Answer:
the moment of inertia is 4.5 × 10⁻⁵ kg.m²
Explanation:
Given that;
point mass m = 0.005 g = ( 0.005 / 1000 ) = 5 × 10⁻⁶ kg
perpendicular distance r = 3m
We know that a point mass doesn't have a moment of inertia around its own axis but, but using the parallel axis theorem, a moment of inertia around a distant axis of rotation can be determined using;
[tex]I_{}[/tex] = mr²
so we substitute
[tex]I_{}[/tex] = (5 × 10⁻⁶ kg) × (3 m)²
[tex]I_{}[/tex] = (5 × 10⁻⁶ kg) × 9 m²
[tex]I_{}[/tex] = 4.5 × 10⁻⁵ kg.m²
Therefore; the moment of inertia is 4.5 × 10⁻⁵ kg.m²
The moment of inertia of given point mass is 4.5 × 10⁻⁵ kgm² at a perpendicular distance of 3 m.
The moment of inertia of given point mass can be determined by,
[tex]I = mr^2[/tex]
Where,
[tex]I[/tex]- moment of inertia
[tex]m[/tex]- mass = 0.005 g = ( 0.005 / 1000 ) = 5 × 10⁻⁶ kg
[tex]r[/tex] - perpendicular distance = 3 m
Put the values in the formula,
[tex]I = (5 \times 10^{-6}{\rm \ kg}) \times (3 {\rm \ m})^2\\\\I = 5 \times 10^{-6}{\rm \ kg} \times 9 {\rm \ m}\\\\I = 4.5 \times 10^{-5} kgm^2[/tex]
Therefore; the moment of inertia of given point mass is 4.5 × 10⁻⁵ kgm².
To know more about moment of inertia,
https://brainly.com/question/6953943
What is a watt a unit of?
light
time
distance
power
Answer:
power...............m....
Answer:
The dimension of power is energy divided by time. In the International System of Units (SI), the unit of power is the watt (W), which is equal to one joule per second.
SI unit: watt (W)
In SI base units: kg⋅m2⋅s−3
Explanation:
A 0.30-m radius car tire rotates how many rad after starting from rest and accelerating at a constant 3.0 rad sa
over a 5.0-s interval?
Answer:
The angular displacement is 37.5 radian.
Explanation:
Given that,
The radius of the car, r = 0.3 m
The acceleration of the car, [tex]\alpha =3\ rad/s^2[/tex]
The initial speed of the car, [tex]\omega_i=0[/tex]
Time, t = 5 s
The angular displacement can be calculated using second equation of motion i.e.
[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\\theta=\dfrac{1}{2}\alpha t^2\\\\\theta=\dfrac{1}{2}\times 3\times (5)^2\\\\\theta=37.5\ rad[/tex]
So, it will make 37.5 radians.
A virtual image larger than the object can be produced by
1) concave mitor 2) convex mutor 3)plane mirror
concave lens
O1
Answer:
concave mirror
Explanation:
Why This is Correct Because, concave mirrors only form a virtual image when a object is larger than the other. when the object is produced between the focus, object and the image object, it becomes a virtual image.
Answer:
The answer is concave mirror
ASAP please thank you !
Find O and P
Explanation:
O + 6V = 9V, so O = 3V
P = 9V as it is parallel to the 9V power supply.
A uniform electric field is produced due to the charge distribution inside the closed cylindrical surface. (a) What type of charge distribution is inside the surface? a positively charged plane parallel to the end faces of the cylinder a positive line charge situated on and parallel to the axis of the cylinder a collection of positive point charges arranged in a line at the center of the cylinder and perpendicular to the axis of the cylinder a collection of negative point charges arranged in a line at the center of the cylinder and perpendicular to the axis of the cylinder a negatively charged plane parallel to the end faces of the cylinder (b) If the radius of the cylinder is 0.66 m and the magnitude of the electric field is 300 N/C, what is the net electric flux through the closed surface? How is the electric flux related to the electric field vector and the normal to the surface? What is the orientation of the electric field relative to the curved surface? N · m2/C (c) What is the net charge inside the cylinder?
Answer with Explanation:
a. Option d is true.
a negatively charged plane parallel to the end faces of the cylinder
b. Radius of cylinder, r=0.66m
Magnitude of electric field, E=300 N/C
We have to find the net flux through the closed surface.
Net electric flux,[tex]\phi=-2 EA=-2E(\pi r^2)[/tex]
[tex]\phi=-2\times 300\times (3.14\times (0.66)^2)[/tex]
[tex]\phi=-820.67 Nm^2/C[/tex]
c.
