Rotation axis. Big disk on the bottom M2 and small disk on the top M1 or mass1. Angular frequency of M₁ is zero. M₂ initial angular frequency is Wi = 6π rad/s. Two objets both can spine freely in the axis someone is holding M1 and M2 is fixed on the axis. So M1 can slide down the axis and M₂ is fixed its rotating. Surface of M₂ is rough like sand paper but bottom of M1 surface is soft and it can rub on M₂ rough surface. then M1 disk is dropped on M2. then both disks are rotating and M1 cant slipe because of M2 rough surface. find wf final anular frequency of the system
Answers
The final angular frequency of the system is 3π/2 rad/s. Moment of inertia increased, frictional torque decreased angular momentum, conservation of energy equation used.
At the point when M1 is dropped onto M2, the two circles will be coupled together and pivot as a solitary framework. The complete precise force of the framework should be saved. At first, just M2 was turning with a precise recurrence of Wi = 6π rad/s, while M1 was very still, so the underlying rakish force was L = I2 * Wi, where I2 is the snapshot of dormancy of M2.
At the point when M1 is dropped onto M2, the snapshot of latency of the framework expands, so to preserve rakish energy, the precise recurrence of the framework should diminish. The last rakish recurrence can be tracked down utilizing the preservation of precise force condition: L = I1 * wf + I2 * wf, where I1 is the snapshot of dormancy of M1 and wf is the last rakish recurrence of the framework.
Since M1 is sliding on the unpleasant surface of M2, there is frictional power that goes against its movement. This frictional power creates a force on the framework, which makes the precise energy decline. The last precise recurrence can be determined utilizing the preservation of energy condition, which compares the underlying active energy of the framework to the last dynamic energy of the framework:
0.5 * I1 * 0^2 + 0.5 * I2 * (6π)^2 = 0.5 * (I1 + I2) * wf^2
Improving on the situation and tackling for wf, we get:
wf = 3π/2 rad/s
Thusly, the last rakish recurrence of the framework is 3π/2 rad/s.
To learn more about angular frequency, refer:
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Related Questions
Which of the following describes the products of a chemical reaction?
A. The original materials
B. The substances that are changed
C. The chemicals on the left side of a chemical equation
Ο Ο
D. The chemicals on the right side of a chemical equation
Answers
Answer:
D The chemicals on the right side of a chamical equation
2. (9 points) A car starts from 10 mph and accelerates along a level road, i.e., no grade change. At 500 ft from its starting point, a radar gun measures its speed as 50 mph. Assuming the car had a constant rate of acceleration, (a) calculate the time elapsed between when the car started at 10 mph to when its speed was measured and (b) what will the speed of the car be another 500 ft downstream of this point
Answers
Answer:
a) t = 11.2 s
b) v = 70.5 mph
Explanation:
a)
Since we need to find the time, we could use the definition of acceleration (rearranging terms) as follows:[tex]t = \frac{v_{f} - v_{o}}{a} (1)[/tex]
where vf = 50 mph, and v₀ = 10 mph.However, we still lack the value of a.Assuming that the acceleration is constant, we can use the following kinematic equation:[tex]v_{f} ^{2} - v_{o} ^{2} = 2*a* \Delta x (2)[/tex]
Since we know that Δx = 500 ft, we could solve (2) for a.In order to simplify things, let's first to convert v₀ and vf from mph to m/s, as follows:[tex]v_{o} = 10 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 4.5 m/s (3)[/tex]
[tex]v_{f} = 50 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 22.5 m/s (4)[/tex]
We can do the same process with Δx, from ft to m, as follows:[tex]\Delta x = 500 ft *\frac{0.3048m}{1ft} = 152.4 m (5)[/tex]
Replacing (3), (4), and (5) in (2) and solving for a, we get:[tex]a = \frac{v_{f} ^{2} - v_{o}^{2}}{2*\Delta x} = \frac{(22.5m/s) ^{2} - (4.5m/s)^{2}}{2*152.4m} = 1.6 m/s2 (6)[/tex]
Replacing (6) in (1) we finally get the value of the time t:[tex]t = \frac{v_{f} - v_{o}}{a} = \frac{(22.5m/s) - (4.5m/s)}{1,6m/s2} = 11.2 s (7)[/tex]
b)
Since the acceleration is constant, as we know the displacement is another 500 ft (152.4m), if we replace in (2) v₀ by the vf we got in a), we can find the new value of vf, as follows:[tex]v_{f} = \sqrt{v_{o} ^{2} +( 2*a* \Delta x)} = \sqrt{(22.5m/s)^{2} + (2*1.6m/s2*152.4m)} \\ v_{f} = 31.5 m/s (8)[/tex]
If we convert vf again to mph, we have:[tex]v_{f} = 31.5m/s*\frac{1mi}{1609m} *\frac{3600s}{1h} = 70.5 mph (9)[/tex]
help please due today
Answers
Answer:
equal and opposite
Explanation:
..........
