Explanation:
nnnnkgcbchg of bbjbjk oh jjzjjzkedkkdkfjjsjdsjfkjfdkkdowkronqfojofj see j FC
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.
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
HELPPPP
The maximum force of sliding friction between a 10 kg rubber box and the concrete
floor is 64 N. How much force should a worker push on the box with if he wants it to
move at a constant velocity?
1) A little less than 64 N
2)A little more than 64 N
3)Exactly 64 N.
4)Exactly 640 N
The force that will move the box at constant velocity must be a little more than 64 N.
The coefficient of sliding friction is obtained from the formula;
μ= F/R
Where;
F = frictional force
μ = coefficient of sliding friction
R = Normal reaction
It is necessary to note that the force that will move the body must be greater than the frictional force acting between the body and the surface in order to move the body. Hence, the force that will move the box at constant velocity must be a little more than 64 N.
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Which can occur during a physical change? (SELECT ALL THAT APPLY!.)
The amount of energy within a substance can change.
The shape of a piece of matter can change.
The chemical reaction produces a new substance.
The atoms of the substance can permanently change.
Answer:
C, B, and D are ones i think i know that can occur during a physical change
Explanation:
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 minimum mass in Mol. of plutonium 234 needed to reach a critical mass nuclear reaction?
Answer:
vddvdvvdvffdfvd
Explanation:
How can wind, coiled wire, and magnets be used together to generate electricity?
Answer:
b
Explanation:
hope helpful
have a great day
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)
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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Which statement describes friction?
Answer:
include the statements pls so i can choose wich one it is and tell you
Explanation:
A machine whose efficiency is 75% is used to lift a load of 100m.calculate the effort put into the machine if it has a velocity ratio of 4.
Explanation:
M.A = load / Effort
efficiency = M.A/V.R X 100
75 = M.A / 4 X 100
75 = 25 X M.A
M.A = 75/25 = 3
M.A = load / effort
3 = 100/E
E = 100/3 = 33.333
A machine whose efficiency is 75% is used to lift a load of 100m. the effort needed to put into the machine is 33.33 meters if the velocity ratio is 4.
What is the mechanical advantage?Mechanical advantage is defined as a measure of the ratio of output force to input force in a system, It is used to analyze the forces in simple machines like levers and pulleys.
Mechanical advantage = output force(load) /input force (effort)
Simple machines like static pulleys, as well as a movable pulleys, provide a great amount of mechanical advantage.
A machine whose efficiency is 75% is used to lift a load of 100m.calc if it has a velocity ratio of 4.
As we know that the mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.
mechanical advantage = velocity ratio ×Efficiency
=4 × 0.75
= 3
The mechanical advantage is 3 , which means the ratio of load to effort is 3
load / effort = 3
100 m/effort =3
effort = 33.33 m
Thus, the effort put into the machine comes out to be 33.33 m.
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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
Is my answer correct
Answer:
it right congratulations
Answer:
A
Explanation:
Simply option A is correct
It's a bit of a common sense question
At the highest point it's velocity will be automatically 0 since it is in rest and so is the acceleration
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
Please help with this question, it’s for physical science
Answer:
s times t
Explanation:
The equation for distance is D = st
Answer:
D is the correct answer (D=S*T)
This is an algebraic law; if you are to move the denominator or numerator to the other side of the equation, you must multiply it by the other side.
S=D/T is the same as D=S*T.
so distance=speed times time.
so if you are going 50 miles per hour, you multiply that times time, say one hour. you get your D distance which is 50 miles.
what is an electrostatic phenomenon?
Answer:
Electrostatic phenomenon is a from the forces that electric charges exert on each other.
A double concave lens is a ________ lens, and a double convex lens is a ______ lens.
a.
converging, diverging
c.
converging, converging
b.
diverging, converging
d.
diverging, diverging
Answer:
b. diverging, converging
Answer:diverging, converging
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
Which statement correctly describes the relationship between thermal energy and particle movement?(1 point)
As thermal energy increases, there is more particle movement.
As thermal energy increases, there is more particle movement.
As thermal energy increases, there is less particle movement.
As thermal energy increases, there is less particle movement.
As thermal energy increases, particle movement does not change.
As thermal energy increases, particle movement does not change.
As thermal energy increases, it is not possible to predict particle movement.
As thermal energy increases, it is not possible to predict particle movement.
Answer:
As thermal energy increases,there is more particle movement
four particles connected by rigid rods of negligible mass where y1 = 5.70 m. the origin is at the center of the rectangle. The system rotates in the xy plane about the z axis with and angular speed of 5.60 rad/s. a) calculate the moment of inertia of the system about the z axis. b) Calculate the rotational kinetic energy
The definition of moment of inertia and kinetic energy we can find the results for the questions are:
a) The moment of inertia is: I = 32.49 kg m²
b) The kinetic energy is: K = 509.4 J
a) The moment of inertia is a scalar quantity that has the same function as mass, it is given by
I = m r²
They indicate that the length of the bar is y₁ = 5.70 m, if the two rigid bars form the square that supports the masses, the distance from the center to the end where each mass is:
r = y₁ / 2
r = 5.70 / 2 = 2.85 m
The moment of inertia of the system is.
I = 4 m r²
Let's calculate.