Net charge,[tex]Q=\epsilon_0\times \phi[/tex]
Where
[tex]\epsilon_0=8.85\times 10^{-12}[/tex]
[tex]Q=-820.67\times 8.85\times 10^{-12}[/tex]
[tex]Q=-7.26\times 10^{-9} C[/tex]
[tex]Q=-7.26nC[/tex]
Where [tex]1nC=10^{-9}C[/tex]
Kinetic energy depends on which two things
Answer:
Explanation: relationship between the object and the observer's frame of reference.
The primary reason for the path of motion of an object being a smooth curve is: Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a the third derivative of parabolas is always zero. b inertia. c tangent direction unit vectors change continuously. d calculus must have continuous derivatives to apply correctly.
Answer:
the correct answer is d
Explanation:
The laws of mechanics are related
F = m a
the acceleration of the body is given by the kinematics
a = [tex]\frac{dv}{dt}[/tex]
v = [tex]\frac{dx}{dt}[/tex]
substituting
a = \frac{d2x}{dt^2}
F = m [tex]\frac{d^2x}{dt^2}[/tex]
Therefore, in order to obtain the force (interaction of a body), continuous curves are needed and derivable from the position and the speed, for which all change in the trajectory of a body must be smooth where smooth is understood to have until the second derived.
Consequently the correct answer is d
How does energy keep the world going? How is energy conserved?
Answer:
The most notable way that reducing energy helps the environment is by decreasing power plant emissions.
Answer:
Explanation:
The conservation of energy law states that energy cant be created or destroyed, only transferred. Since energy cant be created or destroyed, the amount of energy in the universe is completely constant.
Energy keeps the world "going" because we use it every single day in our life. Everything in the universe uses some sort of energy. We digest food for energy, we use electrical energy for electronics, we convert electrical energy into heat energy for heaters and such, we use solar panels to power our houses. There are countless examples of how energy keeps the world "going."
Physical science-current can be increased by...
Option 3.) Increasing the voltage across the wire.
I know that the other answers are incorrect because, for one thing, the more resistance in a substance, the less flow of the current there is. Also, using a longer wire doesn't change anything, it just makes a electrical current go on longer. Lastly decreasing the voltage would make the current decrease in the atoms that flow through it to power an object.
-R3TR0 Z3R0
In the Dorben Company and industrial engineer designed a workstation where the seeing task was difficult because of the size of the components going into the assembly. The desired brightness was 100 fL and the workstation was painted a medium green with a reflectance of 50 percent. What illumination in foot-candles would be required at this workstation to provide the desired brightness? Estimated the required illumination if you repainted the workstation with a light cream paint. what is the luminance of a surface having a 50 % reflectance and a 4 fc illumination?
Answer:
Illumination = 200 fc
luminance = 2FL
Explanation:
given data
desired brightness = 100 fL
painted medium green reflectance = 50 percent
illumination = 4c
solution
we get here Illumination by using the formula that is
Illumination = luminance ÷ reflectance ......................1
Illumination = 100 fL ÷ 0.50
Illumination = 200 fc
With light colored cream paint with a reflectance of 75% the required illumination is 133 fc.
and
luminance will be
luminance = Illumination × reflectance = 4fc × 0.5 = 2FL
A piece of aluminium with mass 1 kg and density 2700 kg/m3 is suspended from a string and then completely immersed in a container of water. Calculate the tension in the string (a) before and (b) after the metal is immersed.
(with steps please)
A piece of aluminium with mass 1 kg and density 2700 kg/m3 is suspended from a string and then completely immersed in a container of water. Calculate the tension in the string (a) before and (b) after the metal is immersed.
ANSWER: 6.296NI am unable to add the workings out, however please do message me and I will be able to provide you with them :)
Secondly, a man is rock climbing when suddenly the rock face he grabs breaks away and he is left with a 1200 lb. rock on his chest. How do scientists say he was able to lift this incredible weight and save himself
Answer:
rush of adrenaline
Explanation:
This is believed to have been possible due to the rush of adrenaline that the individual was experiencing. This has happened a couple times in moments of extreme distress where an individual's body is flooded with adrenaline. This adrenaline allows for super human like strength as well as increased senses. The individual that has lifted the most amount of weight in such a scenario, lifted a total of 3000 lbs. Therefore, it is fair to say that 1200 lb. rock was lifted due to adrenaline.
