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 string is wrapped around a solid cylinder with mass M and radius R. The free end of the string is held in place, allowing the cylinder to fall. Recall that the moment of inertia of a solid cylinder rotated about its center is given by MR2/2. All answers to this problem should be symbolic, purely in terms of the variables M, R, and g. (a) Find the linear acceleration (in m/s2) of the cylinder and the tension in the string (in Newtons) as the cylinder falls. (b) Now suppose the cylinder is hollow instead of solid. The moment of inertia of a hollow cylinder rotated about its center is given by MR2. What is the acceleration and tension in this case?
Answer:
I will use (a / R) for alpha the angular acceleration
T R = I a / R torque equals angular acceleration for cylinder
M g - T = M a linear acceleration of center of mass
T = M (g - a) = I a / R^2 from first equation
If I = 1/2 M R^2 then M ( g - a) = M a / 2 from above
or g = 3 a / 2 and a = 2 g / 3
Also we have T = M (g - a) = M (g - 2 g / 3) = g / 3
Substitute I = M R^2 for the hollow cylinder
Looks like a = g/2 for hollow cylinder
.5A and 4.5V what’s the resistance
Explanation:
[tex]V = IR \Rightarrow R = \dfrac{V}{I}[/tex]
Plugging in the numbers, we get
[tex]R=\dfrac{4.5\:\text{V}}{0.5\:\text{A}} = 9.0[/tex] ohms
PLEASE I NEED HELP WITH 6 and 8 !!!! I will love u sm
Answer:
Explanation:
006
They are acting in opposite directions. Therefore the net force is found by subtraction. The sign is the same as the larger number.
Net Force = 99.6 - 52.8 = 46.8 N acting in the same direction as the 99.6 which is upward.
008
If the two forces act in the same direction, the net force is found by addition.
Net Force = 99.6 + 52.8 = 152.4 N downward.
i just need help plzzzz
Answer:
I think D it's the correct answer!! Let me know if i'm wrong
Explanation:
An astronaut standing on a platform on a foreign planet drops a hammer. If the hammer falls 9.0 meters vertically in 2.5 seconds, what is the acceleration due to gravity on that planet?
Answer: 1.646 m/s²
Explanation:
The distance that is traveled by the astronaut given that the motion is free-fall can be calculated through the equation,
d = Vot + 0.5at²
where d is the distance, Vo is the initial velocity, t is the time, and a is the acceleration. Substituting the known,
6 = (0 m/s)(2.7 s) + 0.5(a)(2.7 s)²
Determining the value of a,
a = 1.646 m/s²
When people do a bungee jump theyare asked how much they weigh. Why?
Answer:
The elasticity of the bungee cord reduces the gravitational forces applied on the body during bungee jumping. For example, if a 100-pound individual jumps from a building and encounters 900 pounds of deceleration force, they will feel 9 "G's" of force.
hopefully this'll help
have a nice day!!! :D
what causes a solid air fresher to lose mass and volume
Answer:
A solid air freshener loses mass and volume as a result of sublimation, where solid particles skip the liquid state and change directly from a solid to a gas. Sublimation requires that the particles in the solid state gain enough energy to immediately become a gas.
Explanation:
Hope this helps.
The seat on a carnival ride is fixed on the end of an 12.60-m-long beam, pivoted at the other end. If the beam sweeps through an angle of 141°, what is the distance through which the rider moves?
The distance through which the rider at the end of the beam moves is;
L = 15.5 m
We are told that the beam on which the carnival ride is fixed is 12.6m in length.
Since the seat is at the end of the beam with the other end pivoted and the beam sweeps through an angle of 141°, then we can say that the radius of this arc formed by the swing is;
Radius; r = 12.6 m
Also, θ = 141°
The distance through which the driver moves will be the length of the arc formed by the beam at angle of 141°.
Formula for length of arc is given as;
L = 2πr(θ/360)
Plugging in the relevant values gives;
L = 2π × 12.6 × 141/360
L = 15.5 m
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The first motor abilities a new born exhibits are
Answer:
here your answer
i am sorry if wrong
What shape is the graph produced by a force vs acceleration graph?
A. Linear
B. Quadratic
C. Circular
D. Triangular
The answer to the question What shape is the graph produced by a force vs acceleration graph is A. Linear
Since Force, F = ma where m = mass and a = acceleration. For constant mass, F ∝ a. That is, F is directly proportional to acceleration, a.
Since this is a linear relationship, the graph of force vs acceleration will be linear.