A boat is moving in a river with a current that has speed vW with respect to the shore. The boat first moves downstream (i.e. in the direction of the current) at a constant speed, vB , with respect to the water. The boat travels a distance D in a time tOut . The boat then changes direction to move upstream (i.e. against the direction of the current) at a constant speed, vB , with respect to the water, and returns to its original starting point (located a distance D from the turn-around point) in a time tIn .
1) What is tOut in terms of vW, vB, and D, as needed?
2) What is tIn in terms of vW, vB, and D, as needed?
3) Assuming D = 120 m, tIn = 170 s, and vW = 0.3 m/s, what is vB, the speed of the boat with respect to the water?
4) Once again, assuming D = 120 m, tIn = 170 s, and vW = 0.3 m/s, what is tOut, the time it takes the boat to move a distance D downstream?
Answers
Answer:
Explanation:
Current has speed vW with respect to the shore and boat has speed vB with respect to water or current so speed of boat with respect to shore
vW + vB .
Distance travelled with respect to shore by boat = D
time ( tout ) = distance / speed with respect to shore
tOut = D / ( vW + vB )
When the boat travels upstream , its velocity with respect to shore
= ( vB - vW ) , vB must be higher .
tin = D / ( vB - vW )
3 ) tin = D / ( vB - vW )
170 = 120 / (vB - 0.3 )
(vB - 0.3 ) = 12 / 17 = .706
vB = 1.006 m / s
4 )
tOut = D / ( vW + vB )
= 120 / ( .3 + 1.006 )
= 92.26 s
Time taken by a body is ratio of the distance traveled by it to the speed.
1)The expression for [tex]t{out}[/tex] is,[tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]
2)The expression for [tex]t{in}[/tex] is,[tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]
3) The speed of the boat with respect to the water is 1.006 m/s. 4) The time it takes the boat to move a distance D downstream is 91.9 seconds.What is upstream and downstream speed?
The net speed of the boat is upstream speed. The difference of the speed of the boat is downstream speed.
Given information-
The speed of the boat with respect to shore is [tex]v_w[/tex].
The speed of the boat in downstream with respect to water is [tex]v_B[/tex].
The distance traveled by the boat is [tex]D[/tex] in time [tex]t_{out}[/tex].
Time taken by a body is ratio of the distance traveled by it to the speed.
1) The net speed of the boat is upstream speed.As the distance traveled by the boat is [tex]D[/tex] in time [tex]t_{out}[/tex]. Thus,[tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]
2) The difference of the speed of the boat is downstream speed.As the distance traveled by the boat is [tex]D[/tex] in time [tex]t_{in}[/tex]. Thus,[tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]
Now the distance is 120 m, the value of [tex]t_{in}[/tex] is 170 s and [tex]v_W[/tex] 0.3 m/s. Thus,
3) The speed of the boat with respect to the water-Put the values in the formula obtains from the 2nd part of the problem,[tex]170=\dfrac{120}{v_B-0.3}\\v_B-0.3=\dfrac{120}{160} \\v_B=0.706+0.3\\v_B=1.006[/tex]
Hence the speed of the boat with respect to the water is 1.006 m/s.
4) The time it takes the boat to move a distance D downstream-Put the values in the formula obtains from the 1st part of the problem,[tex]t_{out}=\dfrac{120}{1.006+0.3}\\t{out}=\dfrac{120}{1.306} \\t{out}=91.9[/tex]
Hence the time it takes the boat to move a distance D downstream is 91.9 seconds.
