Answer:
The elastic limit of a material is the furthest
point it can be stretched or deformed while
being able to return to its previous shape.
Explanation:
Once a material has gone past its elastic
limit, its deformation is said to be inelastic.
The higher the spring constant, the stiffer
the spring.
You start your bicycle ride at the top of a hill moving East. You coast down the hill at a
constant acceleration of 2 m/s^2. When you get to the bottom of the hill you are moving at
18 m/s, and from there you pedal to maintain that velocity for one minute.
Answer:
198 meters
Explanation:
I'm not sure but hope it helps
how can I become a good science student ?
Answer:
Study hard , focus on your studies and alyways ask questions .
Study, revise, write notes, listen in class, don't let yourself be distracted by others, and do the work in class...maybe join stem or science club if you wanna
Hope this helped you- have a good day bro cya)
Two students were pushing a heavy sofa. One student pushed with a force of 200.0 N to the right, while the other pushed with a force of 150.0 N to the right. The floor exerted a frictional force of 100.0 N. If the sofa's mass is 91.0 kg, what is its acceleration? Round your answer to three significant figures.
Answer:
Explanation:
F = ma
a = F/m
a = (200.0 + 150.0 - 100.0) / 91.0
a = 250.0/91.0
a = 2.7472527...
a = 2.75 m/s²
The pulley shown in the attached diagram has a diameter of 30 centimeters and a mass of 19 kilograms. The pulley is a solid disk with an axle through its center.
(a.) What is the moment of inertia of the pulley?
(b.) What is the magnitude of the net torque on the pulley about its axis
(c.) What is the direction of the net torque on the pulley?
(d.) What is the magnitude of the angular acceleration of the pulley?
(e.) What is direction of the angular acceleration?
Answer:
Explanation:
a) I = ½mR² = ½(19)(0.15²) = 0.21375 kg•m²
b) τ = Fnet(r) = (25 - 12)(0.15) = 1.95 N•m
c) CCW
d) a = τ/I = 1.95 / 0.21375 = 9.12280701... = 9.1 rad/s²
e) CCW
A diffraction grating, ruled with 300 lines per mm, is illuminated with a white light source at normal incidence.
(i) What is the angular separation, in the third-order spectrum, between the 400 nm and 600 nm lines? [5]
(ii) Water (of refractive index 1.33) now fills the whole space between the grating and the screen. What is the angular separation, in the first-order spectrum, between the 400 nm and 600 nm lines? [5]
the expression for diffraction grating allows to find the results for the questions for the angular separation are:
i) The third order is Δθ = 0.203 rad.
ii) The first order with water is Δθ = 0.046 rad.
The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.
d sin θ = m λ
Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.
i) Let's start by looking for the separation between two lines
Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.
d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm
d = 3.333 10⁻⁶ m
Let's find the angle of diffraction for the third order (m = 3) for each wavelength.
λ₁ = 400 nm = 400 10⁻⁹ m
sin θ₁ = [tex]\frac{m \ \lambda }{d}[/tex]m λ/ d
sin θ₁ = [tex]\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]
θ₁ = sin⁻¹ 0.3600
θ₁ = 0.368 rad
λ₂ = 600 nm = 600 10⁻⁹ m
sin θ₂ = [tex]\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]
θ₂ = sin⁻¹ 0.5401
θ₂ = 0.571 rad
The angular separation is
Δθ = θ₂ - θ₁
Δθ = 0.571 - 0.368
Δθ = 0.203 rad
ii) In this case, the separation between the network and the observation screen is filled with water.
When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.
[tex]n_i \ sin \theta_1 = n_r \ sin \theta_r[/tex]
The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.
Let's start looking for the incident angles for the first order of diffraction.
m = 1
λ₁ = 400 nm
θ₁ = sin⁻¹ [tex]\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}[/tex]
θ₁ = 0.120 rad
λ₂ = 600 nm
θ₂ = sin⁻¹¹ [tex]\frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}[/tex]
θ₂ = 0.181 rad
we use the equation of refraction.
