When three people with a total mass of 2.00 x 102 kg step into their 1.200 x 103 kg car, the car’s
springs are compressed by 3.0 cm.
1.2.a. What is the spring constant of the car’s springs assuming they act as a single spring?
1.2.b. How far will the car lower if loaded with 3.00 x 102 kg rather than 2.00 x 102 kg
Answer:
a
[tex]k = 457333.3 N/m[/tex]
b
[tex]x_a =0.09\ m[/tex]
Explanation:
From the question we are told that
The total mass of three people is [tex]M = 2.00*10^{2} \ kg[/tex]
The mass of the car is [tex]m_c = 1.200 *10^{3} \ kg[/tex]
The compression of the car spring is [tex]x = 3 \ cm = 0.03 \ m[/tex]
Generally the spring constant is mathematically represented as
[tex]k = \frac{F}{x}[/tex]
Here F is the force exerted by the mass of three people and that of the car , this is mathematically represented as
=> [tex]F = (M +m_c) *g[/tex]
=> [tex]F = ([2.0*10^{2} ]+[ 1.200*10^{3}]) * 9.8[/tex]
=> [tex]F = 13720 \ N[/tex]
So
[tex]k = \frac{13720}{0.03}[/tex]
=> [tex]k = 457333.3 N/m[/tex]
Generally if the mass which the car is loaded with is [tex]m = 3.00*10^{2} \ kg[/tex]
Then the force experienced by the spring is
=> [tex]F_a = (m +m_c) *g[/tex]
=> [tex]F_a = (3.00*10^{3} + 1.200 *10^{3}) * 9.8[/tex]
=> [tex]F_a = 41160 \ N[/tex]
Generally from the above formula the compression is
[tex]x_a = \frac{F_a}{k}[/tex]
=> [tex]x_a = \frac{41160}{457333.3}[/tex]
=> [tex]x_a =0.09\ m[/tex]
A ball of mass 0.500 kg is carefully balanced on a shelf that is 2.70 m above the ground. What is its gravitational potential energy
Answer:
The gravitational potential energy of the ball is 13.23 J.
Explanation:
Given;
mass of the ball, m = 0.5 kg
height of the shelf, h = 2.7 m
The gravitational potential energy is given by;
P.E = mgh
where;
m is mass of the ball
g is acceleration due to gravity = 9.8 m/s²
h is height of the ball
Substitute the givens and solve for gravitational potential energy;
PE = (0.5 x 9.8 x 2.7)
P.E = 13.23 J
Therefore, the gravitational potential energy of the ball is 13.23 J.
Which best describes a characteristic of an adiabatic process?
Answer:
Can you please provide the choices? For now this is all I can give you.Explanation:
In thermodynamics, an adiabatic process is a type of thermodynamic process which occurs without transferring heat or mass between the system and its surroundings.Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work. It also conceptually supports the theory used to explain the first law of thermodynamics and is therefore a key thermodynamic concept.
The system takes in heat but does not release it. If an adiabatic process is reversible, there is no change in entropy; otherwise, there is a rise in entropy, or degree of disorder.
What is adiabatic process?
A thermodynamic process known as an adiabatic process takes place when there is no exchange of mass or heat between the system and its surroundings.
An adiabatic process only transfers energy to the environment as work, in contrast to an isothermal process.
The system's overall heat remains constant since there is no heat exchange with the environment.
It is a fundamental idea in thermodynamics and theoretically supports the theory that underlies the first law of thermodynamics.
Therefore, no heat flow in and out of system, which is a characteristic of an adiabatic process.
Learn more about adiabatic process here:
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Two tectonic plates moving toward one another are at a
ANSWER CHOICES
convergent boundary.
divergent boundary.
subduction boundary.
transform boundary.
Answer:
A. cause i just took the test
Explanation:
Answer:
its A
Explanation:
no explanation is needed, just trust me.
Someone help me pls
Explanation:
The dog was stationary at segment c
Answer:
I don't know if I'm wrong but I'm pretty sure stationary means that the thing is still. I would go with C And maybe D????
To remove a stain using a solvent the stain has to become dissolved in the solvent
True
False
Answer:
True
Explanation:
have a good day:)
Answer: This statement is True
please answer this question
A 150 kg boy and his bike are traveling 12 m/s when he slams on his breaks and stop at his friend’s house. How much impulse is required to produce this change in momentum?
Please someone help me with this I’ll give brainliest
Answer:
J = 1800 kg-m/s
Explanation:
Given that,
Mass of a boy, m = 150 kg
Initial velocity of a boy, u = 12 m/s
Finally, it stops, v = 0
We need to find the impulse is required to produce this change in momentum. We know that impulse is equal to the change in momentum. So,
[tex]J=m(v-u)\\\\=150\times (0-12)\\\\=-1800\ kg-m/s\\\\|J|=1800\ kg-m/s[/tex]
So, the impulse is equal to 1800 kg-m/s
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4. Determine the magnitude of force at point and determine if the ladder will slip.