Thus,
1)The expression for [tex]t{out}[/tex] is,[tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]
2)The expression for [tex]t{in}[/tex] is,[tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]
3) The speed of the boat with respect to the water is 1.006 m/s. 4) The time it takes the boat to move a distance D downstream is 91.9 seconds.Learn more about the upstream and downstream speed here;
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It's time to get a little more specific. Based on the velocity (Vx) graph for the car and the velocity data in the table, divide the total
motion of the car into rough time periods that tell a different "chapter" of the story for this car trip. In each of these time
periods, the car's velocity will be notably different from the previous period. Enter a brief description of the car's motion in each
period. The first one is done for you. Use it as an example to identify and describe the remaining time periods. Note: You can
define as many periods as you think appropriate.
s
B
1
U X
X х.
Font Sizes
А • А
E
E 를 들
E 3
Numbered list
Time Period
Motion Description
0.2 - 4.6 seconds increasing speed in positive direction
Answers
Answer:
0.2 – 4.6 seconds increasing speed in positive direction
4.6 - 7.8 seconds decelerating speed in a positive direction
8 - 17.2 seconds accelerating speed in a negative direction
Explanation:
**Plato** **Edmentum**n~ this question is pretty open ended, so its hard to get it wrong honestly, good luck <3 ~
Answer:
0.2 – 4.6 seconds increasing speed in positive direction
4.6 - 7.8 seconds decelerating speed in a positive direction
8 - 17.2 seconds accelerating speed in a negative direction
Explanation:
Suppose one Sherpa uses a force of 980 N to move a load of equipment to a height of 20 meters in 25 seconds. How much power is used?
Answers
F = 980 N
h = 20 m
t = 25 s
P=? (power)
W=F*h (work)
P=W*t
P=F*h*t
P=980*20*25 =490000 W = 490 kW = 0.49 MW
Displacement vector A points due east and has a magnitude of 1.9 km. Displacement vector B points due north and has a magnitude of 2.08 km. Displacement vector C points due west and has a magnitude of 2.4 km. Displacement vector D points due south and has a magnitude of 2.8 km. Find the magnitude and direction (relative to due east) of the resultant vector A + B + C + D
Answers
Answer:
Explanation:
We shall represent all the four displacement in vector form in terms of unit vector i and j where i represents unit vector towards east , j represents unit vector towards north .
Displacement of A
D₁ = 1.9 i
Displacement of B
D₂ = 2.08 j
Displacement of C
D₃ = - 2.4 i
Displacement of D
D₄ = -2.8 j
Resultant displacement
= 1.9 i + 2.08 j - 2.4 i - 2.8 j
= - 0.5 i - 0.72 j
magnitude of resultant vector
= √ ( .5² + .72² )
=√ ( .25 + .5184 )
= √ .7684
= .876 km
Both i and j are negative of resultant displacement
hence its direction is towards south of west . Angle with west is Ф .
TanФ = .5184 / .25 = 2.0736
Ф = 64.25° .
From east direction is = 180 + 64.25 = 244.25° .
The dielectric constant of the interior of a protein is considerably smaller than that of water. How would this difference in dielectric constants affect the strength of an electrostatic interaction between two opposite charges with the same distance between them if the charged groups were located in the interior of the protein rather than on its surface
Answers
Answer:
the interaction in the protein is greater than the surface with water
\frac{F_i}{F_s} = \frac{\epsilon_s}{ \epsilon_i} \ > 1
Explanation:
The electric force for a charge is
F = [tex]\frac{1}{4\pi \epsilon} \ \frac{q^2}{r^2}[/tex]
In the exercise indicate that the charge is q and the distance r is maintained, the test charge is another
therefore if we use the index i for the dielectric constant ([tex]\epsilon_i[/tex]) in the protein
[tex]F_{i} = \frac{1}{4\pi \epsilon_i} \frac{q^2}{r^2}[/tex]
the electric force in water with dielectric constant ([tex]\epsilon_s[/tex])
[tex]F_s = \frac{1}{4\pi \epsilon_s} \frac{q^2}{r^2}[/tex]
[tex]\epsilon_i < \epsilon_s[/tex]
if we look for the relationship between these forces
[tex]\frac{F_i}{F_s} = \frac{\epsilon_s}{ \epsilon_i} \ > 1[/tex]
therefore the interaction in the protein is greater than the surface with water
Q5. Use Superposition to V. in the circuit below? (5 points)
4 mA
12V
2 ΚΩ
2 mA
1 ΚΩ
2 ΚΩ
Answers
Answer:
4va
12va
2jk
1jk
2jk
Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay aboard the space station to be a full second youngerthan people on the ground? Please show and explain how you would set-up the problem,before you actually try to solve it. If you cannot solve it exactly, please try to offer an estimate.(5 pts)
Answers
Answer:
#_time = 7.5 10⁴ s
Explanation:
In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.