[tex]\theta_r[/tex] = sin⁻¹ ([tex]\frac{n_i}{n_r} \ sin \ \theta_i[/tex] )
λ₁ = 400 nm
θ₁ = sin¹ ([tex]\frac{1 sin 0.120}{1.33}[/tex]
θ₁ = 0.090 rad
λ₂ = 600 nm
θ₂ =sin⁻¹ [tex]\frac{1 sin 0.181}{1.33}[/tex]
θ₂ = 0.1358 rad
The angular separation is
Δθ = 0.1358 - 0.090
Δθ = 0.046 rad.
In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:
i) The third order is Δθ = 0.203 rad.
ii) The first order with water is Δθ = 0.046 rad.
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define potential difference as used in electricity
Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit.
Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit. ... The energy is transferred to the electrical components in a circuit when the charge carriers pass through them. We use a voltmeter to measure potential difference (or voltage).
How much time does it take for a school bus traveling at 15 m/s north to get a distance of 600 meters north?
Possible answers:
40 seconds
9,000 seconds
615 seconds
0.025 seconds
Answer:
40s
Explanation:
Speed depends on how far something travels and
Answer:
Speed depends on how far something is travelling and the time taken for the object to travel that distance
Explanation:
(In reference to the speed formula) Speed=[tex]\frac{Distance(m)}{Time(m/s)}[/tex]
Definition of speed:
Speed is the magnitude (unit) of the rate at which an object is moving.
(pls do correct me if I have any mistakes it makes a big difference to help each other out!)
When a straw with liquid is covered by a finger and raised, the liquid does not leak
from the straw. Explain this using ideas of pressure exerted by particles.
Answer:
lol I don't know think u could help me with my mathematics
Find the speed required to throw a ball straight up and it return 6 seconds later. Neglect air resistance
Answer:
the ball will go up 3s and down 3s
v=gt
where t=3s and g=9.8m/s^2
distance=v0(t)+(1/2)gt^2
where initial velocity (v0)=0
Explanation:
The speed required to throw a ball straight up and returns 6 seconds later would be 29.43 meters/seconds.
What are the three equations of motion?There are three equations of motion given by Newton,
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
As given in the problem we have to find the speed required to throw a ball straight up and it returns 6 seconds later,
S = ut + 1/2*a*t²
0 = u×6 + 0.5×(-9.81)×6²
0 = 6u - 176.8
6u = 176.8
u = 176.8/6
u = 29.43 meters / seconds
Thus, the speed required to throw a ball straight up and returns 6 seconds later would be 29.43 meters/seconds.
To learn more about equations of motion here, refer to the link;
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A flea can jump with an initial velocity of 2.2 m/s at an angle of 21° with respect to the
horizontal.
Answer:
Explanation:
If no one can see it because the lights were out. Did the flea really jump?
What do you want here?
Max height (2.2sin21)²/ 2(9.8) = 3.2 cm
Time of flight 2(2.2sin21)/ (9.8) = 0.16 s
distance of flight (2.2cos21)(0.16) = 33 cm
What is the velocity of an object tht experiences 5.6m/s^2 and a radius of 3 meters?
The velocity of the object is 4.1 m/s. Recall that the centripetal acceleration is the acceleration of a body moving in a circular path.
Given that the centripetal acceleration is obtained using the formula;
ac = v^2/r
ac= centripetal acceleration
v = velocity
r = radius
v^2 = rac
v= √ rac
ac = 5.6m/s^2
r = 3 meters
Substituting values;
v = √3 × 5.6m/s^2
v = √16.8
v = 4.1 m/s
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A ball moving 2 ft/s rolls off a table (on earth) that is 32 inches high. How long will it take the ball to hit the floor answer
Answer:When the ball rolls off the edge of the table, it will continue moving forward at 2.0 m/s until it hits the floor.