This question is incomplete, the complete question;
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.
Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb
Answer:
- the magnitude of force at point A is 79.1033 lb
- since FA < FA_max; Ladder WILL NOT slip
Explanation:
Given that;
∑'MA = 0
⇒ NB [Lsin∅] - W[L/2.cos∅] = 0
NB = W / 2tan∅ -------let this be equation 1
∑Fx = 0
⇒ FA - NB = 0
FA = NB
therefore from equation 1
FA = NB = W / 2tan∅
we substitute in our values
FA = NB = 76 / 2tan(60°) = 21.9393 lb
Now ∑Fy = 0
NA - W = 0
NA = W = 76 lb
Net force at A will be
FA' = √( NA² + FA²)
= √( (W)² + (W / 2tan∅)²)
we substitute in our values
FA' = √( (76)² + (21.9393)²)
= √( 5776 + 481.3328)
= √ 6257.3328
FA' = 79.1033 lb
Therefore the magnitude of force at point A is 79.1033 lb
Now maximum possible frictional force at A
FA_max = μ × NA
so, FA_max = 0.4 × 76
FA_max = 30.4 lb
So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.
Therefore since FA < FA_max; Ladder WILL NOT slip
A wheel turns through 5.5 revolutions while being accelerated from rest at 20rpm/s.(a) What is the final angular speed ? (b) How long does it take to turn the 5.5 revolutions?
Answer:
(a) The final angular speed is 12.05 rad/s
(b) The time taken to turn 5.5 revolutions is 5.74 s
Explanation:
Given;
number of revolutions, θ = 5.5 revolutions
acceleration of the wheel, α = 20 rpm/s
number of revolutions in radian is given as;
θ = 5.5 x 2π = 34.562 rad
angular acceleration in rad/s² is given as;
[tex]\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2[/tex]
(a)
The final angular speed is given as;
[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2\alpha \theta\\\\\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s[/tex]
(b) the time taken to turn 5.5 revolutions is given as
[tex]\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s[/tex]
Starting at 1.3 m/s, a runner accelerates at a constant 0.22 m/s2 for 6.0 s. What is the runner’s displacement during this time interval?
Answer:
answer is 11.76 meter
Explanation:
use 2nd equation of motion
S=ut+1/2at^2
A good alternative to sit ups is
Answer:
Planks, barbell rollouts, and swiss ball pikes
Explanation:
Which is one of Edwin Hubble’s findings that supports the big bang theory?
Answer:
Edwin Hubble found that galaxies are constantly moving away from us. According to his observations with the Hubble Space Telescope, galaxies are moving at different speeds. This shows that the universe is expanding. The farther away a galaxy is, the faster it is moving away. Found this on google hope this helps.
Answer:
A) the universe started at a central point
Explanation:
taking the quiz on eg. :))
what happens to the grvatational fore between two objects as they move fartger apart
Answer:
♡ madeline here ♡
heres your answer
This force of gravitational attraction
is depending on the masses of both
objects and contrarily corresponding
to the square of the distance that divided
their middle parts. This means that as you
move away from an object the gravitational
force lowers. i hope i could help you! ☆
have a great day!
- madeline/madi ✧･ﾟ: *✧･ﾟ:･ﾟ✧*:･ﾟ✧･ﾟ
Explanation:
A hunter aims directly at a target (on the same level) 120 m away. If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target
Answer:
The distance the bullet will miss the target is 1.13 m.
Explanation:
Given;
Initial velocity of the bullet = 250 m/s
Distance of the target = 120 m
Time of motion;
t = 120 / 250
t = 0.48 s
During this time the bullet is under the gravitational pull and the distance it will miss the target is given by;
Y = V₀y + ¹/₂gt²
where;
V₀y is the initial vertical velocity = 0
Y = 0+ ¹/₂gt²
Y = ¹/₂(9.8)(0.48)²
Y = 1.13 m
Therefore, the distance the bullet will miss the target is 1.13 m.
A 2 kg toy car moves at a speed of 5 m/s.
What is the kinetic energy of the car?
Answer:
The answer is 25 JExplanation:
The kinetic energy of an object can be found by using the formula
[tex]KE = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass
v is the velocity
From the question we have
[tex]KE = \frac{1}{2} \times 2 \times {5}^{2} \\ = 1 \times {5}^{2} [/tex]
We have the final answer as
25 JHope this helps you
How much energy is required to move a 1000 kg object from the Earth's surface to an altitude twice the Earth's radius?
An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.
Since the object must be moved away to a distance greater than the radius of the Earth, then change in gravitational potential energy must be based on Newton's Law of Gravitation.