t = [tex]\frac{t_p}{ \sqrt{1- (v/c)^2} }[/tex]
where t_p is the person's own time in an immobile reference frame,
[tex]t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }[/tex]
let's calculate
we assume that the speed of the space station is constant
[tex]t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }[/tex]
[tex]t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}[/tex]
t_ = 0.99998666657 s
therefore the time change is
Δt = t - t_p
Δt = 1 - 0.9998666657
Δt = 1.3333 10⁻⁵ s
this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s
#_time = 1 / Δt
#_time =[tex]\frac{1}{1.3333 \ 10^{-5}}[/tex]
#_time = 7.5 10⁴ s
A motorcycle and rider have a total mass equal to 300 kg. The rider applies the brakes, causing the motorcycle to decelerate at a rate of -5 m/s^2. What is the net force on the motorcycle?
Answers
Answer:
Net force = - 1500 N
Explanation:
We calculate the net force acting using Newton's second Law:
[tex]F_{net}=m*a\\F_{net}=(300 \,kg)*(-5\,m/s^2)\\F_{net}=-1500\,N[/tex]
How much force is needed to accelerate a Kia Soul with a
mass of 1200 kg to 5 m/s2?
Answers
Answer:
[tex]\boxed {\boxed {\sf 6,000 \ Newtons}}[/tex]
Explanation:
Force is the product of mass and acceleration.
[tex]F=ma[/tex]
The mass of the Kia Soul is 1200 kilograms and its acceleration is 5 meters per square second.
[tex]m= 1200 \ kg \\a= 5 \ m/s^2[/tex]
Substitute the values into the formula.
[tex]F= 1200 \ kg * 5 \ m/s^2[/tex]
Multiply.
[tex]F= 6000 \ kg*m/s^2[/tex]
1 kilgram meter per square second is equal to 1 Newton. Our answer of 6000 kg*m/s² equals 6000 N[tex]F= 6000 \ N[/tex]
Answer:
Given :-Mass = 1200 kgAcceleration = 5 m/s²To Find :-Force
Solution :-We know that
F = ma
F = Force
m = mass
a = acceleration
F = 1200 × 5
F = 6000 N
[tex] \\ [/tex]
A man pushes a block of ice across a frozen pond at a constant velocity. While the coefficients of static and kinetic friction for ice are low, they are not zero. Consider this problem to involve friction. If necessary, use Fs for the force of static friction, and Fk as the force of kinetic friction.
Required:
Draw the Free Body Diagram for the block of ice.
Answers
Answer:
F₁> F₂
Explanation:
For this exercise Newton's second law is used, in the adjoint we can see the unapplied forces in this exercise.
Y axis y
N- W = 0
in this axis there is no movement
X axis
F -fr = m a
as they indicate that the velocity is consonant the acceleration is worth zero
F - fr = 0
friction force has the expression
fr = μ N
fr = μ mg
we substitute
F = μ m g
by the time the block is stopped the deferred force is
F₁ = μ_s m g
when it begins to move the force should decrease to
F₂ = μ_k k m g
as the static coefficient is greater than the dynamic coefficient
F₁> F₂
The free body diagram consists of applied force (F) and kinetic frictional force acting in opposite direction.
Net force on the blockThe net force on the block will result constant speed of the block which is zero acceleration.
[tex]\Sigma F= 0\\\\F - F_f = 0\\\\F - \mu_k F_n= 0\\\\F - \mu_k mg = 0\\\\F - \mu k W = 0\\\\F = F_f\ \ \ or \ \ F = \mu_k W[/tex]
Free body diagramThe free body diagram consists of applied force (F) and kinetic frictional force acting in opposite direction.
F → Ф ← Ff
Learn more about free body diagram here: https://brainly.com/question/21691401
Explain how a common housecat gets “worms.”eplain(science)
Answers
Answer:
Cats most commonly contract worms after coming into contact with parasite eggs or infected feces. A cat may walk through an area with eggs or infected feces, and since cats are often such fastidious groomers, they will then ingest the eggs or fecal particles as they clean their fur and feet.