what do you call these sound waves whose frequency is above 20000 hertz
Answer:
Untrasound
Explanation:
Your welcome :)
What is the weight of a girl with a mass of 50 kilograms in a space station with an artificial gravity of 7 N/kg
Answer:
14.5 N I'm pretty sure I think this is the answer
A young gazelle is grazing in a beautifully sunlit corner of the savanna. Suddenly, the gazelle raises herhead and spots a lioness in the tall grass 173 m away, so she turns away running at roughly constantspeed. The lioness immediately chases the gazelle, with an explosive acceleration of 2.57 m/s2. Aninformed source tells us that this lioness is capable of enduring her maximum speed of 21.0 m/s for 25.0seconds at the longest. [Assume that both predator and prey never change their direction or motion inthis case, for the sake of simplicity.]
(a) On average, how fast (at least) does that gazelle need to run to survive? [Show all your steps.]
(b) Produce a qualitative graph of animal's position (vertical axis) versus time (horizontal axis), shared for the lioness's motion and as well as the gazelle's. Place graph labels in a way that fits the narration.
Answer:
a) v₂ = 13.20 m / s
Explanation:
To solve this exercise we will use the kinematic relations
Let's start with the Lioness. Let's find the time to reach top speed
v = v₀ + a t
as part of rest, its initial velocity is zero
t = v / a
t₁ = 21.0 / 2.57
t₁ = 8.17 s
the total time is the acceleration time plus the time (t₂ = 25 s) that the maximum speed can withstand
t = t₁ + t₂
t = 8.17 +25.0 = 33.17 s
Now let's find out what distance the lioness travels in these times
during acceleration
x = v₀ t + ½ a t²
x = ½ a t²
x₁ = ½ 2.57 8.17²
x₁ = 85.77 m
during constant speed part
x₂ = v t₂
x₂ = 21.0 25.0
x₂ = 525 m
therefore the total distance traveled is
x = x₁ + x₂
x = 85.77 + 525
x = 610.77 m
a) the average speed of the gazelle
this must be the distance that the lioness travels minus the initial distance that separates the two animals (xo = 173 m) between the time taken
v₂ = [tex]\frac{x -x_o}{t}[/tex]
v₂ = [tex]\frac{610.77 - 173}{33.17}[/tex]
v₂ = 13.20 m / s
b) in the attachment we can see a graph of the displacement of the two animals
A particle with a charge of -4.3 μC and a mass of 4.4 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 80 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, then give the answer as a negative number.
Answer:
ΔV = - 3274 V
Explanation:
For this exercise we can use conservation of energy
starting point.
Em₀ = U = q ΔV
final point
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
q ΔV = ½ m v²
ΔV = [tex]\frac{m \ v^2 }{q}[/tex]
let's calculate
ΔV = [tex]\frac{4.4 \ 10^{-6} \ 80^2 }{ 2 \ 4.3 10^{-6} }[/tex]
ΔV = 3274.4 1 V
since the charge q is negative, the potential at point B must be less than the potential at point A, so the answers
ΔV = - 3274 V
A 0.1 kg arrow with an initial velocity of 30 m/s hits a 4.0 kg melon initially at rest on a friction-less surface. The arrow emerges out the other side of the melon with a speed of 20 m/s. What is the speed of the melon? Why would we normally not expect to see the melon move with the is speed after being hit by the arrow?
Answer:
Speed of the melon = 0.25 m/s
we would normally don't see the melon moving due to friction with the resting surface.
Explanation:
We use conservation of momentum:
Pi = Pf
with Pi = 0.1 kg * 30 m/s = 3 kg m/s
and Pf = 0.1 kg * 20 m/s + 4.0 kg * V = 2 kg m/s + 4 * V
Then using the equality above, we solve for V (velocity of the melon)
3 kg m/s = 2 kg m/s + 4 V
1 kg m/s = 4 kg * V
Then V = 1 / 4 M/s = 0.25 m/s
So we would normally don't see the melon moving due to friction with the resting surface.
A cube has a density of 1800 kg/m3 while at rest in the laboratory. What is the cube's density as measured by an experimenter in the laboratory as the cube moves through the laboratory at 89.0 % of the speed of light in a direction perpendicular to one of its faces?