Explanation:This is what I would say is the answer bc I had to do reasearch on a lot of this for my work this year so if its not im veery sorry
A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary on a surface with a coefficient of kinetic friction of 0.15 when hit, how far does it move after the bullet emerges?
The distance traveled by the wood after the bullet emerges is 0.16 m.
The given parameters;
mass of the bullet, m = 23 g = 0.023 gspeed of the bullet, u = 230 m/smass of the wood, m = 2 kgfinal speed of the bullet, v = 170 m/scoefficient of friction, μ = 0.15The final velocity of the wood after the bullet hits is calculated as follows;
[tex]m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s[/tex]
The acceleration of the wood is calculated as follows;
[tex]\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2[/tex]
The distance traveled by the wood after the bullet emerges is calculated as follows;
[tex]v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m[/tex]
Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.
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Physical activities teaches an individual sense of responsibility and initiative.
True Or False
Answer:
true
Explanation:
Physical fitness is a personal responsibility so I would say true
Which of the following happens when a substance melts?
Answer:
hola como estas hablas español
Explanation:
A 220g mass is on a frictionless horizontal surface at the end of a spring that has a force constant of 7.0N/m The mass is displaced 5 2m from its equilibrium position and then released to undergo simple harmonic motion.
At what displacement from the equilibrium position is the potential energy equal to kinetic energy?
Answer:
Explanation:
Your numbers seem wonky, so I'll just assume that the initial displacement is a distance A (Amplitude) from the equilibrium position. Spring constant = k
Initial potential energy is
PE = ½kA²
As potential energy and kinetic energy are constantly exchanging in SHM,
the position x where half of the original spring potential exists is found where
½kx² = ½(½kA²)
x² = ½A²
x = (√0.5)A
x ≈ 0.707A
just plug in your actual starting position A
With A = 5.2 cm
x = 3.67695... 3.7 cm
A uniform string of length 0.50 m is fixed at both ends. Find the
wavelength of the fundamental mode of vibration. If the wave
speed is 300 mis, find the frequency of the fundamental and next
possible modes.
Answer:
configuration of string:
Node - Antinode - Node or N-A-N
This is 1/2 wavelength since a full wavelength is N-A-N-A-N
f (fundamental) = V / wavelength
F0 = 300 m/s / 1 m = 100 / sec
F1 = 300 m/s / .5 m = 600 / sec
Each increase is a multiple of the fundamental since the wavelength
increases by 1/2 wavelength to keep nodes at both ends of the string
A 940-g rock is whirled in a horizontal circle at the end of a 1.30-m-long string. (a) If the breaking strength of the string is 120 N, what’s the minimum angle the string can make with the horizontal? (b) At this minimum angle, what’s the rock’s speed?
(a) The minimum angle the string can make with the horizontal is 4.4 ⁰.
(b) The rocks speed at the minimum angle is 165.7 m/s.
The given parameters;
mass of the rock, m = 940 g = 0.94 kglength of the string, L = 1.3 mTension on the string, T = 120 NThe net force on the string is calculated as follows;
[tex]Tsin(\theta) = mg\\\\Tcos(\theta) = \frac{mv^2}{r} \\\\\frac{Tsin(\theta)}{Tcos(\theta)} = \frac{mg r}{mv^2} \\\\tan(\theta) = \frac{rg}{v^2} \\\\v^2 = \frac{rg}{tan (\theta)} \\\\v = \sqrt{\frac{rg}{tan (\theta)}}[/tex]
The minimum angle the string can make with the horizontal is calculated as follows;
[tex]Tsin(\theta) = mg\\\\sin(\theta) = \frac{mg}{T} \\\\sin(\theta) = \frac{0.94 \times 9.8}{120} \\\\sin(\theta) = 0.0767\\\\\theta = sin^{-1} (0.0767)\\\\\theta = 4.4 \ ^o[/tex]
The rocks speed at the minimum angle is calculated as follows;
[tex]v = \sqrt{\frac{rg}{tan(\theta)} } \\\\v = \sqrt{\frac{1.3 \times 9.8}{tan(4.4)} } \\\\v = 165.7 \ m/s[/tex]
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WILL GIVE BRAINLIEST
An object with mass m dropped from height H, after travelling a distance and emerged horizontally from the bottom of the track with a velocity v and height of h. what is The work done due to friction while traveling through the track. could someone also please explain the difference between h and H
Answer:
I think d
Explanation:
can anyone heelp me pls
Answer:
Capsule : solid
Suspension : liquid
lotion : semisolid
Explanation:
A dry cell gives static electricity true or false?
