb is 12-V
Complete question:
A 16-V battery powers a series circuit that contains two lightbulbs. The voltage across the first lightbulb is 12-V.
a. What fractional part of the circuit's total energy is transformed in the first lightbulb? Simply your answer.
b. What fractional part of the circuit's total energy is transformed in the second lightbulb? Simply your answer.
Answer:
(a) Fraction of energy of the first bulb in the total energy of circuit is 3/4
(b) Fraction of energy of the second bulb in the total energy of circuit is 1/4
Explanation:
Given;
total voltage across the series circuit, [tex]V_T[/tex] = 16 V
voltage across the first light bulb, V₁ = 12 V
The voltage across the second light bulb is given by;
V₂ = [tex]V_T[/tex] - V₁
V₂ = 16 V - 12 V
V₂ = 4 V
The total power across the circuit is given by;
P = I[tex]V_T[/tex]
P = 16I
Where;
I is the constant current in the circuit
(a)
power in the first light bulb is given by;
P₁ = 12I
Fraction of this power in the total energy of circuit is given by;
[tex]\frac{P_1}{P} = \frac{12I}{16I} = \frac{3}{4}[/tex]
(b)
power in the second light bulb is given by;
P₂ = 4I
Fraction of this power in the total energy of circuit is given by;
[tex]\frac{P_2}{P} = \frac{4I}{16I} = \frac{1}{4}[/tex]
Does electric and magnetic objects both have north and south poles
Explanation:
hope this answer was helpful
which property of potential energy distinguishes it from kinetic energy
Answer:
Shape and position
Explanation:
Hope this helps! :)