The answer to the question What shape is the graph produced by a force vs acceleration graph is A. Linear
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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.
(c) Changing water into vapour is condensation true or false
Answer:
true
Explanation:
Which of the following happens when a substance melts?
Answer:
hola como estas hablas español
Explanation:
what is 4 differences between saturated unsaturated and supersaturated solutions
Answer:
Unsaturated Solution: Less amount of salt in water, clear solution, no precipitation. Saturated Solution: The maximum amount of salt is dissolved in water, Colour of the solution slightly changes, but no precipitation. Supersaturated Solution: More salt is dissolved in water, Cloudy solution, precipitation is visible.
in order to keep heat in or out, you need a(n)
Answer:
What is instillation
Explanation:
Please HELP!!
Diagram's BELOW↓
1. What is the mass of the object experimented on in this situation?
a. 10 kg
b. 15 kg
c. 20 kg
d. 25 kg
2. What is the net force on this object?
a. 20 N to the left
b. 10 N up
c. 10 N down
d. 35 N to the right
Thank you in advance! Very appreciated! (I will mark brainliest) :)
The mass of the object experimented in the situation will be equal to 20 kg in each case. Hence, option C is correct.
What is Acceleration?Acceleration is the rate at which the speed and orientation of a moving object change over time. When a spot or object moves faster or slower along a straight line, it is said to be accelerated. Even though the speed is constant, motion on a circle accelerates so because the direction is always shifting. Both by and to acceleration for all other types of motion.
It is a vector quantity because it has both magnitude and direction.
The given data as per the question is,
Force, F = 30 N
Acceleration, a = 1.5 m/s²
[tex]Force = mass*acceleration[/tex]
30 N = m × 1.5
m = 30/1.5
m = 20 kg.
Therefore, the mass of the object is 20 kg.
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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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The displacement of an object in SHM is described by the equation
[tex] x = cos\binom{2\pi}{3}t[/tex]
where x is in meters and t in seconds. Determine the velocity of the object at t = 0.6 s.
Answer:
[tex]-1.99\:\mathrm{m/s}[/tex]
Explanation:
Assuming that the equation is intended to be [tex]\displaystyle x=\cos\left(\frac{2\pi}{3}t\right)[/tex], we can find the velocity vs. time equation by taking the first derivative with respect to [tex]t[/tex]:
[tex]\displaystyle \frac{dx}{dt}=\frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)[/tex]
Recall the chain rule:
[tex]\displaystyle f(g(x))'=f'(g(x))\cdot g'(x)[/tex]
Therefore,
[tex]\displaystyle \frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex]
Therefore, the velocity vs. time equation of the object is [tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex].
Substitute [tex]t=0.6\text{ s}[/tex] into this equation to find the velocity at that given time:
[tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}(0.6)\right)\cdot \frac{2\pi}{3}\approx \boxed{-1.99\text{ m/s}}[/tex]
A baseball is hit so that it travels straight upward after being struck by the bat. If its initial velocity is 29 m/s , then what is the maximum height that it will reach?
Answer:
Explanation:
Use kinematic equation v² = u² + 2as
Rearrange for distance
s = (v² - u²) / 2a
Realize that at the top of its flight, the ball has zero velocity and gravity is acting downward in an assumed upward positive reference frame.
s = (0² - 29²) / (2(-9.8))
s = 42.90816...
s = 43 m
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A meteor falls from the sky to the Earth. The meteor already had an initial velocity downward when it was spotted. If it hit the Earth at 335 m/s after being seen for 30 seconds, then what was the initial velocity of the meteor?
Answer:
335 - 30 g
Explanation:
V = Vo + g*T = 335
Vo + 9.8*30 = 335
Vo =
A 0.750 kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board. (a) Calculate the duration of the impact in seconds. (Enter a number.) s (b) What was the average force in newtons exerted downward on the nail
Answer:nh
0.750 kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board.
1.Which term describes the sum of the number of protons and neutrons in an atom?
Your answer is → Mass number
define potential difference as used in electricity
Answer:
This ability of charged particles to do work is called an electric potential.
Explanation:
When two negative charges are brought close to each other, they also repel. But when a positive and a negative charge are brought close together, they attract each other. When these two opposite charges are combined, they can be used to work. This is why we need a positive (+) and a negative (–) to light a bulb or run any electrical tool, equipment, mobile phone or home appliance.
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).