Answer:
3947.7 kg/m³
Explanation:
The relativistic density can be calculated using below expression;
ρ = ρ' /[√( 1 - v²/c²)]
where,
ρ = Change in Density as a result of motion
ρ'= Actual density of cube
ρ' is given as = 1800 kg/m³
v = Velocity that the cube moves = 0.89c
c = Speed of light
c= 3×10^8 m/s
ρ = ρ' /[√( 1 - v²/c²)]
if we substitute the given values we have
ρ= 1800/ √ 〈1- 0.89c²/ c²〉
ρ=3947.7 kg/m³
Hence, the cube's density is 3947.7 kg/m³
A boy throws a rock with an initial horizontal velocity of 17.0 m/s and an initial vertical velocity of 21.0 m/s. How high above the boy's hand is the rock after 2.8 s?
Answer:
53.2
Explanation:
You can use the kinematic equation: displacement of x = (initial velocity + final velocity)*t/2
Subsititing: 17+21 = 38 * 2.8/2 = 53.2
Note: Displacement = distance between the 2 points
Help please ............
Answer:
Explanation:
Electrical current is the movement of charge. The units are Amperes (Amps)
When an object falls, its:
A. PE increases and KE decreases.
B. PE does not change.
C. PE and KE both increase.
D. PE decreases and kE increases
Answer:
Option D. is correct.
Explanation:
The object's mechanical energy refers to the sum of the potential and kinetic energies of the object. When an object falls, its potential energy (PE) decreases, and its kinetic energy (KE) increases. The increase in kinetic energy is exactly equal to the decrease in potential energy.
Option D. is correct.
From the center of the Earth to the moon, what should the orbital radius of such satellite be in order to stay over the same point on the earth’s surface?
In order to have a period that matches the Earth's rotation, a satellite must be in a circular orbit, and 42,164 km from the center of the Earth.
But that's not quite enough to make sure that it always stays over the same point on the Earth's surface (and appears motionless in the sky). For that to happen, the satellite's orbit has to be directly over the Equator.
The Moon has nothing to do with any of this.
6th grade science I mark as brainliest
Answer:
7 would be C, a cell.
Explanation:
Hi.
7 would be C, a cell.
A cell is the basic unit of structure and function in all living things.
If it is living, it is made of cells.
Hope this helps.
Answer:
7. Cell
8. Organelle
in a hydraulic garage the small piston has a radius of 5 cm and the large piston has radius of 15 cm what force must be applied to the small piston in order to lift a car weighting 20000 N on the large piston
The force applied to small piston = 2.2 x 10³ N
Further explanationGiven
a radius of 5 cm and 15 cm
weight 20000 N
Required
Force applied
Solution
Pascal Law :
F₁/A₁=F₂/A₂
A₁ = π.5²
A₂ = π.15²
F₁/ π.5² cm² = 20000/π.15² cm²
F₁ = 2222.22 N⇒2.2 x 10³ N
A sealed container with a lid of area 0.004 m^2 is filled with an ideal gas. The container and gas are allowed to reach thermal equilibrium with the surrounding air. If a 2000 N block is needed to keep the lid from being pushed off the container, what is the absolute pressure inside the container (the pressure compared to vacuum)
Answer:
6 * 10^5 N/m²
Explanation:
Given that :
Area of lid = 0.004m²
Force of block needed to keep the lid from being pushed off the container = 2000 N
Absolute Pressure = atmospheric pressure + force / Area
Force / Area = 2000 N / 0.004 m² = 500,000 = 5 * 10^5
Atmospheric pressure = 1.01325 * 10^5 N/m²
Absolute Pressure = (1.01325 * 10^5) + (5 * 10^5)
Absolute Pressure = 6.01325 * 10^5
= 6 * 10^5 N/m²
A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20 m down the slope, the child enters a flat, slushy region, where she slides for 2.0 s with a constant negative acceleration of −1.5 m/s2 with respect to her direction of motion. She then slides up another icy slope that makes a 20◦ angle with the horizontal.
A) How fast was the child going when she reached the bottom of the first slope? How long did it take her to get there?B) How long was the flat stretch at the bottom?C) How fast was the child going as she started up the second slope?D) How far up the second slope did she slide?
