Answer:
Objects want to keep doing the same thing is a way of stating Newtons First Law.
The correct option is (a) inertia. "Objects want to keep doing the same thing" it implies a connection to the principle of inertia in physics. Inertia is the property of matter that resists changes in its state of motion or rest.
According to Newton's first law of motion, an object will remain at rest or continue moving in a straight line at a constant velocity unless acted upon by an external force. This property is commonly known as "the law of inertia." In other words, objects tend to maintain their current state of motion (whether at rest or in motion) unless influenced by an external force.
When we say "Objects want to keep doing the same thing," we're drawing an analogy between this scientific principle and human behavior. It suggests that objects, like humans, have a natural inclination to resist change and continue their current course of action. This analogy helps convey the idea that objects exhibit a tendency to persist in their existing state.
"Objects want to keep doing the same thing" to inertia emphasizes the notion that objects, like humans, tend to maintain their current state of motion or rest unless influenced by an external force.
Learn more about inertia here:
https://brainly.com/question/30763701
#SPJ2
The correct question is:
Objects want to keep doing the same thing is a way of stating ......
a) inertia
b) force
c) power
d) moment
A 12 kg bowling ball would require what force to accelerate it down an alley at a rate of 2.5 m/s ^ 2
Answer:
hi
Explanation:
hiijjjjjjjjjjjjjjj
Which is true about the temperature of a gas? (2 points)
оа
At a higher temperature, the particles have a higher average kinetic energy.
Oь
At a higher temperature, the volume of the gas is smaller.
Ос
At a lower temperature, there are more particles in the gas.
Od
At a lower temperature, the particles in the gas move faster.
A yo-yo is made of two solid cylindrical disks, each of mass 0.055 kg and diameter 0.070 m , joined by a (concentric) thin solid cylindrical hub of mass 0.0055 kg and diameter 0.011 m . Part A Use conservation of energy to calculate the linear speed of the yo-yo when it reaches the end of its 1.1 m long string, if it is released from rest. Express your answer using three significant figures and include the appropriate units.
Answer: IM 95%sure that the answer is B jus took the test got the answer right
Explanation:
Answer:
sorry I forgot I wish I could help
Time it takes stone to fall from the height of 80 m is approximately equal to *
A. 1 s
B. 2 s
C. 4 s
D. 8 s
Answer:
D
Explanation:
Answer:
c.4s
Explanation:
Find the mass of the two-dimensional object. A disk of radius 7 in with density at distance x from origin given by rho ( x ) = √ 10 x
Answer:
1030.3504 units
Explanation:
Disk radius = 7 inches
Density at distance x = p(x) = [tex]\sqrt{10x}[/tex]
Calculate the mass of the two-dimensional object
M = 2[tex]\pi[/tex][tex]\sqrt{10} * (\frac{2}{5} ) *(7)^{\frac{5}{2} }[/tex] = 1030.3504 units
How many strings of length 10 over the alphabet (a, b, c, d] have at least one b somewhere in the string?
a) 310
b) 410 - 310
c) 10.4
d) 10.39
Complete Question
The complete question is shown on the first uploaded image
Answer:
The correct option is B
Explanation:
The number of alphabet is n= 4 (a , b , c , d )
Generally the total number of string of length 10 over the 4 alphabets is
[tex]N = 4^{10}[/tex]
Gnerally the number of string of length 10 that does not include b is
[tex]T = 3^{10}[/tex]
Generally the number of strings of length 10 over the 4 alphabets that have at least one alphabet b somewhere in the string is
[tex]G = N - T[/tex]
=> [tex]G = 4^{10} - 3^{10}[/tex]
A small child weighs 60 N. If mommy left him sitting on top of the stairs, which are 12 m high, how much energy does the child have!
Please help ASAP
Answer:
6000 joules
Explanation:
I jus learned dis
Answer:6000j
Explanation:
Hope that helps
What observations did the solar system this geocentric models of the solar system help to explain
Answer:
Geocentric model of the solar system helped to explain retrograde motions of planets. Explanation: Geocentric model of planets was proposed by Ptolemy. It stated that all sun, planets and stars revolve round the earth in circular orbits.
Answer:
retrograde motion
Explanation:
i reverse searched the image
If two identical objects are dropped one after 1 s delay with respect to another then in the absence of the air resistance *
A. distance between two falling objects will keep increasing
B. distance between two falling objects will keep decreasing
C. will stay the same
D. All of the above
Answer:
A.No. Assuming no other factors (such as air resistance) the first object will have a velocity of 32 feet/second when the other is dropped. Since they will both have the same acceleration, the first distance between them will increase by 32 feet per second.