I =4 m 2.85²
I = 32.49m kg m²
If we assume that the mass of each point particle is:
m = 1 kg
I = 32.49 kg m²
b) They ask for the rotational kinetic energy, if the angular velocity is.
w = 5.60 rad / s
The kinetic energy for rotational motion is.
K = ½ I w²
Let's calculate.
K = ½ 32.49 5.60²
K = 509.4 J
In conclusion with the definition of moment of inertia and kinetic energy we can find the results for the questions are:
a) The moment of inertia is: I = 32.49 kg m²
b) The kinetic energy is: K = 509.4 J
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in a closed system of a cannon and cannonball, which changes would both results in an increase in the kinetic energy of the cannonball when fired from the cannon?
A) decrease the mass of the cannon or decrease the mass of the cannonball
B) increase the mass of the cannon or decrease the mass of the cannonball
C) decrease the mass of the cannon or increase the mass of the cannonball
D) increase the mass of the cannon or increase the mass of the cannonball
Changes that would results in an increase in the kinetic energy of the cannonball in a closed system, when it is fired from the cannon is B:increase the mass of the cannon or decrease the mass of the cannonball.
A closed system can be considered one that is not been affected by external forces. With the conservation of momentum, there would be collisions as well inexplosions.This explains that when a cannon is fired, there would be gain of forward momentum by the cannon ball, while the cannon experience backward momentum.For the kinetic energy of the cannonball to increase, when fired from the cannon, then the mass is the cannon would be increased or the mass of the cannonball would be decreased.
Therefore, option B is correct.
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A massive launcher sends a projectile vertically upwards from the surface of a planet of mass M and radius R. You may assume that this planet has no atmosphere. Part A If the projectile is launched at the escape speed, how do the magnitudes of its initial KE and gravitational potential energy (GPE) compare
Answer:
It can be shown that the potential energy of an object at the surface of the planet would be -G M / R if the potential at infinity is chosen to be zero.
Kinetic energy of G M / R would be required for the escape speed of such an object. The total energy in all such cases is zero.
This can easily be seen by considering the speed of an object falling from infinity towards the planet - the total energy will remain zero if it was zero when the object started to fall.
The potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.
What is escape speed?Escape speed is the minimum speed required for a free, non-propelled object to escape from the gravitational pull of the main body and reach an infinite distance from it in celestial physics.
It is proven that if the potential at infinity is set to zero, the potential energy of an item on the planet's surface is [tex]\rm \frac{-GM}{R}[/tex].
The escape speed of such an item would necessitate kinetic energy will be [tex]\rm \frac{GM}{R}[/tex]. In all of these circumstances, the total energy is zero.
Consider the speed of an item falling from infinity towards the planet: if the total energy was zero before the thing began to descend, the total energy will stay zero.
Hence the potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.
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2. a) A disc rotates about its axis at speed 25 revolutions per minute and takes 15 s to stop. Calculate the
i) angular acceleration of the disc.
ii) number of rotation of the dise makes before it stops.
The statement shows a case of rotational motion, in which the disc decelerates at constant rate.
i) The angular acceleration of the disc ([tex]\alpha[/tex]), in revolutions per square second, is found by the following kinematic formula:
[tex]\alpha = \frac{\omega_{f}-\omega_{o}}{t}[/tex] (1)
Where:
[tex]\omega_{o}[/tex] - Initial angular speed, in revolutions per second.[tex]\omega_{f}[/tex] - Final angular speed, in revolutions per second. [tex]t[/tex] - Time, in seconds.If we know that [tex]\omega_{o} = \frac{5}{12}\,\frac{rev}{s}[/tex], [tex]\omega_{f} = 0\,\frac{rev}{s}[/tex] y [tex]t = 15\,s[/tex], then the angular acceleration of the disc is:
[tex]\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}[/tex]
[tex]\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}[/tex]
The angular acceleration of the disc is [tex]\frac{1}{36}[/tex] radians per square second.
ii) The number of rotations that the disk makes before it stops ([tex]\Delta \theta[/tex]), in revolutions, is determined by the following formula:
[tex]\Delta \theta = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha}[/tex] (2)
If we know that [tex]\omega_{o} = \frac{5}{12}\,\frac{rev}{s}[/tex], [tex]\omega_{f} = 0\,\frac{rev}{s}[/tex] y [tex]\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}[/tex], then the number of rotations done by the disc is:
[tex]\Delta \theta = 3.125\,rev[/tex]
The disc makes 3.125 revolutions before it stops.
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A bowling ball, basketball, and tennis ball are all raised to the same height above the ground. Give the order of objects from the least potential energy to the most
Answer:
bowling ball, basket ball, tennis ball
Explanation:
Describe what causes the planets to stay in orbit around the Sun
Answer:
Gravitional pull or gravity
Explanation:
Its that simple
Need help on this thank you
Answer:
TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.
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
A mover pushes a 45 kg crate across a level floor with a force of 300 N, but the crate accelerates at a rate of only 4.44 m/s2 because a friction force opposes the crate's motion. What is the magnitude of this force of friction? 300 N Force of Friction O A. 50 N N OB. 25 N C. 15 N 0 D. 100 N
By Newton's second law, the net force on the crate is
∑ F = 300 N - f = (45 kg) (4.44 m/s²)
where f is the magnitude of friction. Solve for f :
300 N - f = (45 kg) (4.44 m/s²)
f = 300 N - (45 kg) (4.44 m/s²)
f = 100.2 N ≈ 100 N (D)
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.
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