By the Work-Energy Theorem, the work ([tex]W[/tex]), in joules, done on the object is equal to the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:
[tex]W = U_{g}[/tex] (1)
[tex]W = -G\cdot m\cdot M\cdot \left(\frac{1}{r_{f}}-\frac{1}{r_{o}} \right)[/tex] (1b)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.[tex]m[/tex] - Mass of the object, in kilograms.[tex]M[/tex] - Mass of the Earth, in kilograms.[tex]r_{o}[/tex] - Initial distance, in meters.[tex]r_{f}[/tex] - Final distance, in meters.If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]m = 1000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]r_{o} = 6.371\times 10^{6}\,m[/tex] and [tex]r_{f} = 19.113\times 10^{6}\,m[/tex], then the energy required to move the object from the Earth's surface is:
[tex]W = -\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1000\,kg)\cdot (5.972\times 10^{24}\,kg)\cdot \left[\frac{1}{19.113\times 10^{6}\,m} - \frac{1}{6.371\times 10^{6}\,m} \right][/tex][tex]W = 4.171\times 10^{10}\,J[/tex]
An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.
We kindly invite to check this question on gravitational potential energy: https://brainly.com/question/19768887
A boy on a bicycle rides in a circle of radius ro at speed vo. If the boy now rides at a radius equal to half the initial radius ro, by what approximate factor must he change his speed in order to have the same radial acceleration
Answer:
The speed must change by [tex]\frac{1}{\sqrt{2}}[/tex] factor in order to have the same radial acceleration.
Explanation:
The radial or centripetal acceleration is given by:
[tex] a_{c} = \frac{v^{2}}{r} [/tex]
Where:
v: is the speed = v₀
r: is the radius = r₀
[tex] a_{c} = \frac{v_{0}^{2}}{r_{0}} [/tex] (1)
If the radius is now equal to half the initial radius the speed must be:
[tex]a_{c} = \frac{v^{2}}{r_{0}/2}[/tex] (2)
By equating equation (1) and (2):
[tex] \frac{v_{0}^{2}}{r_{0}} = \frac{v^{2}}{r_{0}/2} [/tex]
[tex]v^{2} = \frac{v_{0}^{2}}{2}[/tex]
[tex] v = \frac{v_{0}}{\sqrt{2}} [/tex]
Therefore, the speed must change by [tex]\frac{1}{\sqrt{2}}[/tex] factor in order to have the same radial acceleration.
I hope it helps you!
answer this plzzzzzzzzzzzzz
Help!!
A 30-N force is applied to a 4-kg object to move it with a constant
velocity of 2 m/s across a level surface. The coefficient of friction
between the object and the surface is approximately (Use the
approximation: g - 10 m/s/s.)
A 0.20
B O 0.50
C 0.55
D 0.75
Answer:
[tex]\mu=0.75[/tex]
Explanation:
Given that,
Force acting on an object, F = 30 N
Mass of the object, m = 4 kg
It is moving with a constant velocity of 2 m/s across a level surface.
We need to find the coefficient of friction between the object and the surface. Let it is μ. Force in terms of coefficient of friction is given by :
F = μ N, Where N is normal force, N = mg
[tex]\mu=\dfrac{F}{mg}\\\\\mu=\dfrac{30}{4\times 10}\\\\\mu=0.75[/tex]
So, the coefficient of friction between the object and the surface is 0.75.
a force of 100.0N is appliesd to an object of mass 85.0kg. what is the object acceleration
Answer:
1.18 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
f is the force
m is the mass
From the question we have
[tex]a = \frac{100}{85} = \frac{20}{17} \\ = 1.176470...[/tex]
We have the final answer as
1.18 m/s²Hope this helps you
A pulley is in the form of a uniform solid cylinder of radius 7cm and mass 2kg. One end of a very light rope is fixed to wind the pulley and the other end is
free. When we pull the free end of the rope the pulley starts rotating from rest and accelerates uniformly. If the angular acceleration is 100rad/s so the
tangential acceleration of the rope is:
Chọn một:
a.
200 N
b.
49 N
C.
0.49 N
d.
7N
I need help with this science work
An object that is oscillating on a spring is given by the equation x = (10.0 cm) cos[(6.00 s-1)t]. At what value of t after t = 0.00 s is the cart first located at x = 8.00 cm?