Explanation:
A 4.45 g object moving to the right at 18.6 cm/s makes an elastic head-on collision with an 8.9 g object that is initially at rest.
18.6 cm/s
4.45 g
8.9 g
Find the velocity of the first object immediately after the collision. The acceleration of gravity is 9.8 m/s 2 . Answer in units of cm/s.
Answer:
v₁f = -6.2 cm/s
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved, as follows:[tex]m_{1} *v_{1o} = m_{1}* v_{1f} + m_{2}* v_{2f}[/tex]
As the collision is elastic, total kinetic energy must be conserved also:[tex]\frac{1}{2}*m_{1}*v_{1o}^{2} = \frac{1}{2}*m_{1}*v_{1f} ^{2} + \frac{1}2}*m_{2}*v_{2f}^{2}[/tex]
From the givens, we know that m₂ = 2* m₁Replacing in the above equations, rearranging both sides and simplifying, we can find the following expression for v₁f:[tex]v_{1f} = \frac{-m_{1} }{3*m_{1}} *v_{1o} =\frac{-v_{1o}}{3} = -\frac{18.6 cm/s}{3} = -6.2 cm/s[/tex]
v₁f = -6.2 cm/s (which means that it bounces back after the collision).An ideal gas increases in temperature from 22°C to 42°C by two different processes. In one process, the temperature increases at constant volume, and in the other process the temperature increases at constant pressure. Which process requires more heat or are the required amount of heat same in both?
Answer:
a- More heat is required for the constant-pressure process than for the constant-volume
Explanation:
we have to solve using the thermodynamic first law. this is the heat applied to the system
dQ = dU + dW
definition of terms:
dU = change in internal energy
dW = work done
we have it that
change in internal energy dU is directly proportional to work done dW
but when we are in constant volume process, work done of the gas is zero
therefore
dQ of constant pressure is > than that of constant volume
so constant pressure process requires more heat
The process that requires more heat is the constant-pressure process than the constant-volume process.
According to the first law of thermodynamics, the heat that's applied to the system will be the addition of the change in internal energy and the work done.
In a constant-volume process, the work done on the gas is equal to zero. More heat will be required for the constant-pressure process than for the constant-volume process.
Also, it should be noted that the change in the thermal energy of the gas will be the same for the constant-pressure process and the constant-volume process.
Read related link on:
https://brainly.com/question/16951562
. A cathode ray tube (CRT) is a device that produces a focused beam of electrons in a vacuum. The electrons strike a phosphor-coated glass screen at the end of the tube, which produces a bright spot of light. The position of the bright spot of light on the screen can be adjusted by deflecting the electrons with electrical fields, magnetic fields, or both. Although the CRT tube was once commonly found in televisions, computer displays, and oscilloscopes, newer appliances use a liquid crystal display (LCD) or plasma screen. You still may come across a CRT in your study of science. Consider a CRT with an electron beam average current of 25.00μA25.00μA . How many electrons strike the screen every minute?
Answer:
The value is [tex]n= 9.375 *10^{15} \ electrons [/tex]
Explanation:
From the question we are told that
The average current is [tex]I = 25.0 \mu A = 25.0 *10^{-6} \ A[/tex]
Generally the quantity of charge (electron ) is mathematically represented as
[tex]Q = ne[/tex]
Here e is the charge on a single electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
Generally current is mathematically represented as
[tex]I = \frac{Q}{t}[/tex]
=> [tex]I = \frac{ne}{t}[/tex]
Here t is time which is given as 1 minutes = 60 seconds
and n is the number of electrons
So
[tex]25.0 *10^{-6} = \frac{ n* 1.60 *10^{-19}}{60}[/tex]
=> [tex] 60 * 25.0 *10^{-6} = n* 1.60 *10^{-19} [/tex]
=> [tex]n= \frac{60 * 25.0 *10^{-6} }{ 1.60 *10^{-19} }[/tex]
=> [tex]n= 9.375 *10^{15} \ electrons [/tex]
The number of electrons that strike the screen every minute is; n = 9.375 × 10¹⁵ electrons
What is the number of electrons?
We are given;
Average Current; I = 25 μA = 25 × 10⁻⁶ A
Formula for the current is;
I = Q/t = ne/t
where;
n is number of electrons
e is electron charge = 1.6 * 10⁻¹⁹ C
t is time = 1 minute = 60 seconds
Thus making n the subject gives;
n = It/e
n = (25 × 10⁻⁶ * 60)/(1.6 * 10⁻¹⁹)
n = 9.375 × 10¹⁵ electrons
Read more about number of electrons at; https://brainly.com/question/11406294
An 80-kg bungee jumper jumps off a bridge. Rubber bungee cords act as a large spring attaching the jumper to the bridge. A bear standing in the river below catches the jumper. If the spring constant of the bungees is 20 N/m and they stretch 50 m. How much force must the bear apply to keep the jumper from moving?
Answer:
The force is [tex]F_b = 216 \ N [/tex]
Explanation:
From the question we are told that
The mass of the bungee jumper is m = 80 kg
The spring constant is [tex]k = 20 \ N/ m[/tex]
The extension of the rubber bungee cords is x = 50 m
Generally the weight of the jumper is
[tex]W = m * g[/tex]
=> [tex]W = 80 * 9.8 [/tex]
=> tex]W = 784 \ N [/tex]
Generally the returning force of the rubber bungee cords is mathematically represented as
[tex]F = k * x[/tex]
=> [tex]F = 20 * 50 [/tex]
=> [tex]F = 1000 \ N [/tex]
The force to be applied by the bear is
[tex]F_b = F - W[/tex]
=> [tex]F_b = 100 - 784[/tex]
=> [tex]F_b = 216 \ N [/tex]
I need help with this answer
A motorcyclist goes around an un-banked (i.e., flat) circular turn of radius 31m, at a constant speed of 110km/hr (convert this to m/s). What is the minimum coefficient of static friction needed to keep the tires from slipping? Explain why the answer is (or is not) plausible.
Answer:
[tex]\mu_{s} = 3.071[/tex]
This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.
Explanation:
From Second Newton's Law we understand that centripetal acceleration experimented by motocyclist is due to force derived from static friction. And normal force of the ground on motocyclist equals weight of motocyclist due to the flatness of circular turn. The equations of equilibrium of the motocyclist is:
[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (Eq. 1)
[tex]\Sigma F_{y} = N-m\cdot g = 0[/tex] (Eq. 2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
[tex]m[/tex] - Mass of the motocyclist, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v[/tex] - Speed of the motorcyclist, measured in meters per second.
[tex]R[/tex] - Radius of the circular turn, measured in meters.
The static coefficient of friction is cleared in (Eq. 1):
[tex]\mu_{s} = \frac{m\cdot v^{2}}{N\cdot R}[/tex]
From (Eq. 2) we get that normai force is:
[tex]N = m\cdot g[/tex]
And we expand the resulting expression in (Eq. 1):
[tex]\mu_{s} = \frac{m\cdot v^{2}}{m\cdot g\cdot R}[/tex]
[tex]\mu_{s} = \frac{v^{2}}{g\cdot R}[/tex] (Eq. 3)
If we know that [tex]v = 30.556\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 31\,m[/tex], the expected static coefficient of friction is:
[tex]\mu_{s} = \frac{\left(30.556\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (31\,m)}[/tex]
[tex]\mu_{s} = 3.071[/tex]
This result represents an absurd, not plausible, as coefficient of frictions from materials have values between 0 and 1.
Given that water at standard pressure freezes at 0∘C, which corresponds to 32∘F, and that it boils at 100∘C, which corresponds to 212∘F, calculate the temperature difference ΔT in degrees Fahrenheit that corresponds to a temperature difference of 1 K on the Kelvin scale. Give your answer to two significant figures.
Answer:
In two significant figure 360K
Explanation:
The temperature difference (ΔT) can be calculated as the boiling temperature minus the freezing temperature in Fahrenheit.
Hence,
ΔT = 212 - 32
ΔT = 180°F
To convert to °F to kelvin, we use the formula below
= (°F - 32) × 5/9 + 273.15
= (180°F - 32) × 5/9 + 273.15
= 355.37K ⇔ 360K
I WILL MARK YOU AS BRAINLIEST IF RIGHT
A 75 kg skier travels downhill 1200 meters in 56 seconds. What is the velocity of the skier?
If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the CANNON will be ("E4" means "*10^4") *
1 point
4 J
40 J
400 J
None of the above
Answer:
M V = m v conservation of momentum (Caps-cannon Small-projectile)
V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s recoil velocity of cannon
KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2 = 40 kg m^2/s^2 = 40 J
6) The magnitude of the force the Sun exerts on Uranus is 1.41 x 1021 newtons. Explain how it is possible for the Sun to exert agreater force on Uranus than Neptune exerts on Uranus.
Answer and Explanation:
TL: DR The Sun is much more massive than Neptune — more than enough to make up for the somewhat smaller distance between the two planets at the closest approach.
[The surprise in this answer (to me, a non-astronomer), is that the gap between the orbits of Neptune and Uranus is large — half the distance from Uranus to the Sun.]
The ratio of gravitational attraction of the Sun on Uranus versus Neptune on Uranus is directly proportional to the ratio of the Sun’s mass to Neptune’s and inversely proportional to the ratio of the square of the distances (let’s use the closest approach of the two planets to one another to calculate a maximum attraction).
Numbers:
Sun’s mass: 2 x 10^30 kg
Neptune’s mass: 1 x 10^26 kg
Distance of Sun to Uranus: 3 x 10^9 km
Closest approach of Uranus and Neptune: 1.5 x 10^9 km
Without doing any arithmetic, we see that even at their closest approach, Uranus and Neptune are separated by about one-half of the Uranus to Sun distance. Squaring that ratio, we see that if the Sun and Neptune had the same mass, the attraction between the Sun and Uranus would only be about 1/4 of that between the Sun and Neptune; however, the Sun has 20000 times the mass of Neptune, so the attraction between Uranus and the Sun is about 5000 times stronger than the maximum attraction between Uranus and Neptune.
The explanation of the possibility of why sun exerts a greater force on Uranus than Neptune exerts on Uranus is; because the force was calculated to be greater.
The formula for calculating the Force of Gravity between two masses is:
F = G*m₁*m₂/r²
Where;
F = force of gravity
G = gravitational constant = 6.674 × 10⁻¹¹ N•m²/kg²
m₁ = mass of the larger object
m₂ = mass of the smaller object
r = the distance between the centers of the two masses
Now, from online values, we have the following;
mass of Neptune; m₁ = 102.413 × 10²⁴ kg
mass of Uranus; m₂ = 86.813 × 10²⁴ kg
average distance between the centers of Neptune and Uranus; r = 1.62745 × 10¹² m
Thus, force exerted by Neptune on Uranus is;
F = (6.674 × 10⁻¹¹ × 102.413 × 10²⁴ × 86.813 × 10²⁴)/(1.62745 × 10¹²)²
F = 2.240 × 10¹⁷ N
We are told that the force the Sun exerts on Uranus is 1.41 the force the Sun exerts on Uranus is 1.41 × 10²¹ N.
That is greater than the force Neptune exerts on Uranus.
Read more about Force of Gravity at; https://brainly.com/question/7281908
A region around the nucleus of an atom where electrons are likely to be found
Answer:
The region where an electron is most likely to be is called an orbital. Each orbital can have at most two electrons. Some orbitals, called S orbitals, are shaped like spheres, with the nucleus in the center.
Explanation:Hope this helps :)
By definition, a region around the nucleus of an atom where electrons are likely to be found is called an orbital.
First of all, an atom is the smallest constituent unit of ordinary matter that has the properties of a chemical element.
All atoms are made up of subatomic particles: protons and neutrons, which are part of their nucleus, and electrons, which revolve around them. Protons are positively charged, neutrons are neutrally charged, and electrons are negatively charged.
In other words, every atom consists of a nucleus in which neutrons and protons meet and energy levels where electrons are located.
This is, the atomic nucleus is the central part of the atom that is made up of protons and neutrons, while the orbitals or peripheral region is an area where electrons are found.
In summary, a region around the nucleus of an atom where electrons are likely to be found is called an orbital.
Learn more:
https://brainly.com/question/10866484?referrer=searchResultshttps://brainly.com/question/1275002?referrer=searchResultsbrainly.com/question/1814899?referrer=searchResults brainly.com/question/2449569?referrer=searchResultsas a result, the net electric force experienced by each negatively charged particle is reduced to F/2. The value of q is
Answer:
The value of q is [tex]\dfrac{Q}{8}[/tex]
Explanation:
Given that,
Each charge = -Q
Distance between charges = L
Reduced force = [tex]\dfrac{F}{2}[/tex]
Suppose, Two particles each with a charge -Q are fixed a distance L apart as shown above. Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the original two particles.
We know that,
The force on each end is
[tex]F=\dfrac{kQ^2}{L^2}[/tex]...(I)
If the charge q is placed at mid point then
The force on each end charge is
[tex]\dfrac{F}{2}=F+F'[/tex]....(II)
We need to calculate the value of q
Using equation (II)
[tex]\dfrac{F}{2}=F+F'[/tex]
Put the value of F into the formula
[tex]\dfrac{\dfrac{kQ^2}{L^2}}{2}=k\dfrac{Q^2}{L^2}+k\dfrac{q\times(-Q)}{(\dfrac{L}{2})^2}[/tex]
[tex]\dfrac{kq(-Q)}{(\dfrac{L}{2})^2}=-\dfrac{kQ^2}{2L^2}[/tex]
[tex]\dfrac{q}{\dfrac{1}{4}}=\dfrac{Q}{2}[/tex]
[tex]q=\dfrac{Q}{8}[/tex]
Hence, The value of q is [tex]\dfrac{Q}{8}[/tex]
A student studies the effect of an object's speed on its amount of kinetic energy. This graph summarizes the data from the study Kinetic energy Speed Which statement best describes what the graph shows?
A. As speed increases, kinetic energy increases exponentially
B. As speed increases, kinetic energy stays the same
C. As speed decreases, kinetic energy doubles each time.
D. As kinetic energy increases, speed decreases exponentially
The answer is A I Hope this answer helps you (i got the question right)
Answer:
A. As speed increases, kinetic energy increases exponentially.
Explanation:
The amount of kinetic energy an object has depends on the speed. Kinetic energy is also known as "motion energy." This being said, if speed is increasing, decreasing, or staying constant, the kinetic energy of the object will too.
Alejandro made 6.4 liters of punch using half apple juice and half orange juice. How many milliliters of apple juice are in the punch?
Answer:
3.2
Explanation:
I hope that this helps! Have a good day!!
In a railroad switchyard, a rail car of mass 41,700 kg starts from rest and rolls down a 2.65-m-high incline and onto a level stretch of track. It then hits a spring bumper, whose spring compresses 79.4 cm. Find the spring constant.
Answer:
The spring constant is 3.44x10⁶ kg/s².
Explanation:
We cand find the spring constant by conservation of energy:
[tex] E_{i} = E_{f} [/tex]
[tex] mgh = \frac{1}{2}kx^{2} [/tex] (1)
Where:
m is the mass = 41700 kg
g is the gravity = 9.81 m/s²
h is the height = 2.65 m
x is the distance of spring compression = 79.4 cm
k is the spring constant =?
Solving equation (1) for k:
[tex]k = \frac{2mgh}{x^{2}} = \frac{2*41700 kg*9.81 m/s^{2}*2.65 m}{(0.794 m)^{2}} = 3.44 \cdot 10^{6} kg/s^{2}[/tex]
Therefore, the spring constant is 3.44x10⁶ kg/s².
I hope it helps you!
Metals that have shine and luster?
Answer:
luster
Explanation:
A single living thing.
Answer:
What do you mean ma´am/sir?
Explanation:
The word acid comes from the Latin word
Find the work done by a 75.0 kg person in climbing a 2.50 m flight of stairs at a constant speed.
Answer:
1,839.375 Joules
Explanation:
Work is said to be done is the force applied to an object cause the object to move through a distance.
Workdone = Force * Distance
Workdone = mass * acceleration due to gravity * distance
Given
Mass = 75.0kg
acceleration due to gravity = 9.81m/s²
distance = 2.50m
Substitute the given parameters into the formula:
Workdone = 75.0*9.81*2.50
Workdone = 1,839.375Joules
Hence the workdone is 1,839.375 Joules
How much work is required to move it at constant speed 5.0 m along the floor against a friction force of 290 N?
Answer:
The answer is 1450 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
force = 290 N
distance = 5 m
We have
workdone = 290 × 5
We have the final answer as
1450 JHope this helps you
Is it true or false that the displacement always equals the product of the average velocity and the time interval?
Answer:
True.
Explanation:
Applying the definition of average velocity, we know that we can always write the following expression:[tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)
By definition, Δx is just the displacement, and Δt is the time interval.So, just rearranging terms in (1), we get:[tex]\Delta x} = v_{avg}* {\Delta t}[/tex]