Explanation:
this is the only thing in my book hope it helps
Someone help please
Answers
Answer:
it would be downwards due to gravitational force
Which of the following will be attracted toward a positively charged cloth?
Positively charged sock
Negatively charged pipe
Sound waves
Light energy
Answers
Postive and negatives attract, positive and positive repel. answer is negatively charged pipe.
sound waves and light energy are not "affected" by static electricity
when material allow to heat to pass though them rapidly they are known as __________
Answers
Answer:
They are conductors/conductive. Materials that can transfer thermal energy well are conductive.
Explanation:
Answer:when material allow to heat to pass though them rapidly they are known as Conductors
What are conductors
In physics /electrical engineering They allow the flow of charge (electrical current) in one or more directions. Materials made of metal are common electrical conductors.
What are examples of conductors
.Silver
.copper
.iron
.aluminum
.brass
Difference between conductors and insulators?
Insulators a material that is a poor conductor (as of electricity or heat)Whereas Conductors are fantastic at allowing the flow of the heat.
Hope this helps
-Tobie
Which of the following does NOT have a positive impact on your position on the
health continuum?
avoiding risk behaviors
having a positive social environment
eating nutritious foods
O having a chronic disease
Answers
Answer:
Having a chronic disease
Explanation:
Imagine a third particle, which we will call a cyberon. It has three times the mass of an electron (3_m). It has a positive charge that is three times the magnitude (3_(qe)) of the charge on an electron. What is the ratio of the speed v_c that the cyberon would have when it reaches the upper plate after being released from rest at position h_0 to the speed ve that the electron would have?
Answers
Answer:
The answer is "The last choice".
Explanation:
Please find the complete question in the attachment.
In an external electric field, its electrical energy at positive charge becomes directed to just the electrical domain. Therefore it will speed towards its base plate whenever cyber one is released to rest at h0. It was never going to reach the top plate. Thus, the last choice corrects because in this the cyber-on never reaches its upper stage.
Bartender slides a beer mug at 1.1 m/s towards a customer at the end of the bar which is 1.8 m tall. The customer makes a grab for the mug and misses and mug sails at the end of the bar. a) How far away from the end of the bar does the mug hit the floor
Answers
Answer:
Δx = 0.7 m
Explanation:
Once the mug is moving in the horizontal direction, it keeps moving at the same speed of 1.1 m/s, due to no other force acts on it in this direction.Since the horizontal and vertical movements are independent each other (due to they are mutually perpendicular), in the vertical direction, the initial speed is just zero.In the vertical direction, the mug is accelerated by the force of gravity at all times, with a constant value of 9.8 m/s2, aimed downward.So, we can use the following kinematic equation in order to get the time passed from the instant that the mug left the bar, until it hit the floor, as follows:[tex]\Delta y = \frac{1}{2} * g* t^{2} = (1)[/tex]where Δy = 0-1.8m = -1.8m, g= -9.8m/s2.Replacing these values in (1) and solving for t, we get:[tex]t = \sqrt{\frac{2*1.8m}{ 9.8m/s2} } = 0.6 s (2)[/tex]
Now, since the mug obviously finishes its horizontal trip at this same time (hitting ground), we can find the horizontal distance traveled, just applying the definition of average speed, as follows:[tex]\Delta x = v_{o} * t = 1.1 m/s* 0.6 s = 0.7 m (3)[/tex]
Diagram B D c с Which car has: Ke = 100 PE=0? * 1 point A B C D
Answers
Answer:
The car C has KE = 100, PE = 0
Explanation:
The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.
The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.
Kinetic energy is the energy that a body possesses due to its motion.
Potential energy is the energy a body possesses due to its position.
From the principle of conservation of energy, kinetic energy can be transformed into potential energy and vice versa, but in all cases the energy is conserved or constant.
In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.
Therefore, the car C has KE = 100, PE = 0
Name the state of matter that diffusion happens the fastest in.
Answers
Answer:
Liquids
Explanation:
Diffusion occurs fastest in liquids.
what type of reaction is being shown in this energy diagram?