Answer:
[tex]t=0.0107\ \text{s}[/tex]
Explanation:
[tex]x=10\cos(6t)[/tex]
Now [tex]x=8\ \text{cm}[/tex]
Substituting the value of [tex]x[/tex] in the equation we get
[tex]8=10\cos6t[/tex]
[tex]\Rightarrow 0.8=\cos6t[/tex]
[tex]\Rightarrow \cos^{-1}0.8=6t[/tex]
[tex]\Rightarrow t=\dfrac{\cos^{-1}0.8}{6}[/tex] the values here are used found in radians
[tex]\Rightarrow t=0.0107\ \text{s}[/tex]
So, at [tex]t=0.0107\ \text{s}[/tex] the value of [tex]x=8\ \text{cm}[/tex].
please help i’ll mark u branliest
Answer:
62
Explanation:
it doesn't need explanation
where does a solid material go when a solution is made
Answer:
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Explanation:
Hey, wanna pair our account on Brainly so we can share the perks? https://brainly.com/invite/a88a52bd8e2ff3030b43b9056dfb032b?utm_source=invite&utm_medium=link&utm_campaign=bfp_parental_feed&utm_term=parent
An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
Required:
Find the minimum uncertainty in the momentum of the object. Find the minimum uncertainty in the object's velocity, assuming that the object is (b) a golf ball (mass=0.045 kg) and (c) an electron.
Answer:
[tex]2.78\times 10^{-35}\ \text{kg m/s}[/tex]
[tex]6.178\times 10^{-34}\ \text{m/s}[/tex]
[tex]0.31\times 10^{-4}\ \text{m/s}[/tex]
Explanation:
[tex]\Delta x[/tex] = Uncertainty in position = 1.9 m
[tex]\Delta p[/tex] = Uncertainty in momentum
h = Planck's constant = [tex]6.626\times 10^{-34}\ \text{Js}[/tex]
m = Mass of object
From Heisenberg's uncertainty principle we know
[tex]\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}[/tex]
The minimum uncertainty in the momentum of the object is [tex]2.78\times 10^{-35}\ \text{kg m/s}[/tex]
Golf ball minimum uncertainty in the momentum of the object
[tex]m=0.045\ \text{kg}[/tex]
Uncertainty in velocity is given by
[tex]\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}[/tex]
The minimum uncertainty in the object's velocity is [tex]6.178\times 10^{-34}\ \text{m/s}[/tex]
Electron
[tex]m=9.11\times 10^{-31}\ \text{kg}[/tex]
[tex]\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}[/tex]
The minimum uncertainty in the object's velocity is [tex]0.31\times 10^{-4}\ \text{m/s}[/tex].
Can someone tell me what a free fall is
Answer:
free fall is any motion of a body where gravity is the only force acting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it
Explanation:
hope this helped
Answer:A freefall is what you experience when you go skydiving and that is because you are falling out of the sky
Explanation:
A circular coil of 185 turns and radius 4.50 cm is placed inside and coaxial to a solenoid with 350 turns/m. If the current in the solenoid is represented by the equation I = a(1 − e−bt) where a = 12.5 A and b = 2.10 s−1, determine the magnitude of the induced emf in the coil when t = 1.50 s.
Answer:
Magnitude of the induced emf is 11.62 V
Explanation:
Given:
No. of turns of the circular coil, N=185
Radius, R=4.50 cm=0.045 m
No. of turns of solenoid per meter(m), n=350
a=12.5 A
b=2.10 [tex]s^{-1}[/tex]
Now,
To determine the emf induced in the coil at t = 1.50 s:
The given equation is:
[tex]I=a(1-e^{-bt})[/tex]
Now,
[tex]B=\mu_{o}nI=\mu_{o}na(1-e^{-bt})[/tex]
[tex]\frac{dB}{dt}=\mu_{o}nabe^{-bt}[/tex]
Now, substituting the respective values:
[tex]\frac{dB}{dt}=-4\pi\times 10^{-7}\times 350\times 12.5\times 2.10e^{-2.10\times 1.50}[/tex]
[tex]\frac{dB}{dt}=0.988\ T/s[/tex]
Now,
[tex]emf=NA\frac{dB}{dt}[/tex]
where,
A=Area=[tex]\pi R^{2}[/tex]
Thus,
[tex]emf=185\times \pi\times (4.50\times 10^{-2})^{2}\times 11.545e^{-3.15}[/tex]
[tex]emf=11.62 V[/tex]
As shown in the diagram below, a rope attached to a 500.-kilogram crate is used to exert a force of 45 newtons at an angle of 65 degrees above the horizontal 45 N at an angle of 65 degrees above the horizontal.
500 kg
The horizontal component of the force acting on the crate is?
Answer:
19.01 N
Explanation:
F = Force being applied to the crate = 45 N
[tex]\theta[/tex] = Angle at which the force is being applied = [tex]65^{\circ}[/tex]
Horizontal component of force is given by
[tex]F_x=F\cos\theta\\\Rightarrow F_x=45\times \cos65^{\circ}\\\Rightarrow F_x=19.01\ \text{N}[/tex]
The horizontal component of the force acting on the crate is 19.01 N.