Answer:
False
Explanation:
QUESTION 5 When 235U is bombarded with one neutron, fission occurs and the products are three neutrons, 94Kr and?
a. 139Ba
b. 141Ba
c. 139Ce
d. 139Xe
Which statement describes friction?
Answer:
include the statements pls so i can choose wich one it is and tell you
Explanation:
what is an electrostatic phenomenon?
Answer:
Electrostatic phenomenon is a from the forces that electric charges exert on each other.
A spring of spring constant k = 200 N m−1 is slowly extended from an extension of 3.0 cm to an extension of 5.0 cm. Calculate the work done by the extending force. 60
Answer:
31
Explanation:
No need
True or False
In newton's 2nd law, f=ma there is a correlation between the force and acceleration. This supports that there is a causal relationship that acceleration has on the net force.
Answer: See explanation
Explanation:
Please see attached picture. Due to the fact one of the words in the equation F=ma contains a word not allowed.
4. What is the density of a block with a mass of 36 g and a volume of 9 cm?
O A 45 g/cm3
O B.27 g/cm
O 0.4 g/cm
O D. 0.25 g/cm
Answer:
0.4 g/cm
Explanation:
density (g cm ³) = mass (g)
÷
volume (cm³)
Answer:
C. 4 g/cm
Explanation:
Use the formula:
density = mass ÷ volume
mass = 36volume = 9Sub in the values:
density = 36 ÷ 9 = 4 g/cm
Answer = 4 g/cm
What is the mass of an object that is experiencing a net force of 225 N and an acceleration of 3.0 m/s^2?
Answer:
Mass of a object 75 Kilograms
Explanation:
Net force acting on an object, [tex]Fnet = 225N[/tex]
Acceleration produced, [tex]a = 3.0m/s^2[/tex]
According to Newton's second law :
F = m a
[tex]M =\frac{F}{a}[/tex]
[tex]m =\frac{225N}{300m/s^2}[/tex]
[tex]m= 75 Kg[/tex]
So, the mass of an object is 75kg.
Hence, this is the required solution.
Answer:
75 kg.
Explanation:
Net Force
[tex]\sf \longmapsto F_{net} = 225N[/tex]
Acceleration produced
[tex]\sf \longmapsto \: a = 3 .0 m / {s}^{2}[/tex]
According to Newton's 2nd Law –
F = m•a
[tex]\sf \longmapsto \: M = \frac{F}{A} [/tex]
[tex]\sf \longmapsto \: M = \frac{225N}{300m/s ^{2} } [/tex]
[tex]\sf \longmapsto \: M = \: 75 \: kg[/tex]
Therefore, the mass of an object is 75 kg.
The Canadian bobsled team hit the brakes on the sled they are traveling East in so that it decelerates at a rate of 0.43
m/s^2.
a) How long does it take to stop if it travels 85 m before coming to rest?
b) Draw the velocity vector.
c) Draw the acceleration vector.
Answer:
Explanation:
85 = ½(0.43)t²
t = √(2(85)/0.43)
t = 19.883380...
t = 20 s
v→ 8.55 m/s initial, 0 m/s final
a← 0.43 m/s²