Answer:
A) v₁ = 10.1 m/s t₁= 4.0 s
B) x₂= 17.2 m
C) v₂=7.1 m/s
D) x₂=7.5 m
Explanation:
A)
Assuming no friction, total mechanical energy must keep constant, so the following is always true:[tex]\Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0 (1)[/tex]
Choosing the ground level as our zero reference level, Uf =0.Since the child starts from rest, K₀ = 0.From (1), ΔU becomes:[tex]\Delta U = 0- m*g*h = -m*g*h (2)[/tex] In the same way, ΔK becomes:[tex]\Delta K = \frac{1}{2}*m*v_{1}^{2} (3)[/tex] Replacing (2) and (3) in (1), and simplifying, we get:[tex]\frac{1}{2}*v_{1}^{2} = g*h (4)[/tex]
In order to find v₁, we need first to find h, the height of the slide.From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:[tex]h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)[/tex]
Replacing (5) in (4) and solving for v₁, we get:
[tex]v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s (6)[/tex]
As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.We can do this in more than one way, but a very simple one is using kinematic equations.If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:[tex]v_{1}^{2} - v_{o}^{2} = 2*a* x_{1} (7)[/tex]
Since v₀ = 0 (the child starts from rest) we can solve for a:[tex]a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2 (8)[/tex]
Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:[tex]t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s (9)[/tex]
B)
Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:[tex]x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2} (10)[/tex]
Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):[tex]x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2} = 17.2 m (11)[/tex]
C)
From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:[tex]v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)[/tex]
D)
Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:[tex]\frac{1}{2}*v_{2}^{2} = g*h_{2} (13)[/tex]
Replacing from (12) in (13), we can solve for h₂:[tex]h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m (14)[/tex]
Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:[tex]x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)[/tex]
2. Am 80.0 kg astronaut is training for accelerations that he will experience upon re-entry to earth’s gravity from space. He is placed in a centrifuge (r = 25.0 m) and spun at a constant angular velocity of 10.0 rpm (revolutions per minute).
a. Find the linear velocity of the centrifuge in m/s. Show your work
b. Find the magnitude and direction of the centripetal acceleration when he is spinning at this constant velocity.
c. How many g’s is the astronaut experiencing? (at constant velocity)
d. Find the linear deceleration and torque required to bring the centrifuge (5000.0 kg) to a stop over a 5 minute time period.
Answer:
a) v = 26.2 m / s, b) acceleration is radial, a = 27.4 m / s², c) a = 2.8 g and
d) a = - 8.73 10⁻² m / s², τ = 1.09 10⁴ N m
Explanation:
a) For this exercise we can use the relationships between rotational and linear motion
v = w r
let's reduce the magnitudes to the SI system
w = 10 rpm (2pi rad / 1 rev) 1 min / 60s) = 1,047 rad / s
r = 25.0 m
let's calculate
v = 1.047 25.0
v = 26.2 m / s
b) When the body is rotating at constant speed, the relationship must be perpendicular to the speed, therefore the direction of acceleration is radial, that is, towards the center of the circle and its magnitude is
a = v² / r
a = 26.2²/25
a = 27.4 m / s²
c) Let's look for the relationship between the centripetal acceleration and the acceleration due to gravity
a / g = 27.4 / 9.8
a / g = 2.8
a = 2.8 g
d) let's find the deceleration and torque to stop the centripette in 5 min
t = 5 min (60 s / 1min) = 300 s
let's use the rotational kinematics relations
w = w₀ + α t
initial angular velocity is wo = 1,047 rad / s and the final as is stop do w = 0
α = - w₀ / t
α = - 1,047 / 300
α = -3.49 10⁻³ rad / s²
angular and linear are related
a = α r
a = -3.49 10⁻³ 25
a = - 8.73 10⁻² m / s²
the negative sign indicates that the acceleration is stopping the movement
torque is
τ = F r
The force can be found with Newton's second law
F = m a
we substitute
τ = m a r
τ = 5000.0 8.73 10⁻² 25
τ = 1.09 10⁴ N m
Students perform an experiment in which they drop two eggs with equal mass from a balcony. In the first trial, the egg hits the ground and breaks. In the second trial, the egg hits a foam cushion and does not break or bounce.
Answer: C
Explanation: Because it is just got the same question on the Impulse and Momentum quiz
regular reflection is the reflection of light on a surface
on smooth surface like mirror.
Later in the game, the quarterback throws a pass to the wide receiver with a defender in hot pursuit. If the pass does not arrive to the wide receiver in two seconds, the pass will be intercepted. If the receiver is 30 yards away and the pass is thrown at a 10 degree angle from the ground, how fast must the ball be thrown to reach the receiver
Answer:
Explanation:
In projectile motion , formula for range is as follows
R = u² sin 2 α / g , where u is initial velocity of throw , α is angle of throw
Given R , range = 30 yards , α = 10°
30 = u² sin 20 / 9.8
u² x .342 = 294
u² = 859.65
u = 29.32 m / s