X exothermic, because energy is absorbed from the surroundings
O exothermic, because energy is released into the surrounding
X endothermic, because energy is released into the surrounding
X endothermic, because energy is absorbed from the surroundings
best of luck nerds
Answers
Answer:
O exothermic, because energy is released into the surrounding
Explanation:
From the diagram the energy of the reactant is higher than the energy of the product, thereby making it exothermic. If you study diagram well, exothermic reaction means that the reactions releases energy into the surroundings.
Consider the air over a city to be a box that measures 100 km per side that reaches up to an altitude of 1.0 km. Wind (clean air) is blowing into the box along one of its sides with a speed of 4 m/s. An air pollutant is emitted into the box at a rate of 10.0 kg/s; the pollutant degrades with a rate constant k = 0.20/hr. a. Find the steady state concentration of the pollutant (µg/m3 ) in the box if the air is assumed to be completely mixed. b. If the wind speed suddenly drops to 1 m/s, estimate the concentration of the pollutant (µg/m3 ) two hours later.
Answers
Answer:
a) ρ = 6.25 10⁵ μg / m³, b) ρ = 1 10⁷ μg / m³
Explanation:
Let's analyze the exercise a little before starting, we must know the amount of pollutant in the box, that the one that enters less the one that degrades and with this value find the density or concentration.
Let's start by finding the volume of air that goes into the box
V = Lh x
Let's find the distance of air that enters per unit of time, as it goes at constant speed
x = v₀ t
we substitute
V₀ = Lh v₀ t
At this same time, a quantity of pollutant is distributed
Q₀ = r t
the contaminant that is entering reaches the entire box, therefore the total amount of contaminant is
Q = Qo t
we substitute
Q = r t²
the net amount of pollutant that remains is that less enters the one that degraded in the same time, as they ask for the steady state
[tex]Q_{net}[/tex]= Q - k t
the pollutant concentration is
ρ = Q_net / V
V = L L h
ρ =[tex]\frac{r \ t^2 - k \ t}{ L^2 h}[/tex]
ρ = [tex](r \frac{ L^2}{v_o^2} - k \frac{L}{v_o} ) \frac{1}{L^2 h}[/tex]
ρ = [tex]\frac{r}{ v_o h} -\frac{k}{v_o L h}[/tex]
let's reduce the magnitudes to the SI system
r = 10 kg / s
L = 100 km = 100 10³ m
h = 1 km = 1 10³ m
k = dq / dt = 0.20 1/h ( 1h/3600 s) = 5.5555 10⁻⁵ 1/s
v₀ = 4 m / s
let's calculate
The volume of the box
V = (100 100 1) 109
V = 1 10¹³ m³
ρ = [tex]\frac{10}{ 4^2 \ 1\ 10^3 } - \frac{5.5556 \ 10^{-5}}{ 4 \ 100 \ 10^3 1 \ 10^3}[/tex]
ρ = [tex]6.25 10^{-4} - 1.389 ^{-13}[/tex]
ρ = 6.25 10⁻⁴ kg / m³
let's reduce to μg / m³
ρ = 6.25 10⁻⁻⁴ kg / m³ (10⁹ μg / 1kg)
ρ = 6.25 10⁵ μg / m³
b) in case the air speed decreases to v₀ = 1 m / s
ρ= \frac{10}{ 1^2 \ 1\ 10^3 } - \frac{5.5556 \ 10^{-5}}{ 1 \ 100 \ 10^3 1 \ 10^3}
ρ = 1 10⁻² - 5.5556 10⁻¹³
ρ = 1 10⁻² kg / m³
ρ = 1 10⁷ μg / m³
g Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 46.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.834 m/s2. Calculate her mass.
Answers
F divided by acceleration equals her mass.
46/0.834 = 55.156 kg
What is the acceleration of an object going from O m/s to 25 m/s in 5s?
Answers
Answer:
5m/s^2 is the acceleration.
Answer:
[tex]\boxed {\boxed {\sf a= 5 \ m/s^2}}[/tex]
Explanation:
Acceleration is the change in speed over time.
[tex]a=\frac{ v_f-v_i}{t}[/tex]
The object accelerates from 0 meters per second to 25 meters per second in 5 seconds.
[tex]v_f= 25 \ m/s\\v_i= 0 \ m/s \\t= 5 \ s[/tex]
Substitute the values into the formula.
[tex]a=\frac{ 25 \ m/s -0 m/s }{ 5 \ s}[/tex]
Solve the numerator.
[tex]a=\frac{25 \ m/s}{5 \ s}[/tex]
Divide
[tex]a= 5 \ m/s/s= 5 \ m/s^2[/tex]
The object's acceleration is 5 meters per square second.
You are driving a car behind a truck. Both your car and the truck are moving at a speed of 80km/hr. If the driver of the truck suddenly slams on the brakes, what minimum distance betweenyour car and the truck is needed so that your car does not crash into the truck’s rear end? (This is called the "​minimum trailing distance​".) To simplify this problem, assume that the truck andthe car have the same braking acceleration.
a. In order to simplify the calculations for this problem, you are told to assume that the braking acceleration of the car and the truck are the same. What other reasonable assumptions do you need to make in order to solve this problem?
b. For both the truck and the car, draw an acceleration- and velocity-versus-time graph.
c. Find an expression for the minimum trailing distance. (Your expression should only contain symbols of physical quantities. No numbers are needed here.)
d. Find the numerical value for the minimum trailing distance (Plug the values of physical quantities into your expression from part A (do not forget units!))
Answers
Answer:
Explanation:
Let the velocity of car and truck be u and breaking acceleration be a .
We shall have to assume the reflex time of the driver of the car . By the time he applies brake , his car will cover some distance . There will be some time tag between the time the truck starts decelerating and the driver of the car responding to that . During this period the car will not start decelerating . It will keep on moving with uniform velocity of u .
Let this time lag be t .
b )
For answer see the attached file
c )
The minimum trailing distance will be the distance covered by car before it starts decelerating in response to truck's deceleration .
minimum trailing distance d = u x t
d ) u = 80 km / h = 22.22 m /s
reflex action time t = 0.1 s ( assumed time )
d = 22.22 x .1
= 2.2 m
Dereck is looking at how electrically charged objects can attract other objects without touching. What control would he need to use?
An electrically charged object
An uncharged object
A positively charged object
A negatively charged object
Answers
Answer:
its An uncharged object.
if its not charged the electrically wont go on it
Answer:
uncharged object
Explanation:
Mischievous Joey likes to play with his family's lazy susan (this drives Mom crazy because it is an antique). He puts the salt shaker near the edge and tries to spin the tray at a speed so that the shaker just barely goes around without slipping off. Joey finds that the shaker just barely stays on when the turntable is making one complete turn every two seconds. Joey's older sister measures the mass of the shaker to be 79 grams. She also measures the radius of the turntable to be 0.23 m, and she is able to calculate that the speed of the shaker as it successfully goes around in a circle is 0.7222 m/s.
Required:
What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?
Answers
Answer:
0.179 N
Explanation:
What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?
The horizontal part of the constant force of the turntable on the shaker is the centripetal force of the turntable on the shaker, F.
So, F = mv²/r where m = mass of shaker = 79 g = 0.079 kg, v = speed of shaker = 0.7222 m/s and r = radius of turntable = 0.23 m
So, substituting the values of the variables into the equation, we have
F = mv²/r
F = 0.079 kg (0.7222 m/s)²/0.23 m
F = 0.0412 kgm/s² ÷ 0.23 m
F = 0.179 kgm/s²
F = 0.179 N
2) The track for a racing event was designed so that riders jump off the slope at 37 degrees from a height of 1 m. During a race it was observed that the rider remained in mid air for 1.5 seconds. Determine the speed at which he was traveling off the slope, the horizontal distance he travels before striking the ground and the maximum height he attains. Neglect the size of the bike and rider.
Answers
Answer:
3.277 m
Explanation:
Given :
Maximum Height (Hmax) = (u²sin²θ) / 2g
Xv = Xh + Uv * t + 0.5gt²
Xv and Xh are vertical and horizontal distances
-1 = 0 + sin37 * 1.5 Uv + 0.5*-9.8*1.5^2
-1 = 0 + 0.903Uv - 11.025
-1 + 11.025 = 0.903Uv
10.025 = 0.903Uv
Uv = 10.025 / 0.903
Uv = 11.10 m/s
Hmax = 1 + (u²sin²θ) / 2g
= (11.10^2 * (sin37)^2) / 2*9.8
= 44.624360 / 19.6
= 2.277
Hmax = 1 + 2.277
Hmax = 3.277 m