Content from magnets
Answer:
magnets are magnetic :)
Explanation:
Explain why two acetate rods, both charged with silk repel
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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Two spherically symmetric planets with no atmosphere have the same average density, but planet B has twice the radius of planet A. A small satellite of mass mA has period TA when it orbits planet A in a circular orbit that is just above the surface of the planet. A small satellite of mass mB has period TB when it orbits planet B in a circular orbit that is just above the surface of the planet.
A period of a satellite is the time taken by the satellite to travel round a
body.
The comparison between the periods [tex]T_B[/tex], and [tex]T_A[/tex] is [tex]\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}[/tex]
Reason:
The period, T, of a satellite is given as follows;
[tex]T = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \cdot M} }[/tex]
Volume of the planet A = [tex]\dfrac{4}{3} \cdot \pi \cdot r^3[/tex]
Mass of planet A, [tex]m_A[/tex] = [tex]\dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho[/tex]
Volume of the planet B = [tex]\dfrac{4}{3} \cdot \pi \cdot (2 \cdot r)^3 = \dfrac{32}{3} \cdot \pi \cdot r^3[/tex]
Mass of planet B, [tex]m_B[/tex] = [tex]\dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho[/tex]
Period of the satellite on planet A, [tex]T_A[/tex], is given as follows;
[tex]T_A = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{4}{3} \cdot \pi \times \rho} }[/tex]
Period of the satellite on planet B, [tex]T_B[/tex], is given as follows;
[tex]T_B = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{32}{3} \cdot \pi \times \rho} }[/tex]
Therefore, get;
[tex]\dfrac{T_A}{T_B} = \dfrac{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 4 \cdot \pi \times \rho} }}{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 32 \cdot \pi \times \rho} }} = \sqrt{\dfrac{32}{4} } = \sqrt{8} = 2 \cdot \sqrt{2}[/tex]
Therefore, [tex]T_A[/tex] = (2·√2)·[tex]T_B[/tex]
[tex]T_B = \dfrac{T_A}{2 \cdot \sqrt{2} } = \dfrac{\sqrt{2} \cdot T_A}{4 }[/tex]
The comparison between [tex]T_A[/tex] and [tex]T_B[/tex] is therefore;
[tex]\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}[/tex]Learn more here:
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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)
1. As the angle of the ramp is increased, the normal force increases /decreases / remains the same and the friction-force increases /decreases / remains the same. [1 Point] 2. As the angle of the ramp is increased, the force parallel increases /decreases / remains the same. [1 Point] 3. The angle at which the force down the plane was equal to the force of friction (for the cabinet) was _____________. [1 Point] 4. Consider a very low (~ zero) friction, 5.0 kg skateboard on a ramp at an angle of 15o to the horizontal. What would be the net force that would cause acceleration when the skateboard is allowed to move
(1) As the angle of the ramp is increased, the normal force decreases.
(2) As the angle of the ramp is increased, the parallel force increases.
(3) The angle at which the force down the plane was equal to the force of friction is zero degree.
(4) The net force that would cause acceleration is 47.33 N.
Let the angle of inclination of the ramp = θ
(1)
The normal force on an object on the ramp inclined to the ramp is calculated as follows;
[tex]F_n = mgcos (\theta)[/tex]
when θ is 0;
[tex]F_n = mgcos (0)\\\\F_n = mg[/tex]
when θ is 90;
[tex]F_n = mgcos(90)\\\\F_n = 0[/tex]
Thus, as the angle of the ramp is increased, the normal force decreases.
(2)
The parallel force on an object on the ramp inclined to the ramp is calculated as follows;
[tex]F_x = mgsin(\theta)\\\\[/tex]
when θ is 0;
[tex]F_x = mgsin(\theta)\\\\F_x = mgsin(0) \\\\F_x = 0[/tex]
when θ is 90;
[tex]F_x = mgsin(90)\\\\F_x = mg[/tex]
Thus, as the angle of the ramp is increased, the parallel force increases.
(3)
The force of friction is calculated as follows;
[tex]F_n = \mu F_n[/tex]
[tex]F_k = \mu mgcos(\theta)[/tex]
[tex]F_k = \mu mg cos(0)\\\\F_k = \mu mg[/tex]
Thus, the angle is zero degree
(4)
The net force that would cause acceleration is calculated as follows;
[tex]F_k = Fn\\\\F_k = mg cos(\theta)\\\\F_k = 5 \times 9.8 \times cos(15)\\\\F_k = 47.33 \ N[/tex]
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Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ tốc độ của vật lúc chạm đất?
Answer:
Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ cao cực đại của vật so với mặt đất?
Explanation: