Answer:
The maximum speed of the car is 35 m/s
The total distance traveled by the car is 658.33 m
Explanation:
Given;
initial velocity of the car, u = 15 m/s
acceleration of the car, a = 2 m/s²
time of car motion, t = 10 s
(i)
Initial distance traveled by the car is given by;
d₁ = ut + ¹/₂at²
d₁ = (15 x 10) + ¹/₂(2)(10)²
d₁ = 150 + 100
d₁ = 250 m
The maximum speed of the car during this is given by;
v² = u² + 2ad₁
v² = (15)² + (2 x 2 x 250)
v² = 1225
v = √1225
v = 35 m/s
(ii)
The final distance cover by the car during the deceleration of 1.5 m/s².
Note: the final or maximum speed of the car becomes the initial velocity during deceleration.
v² = u² + 2ad₂
where;
v is the final speed of the car when it stops = 0
0 = u² + 2ad₂
0 = (35²) + (2 x - 1.5 x d₂)
0 = 1225 - 3d₂
3d₂ = 1225
d₂ = 1225 / 3
d₂ = 408.33 m
The total distance traveled by the car is given by;
d = d₁ + d₂
d = 250 m + 408.33 m
d = 658.33 m
Which two formulas are used to calculate potential and kinetic energy
Answer:
gravitational potential energy:
GPE = m g h
kinetic energy:
KE = 1/2 m v^2
Answer:
[tex]\boxed{\bold { \large { \boxed {KE=\frac{1}{2} mv^2 \ , \ PE=mgh}}}}[/tex]
Explanation:
Kinetic energy formula
[tex]\displaystyle KE=\frac{1}{2} mv^2[/tex]
Potential energy formula
[tex]\displaystyle PE=mgh[/tex]
[tex]\displaystyle KE \Rightarrow \sf kinetic \ energy \ (J)[/tex]
[tex]\displaystyle PE \Rightarrow \sf potential \ energy \ (J)[/tex]
[tex]\displaystyle m \Rightarrow \sf mass \ (kg)[/tex]
[tex]\displaystyle v \Rightarrow \sf velocity \ (m/s)[/tex]
[tex]\displaystyle g \Rightarrow \sf acceleration \ of \ gravity\ (m/s^2)[/tex]
[tex]\displaystyle h \Rightarrow \sf height \ (m)[/tex]
waves disturb ____, but do not transmit it
a. energy
b. matter
c. sound
d. none of the above
Answer:
energy
Explanation:
im sure
An elastic conducting material is stretched into a circular loop of 13.6 cm radius. It is placed with its plane perpendicular to a uniform 0.871 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 73.9 cm/s. What emf is induced in volts in the loop at that instant?
Answer: The Answer is attached to the image below
9)A skier starts from rest from the top of a 40 m high slope which makes 40 degrees with the ground. Coefficient of friction is 0.1 What is the velocity of the skier at the bottom of the ramp?
Answer:
The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.
Explanation:
We can determine the final velocity of the skier at the bottom of the ramp by Principle of Energy Conservation and Work-Energy Theorem, whose model is:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+W_{disp}[/tex] (1)
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.
[tex]W_{disp}[/tex] - Work dissipated by friction, measured in joules.
By definitions of gravitational potential and translational kinetic energy and work, we expand and simplify the model:
[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2}) =\mu_{k}\cdot N\cdot \Delta s[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the skier, measured in meters.
[tex]N[/tex] - Normal force from the incline on the skier, measured in newtons.
[tex]\Delta s[/tex] - Distance covered by the skier, measured in meters.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
The normal force exerted on the skier and the covered distance are, respectively:
[tex]N = m\cdot g\cdot \cos \theta[/tex] (3)
[tex]\Delta s = \frac{z_{1}-z_{2}}{\sin \theta}[/tex] (4)
Where [tex]\theta[/tex] is the angle of the incline above the horizontal, measured in sexagesimal degrees.
By applying (3) and (4) in (2), we get that:
[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot m\cdot g \cdot \cos \theta \cdot \left(\frac{z_{1}-z_{2}}{\sin \theta} \right)[/tex]
[tex]g\cdot (z_{1}-z_{2}) +\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})= \mu_{k}\cdot g \cdot \left(\frac{z_{1}-z_{2}}{\tan \theta} \right)[/tex] (5)
Then, we clear the velocity of the skier at the bottom of the ramp is: ([tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.1[/tex], [tex]\theta = 40^{\circ}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}-z_{2} = 40\,m[/tex])
[tex]\left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = \frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex]
[tex]2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = v_{1}^{2}-v_{2}^{2}[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}-2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2})}[/tex] (6)
[tex]v_{2} = \sqrt{\left(0\,\frac{m}{s} \right)^{2}-2\cdot \left(\frac{0.1}{\tan 40^{\circ}} -1\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m)}[/tex]
[tex]v_{2} \approx 26.288\,\frac{m}{s}[/tex]
The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.
Ball 1 (1.5 kg) moves to the right at 2 m/s and ball 2
(2.5 kg) moves to the left at 1.5 m/s. The balls stick together after collision. What is the speed and direction of ball 2 after the collision?
Answer:
0.1875 m/s leftward
Explanation:
Taking rightwards as positive
We are given:
Ball 1:
Mass (m1) = 1.5 kg
velocity (u1) = 2 m/s
Ball 2:
Mass (m2) = 2.5 kg
velocity (u2) = -1.5 m/s [negative because it is in the opposite direction]
Speed and Direction of Ball 2:
We are told that the balls stick together after the collision
We can say that the balls have the same velocity since they are sticking together
So, Final velocity of Ball 1 (v1) = Final velocity of Ball 2 (v2) = V m/s
According to the law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
replacing the variables
1.5(2) + (2.5)(-1.5) = V (1.5 + 2.5) [v1 = v2 = V]
3 + (-3.75) = 4V
-0.75 = 4V
V = -0.75/4 [dividing both sides by 4]
V = -0.1875 m/s
Hence, the balls will move at a velocity of 0.1875 m/s in the Leftward direction
If you doubled the load resistor in a Wheatstone bridge, the load current would not be half as much. Why not?
Question 2 of 10
What is the current in the 10.0 2 resistor?
10.00
w
120.0 V
20.00
w
30.00
A. 10.0 A
B. 12.0 A
C. 0.0833 A
O D. 2.00 A
Answer:
2.00 A
Explanation:
Total resistance = 10 + 20 + 30 = 60 ohms
Potential difference (V) = 120volts
Current (I) =?
from Ohms law V = IR
==> I = V/R = 120/60 = 2 A
Note: for resistors in series equal amount of electric current flows through the circuit
3
A ball rolls at a speed of 4 km/h towards the back of a bus traveling forward at a speed of 35
km/h. What is the magnitude of the ball's velocity relative to the ground?
(33 Points)
O A: 30 km/h
B: 31 km/h
O c: 35 km/h
OD: 39 km/h
Answer:
31km/hExplanation:
Step one:
given data
speed of bus= 35km/h forward
speed of ball= 4km/h backwards
`Step two:
Required
the magnitude of the velocity of the ball relative to the ground is the net velocity
that is
=35-4
=31km/h
The velocity of the ball relative to the ground is 31km/h
1. A stone of mass 0.8 kg is attached to a 0.9 m long string. The string will break if the tension exceeds 60 N. The stone is whirled in a horizontal circle on a frictionless tabletop while the other end of the string remains fixed. What is the maximum speed the stone can attain without breaking the string?
A. 8.22 m/s
B. 7.30 m/s
C. 9.34 m/s
D. 7.76 m/s
2. A highway curve with radius 900 ft is banked so that a car traveling at 55 mph will not skid sideways even in the absence of friction. At what angle should the curve be banked to prevent skidding?
A. 14.6°
B. 18.9°
C. 10.9°
D. 12.7°
3. A button will remain on a horizontal platform rotating at 40 rev/min as long as it is no more than 0.15 m from the axis. How far from the axis can the button be placed without slipping if the platform rotates at 60 rev/min?
A. 0.365 m
B. 0.338 m
C. 0.225 m
D. 0.294 m
4. What is the tension in a cord 10 m long if a mass of 5 kg is attached to it and is being spun around in a circle at a speed of 8 m/s?
A. 67 N
B. 28 N
C. 32 N
D. 50 N
5. A 0.5 kg mass attached to a string 2 m long is whirled around in a horizontal circle at a speed of 5 m/s. What is the centripetal acceleration of the mass?
A. 11.3 m/s2
B. 12.5 m/s2
C. 5.9 m/s2
D. 10.2 m/s2
6. Find the maximum speed with which a car can round a curve that has a radius of 80 m without slipping if the road is unbanked and the coefficient of friction between the road and the tires is 0.81.
A. 44.3 m/s
B. 20.8 m/s
C. 25.2 m/s
D. 30.6 m/s
Answer:
1. A. 8.22. m/s
Explanation:
What is the ratio of thicknesses of crown glass and water that would contain the same number of wavelengths of light?
Answer:
the thickness of the glass divided by thickness of water is going to be 1.333 divided by 1.52, which is 0.877. So, the height of this glass, in order to have the same number of wavelengths as in water, the height of the glass will be 0.877 times the height of the water, and so it will be smaller.
Use the graph to determine the object's average velocity
What is the average velocity of the object?
m/s
Position vs Time
Position (m)
N A 0 0
0
10
20
30
40
50
Time (s)
Answer:
10
Explanation:
10 is answer because velocity is time /position. So time is 50 and position is 5
The 600-N ball shown is suspended on a string AB and rests against the frictionless vertical wall. The string makes an angle of 30° with the wall. The line AB goes through the center of the ball, and the contact point with the wall is at the same vertical height as the center of the ball. The ball presses against the wall with a force of magnitude:
Answer: T = 692.82 and 346.4 N
Explanation:
Given that;
w = 600 N
∅ = 30°
ΣFy = ma
a = 0 m/s²
ΣF = T(cos30°) - W = 0
T(cos30°) = W
we Divide both sides by cos30°
T = W / cos30o
T= 600N / cos30°
T = 692.82
and ∑fx
F = T sin∅
F = 692.82 × (sin30°)
F = 346.4 N
The equilibrium condition allows finding the result for the force of the ball against the wall is:
The force of the ball directed towards the wall is 346.4 N
Newton's second law gives a relationship between force, mass and acceleration of bodies. In the case where the acceleration is zero, it is called the equilibrium condition.
∑ F = 0
A free-body diagram is a diagram of the forces without the details of the bodies. In the attached we have a free-body diagram of the system.
Let's use trigonometry to break down stress.
sin 30 = [tex]\frac{T_x}{T}[/tex]
cos 30 = [tex]\frac{T_y}{T}[/tex]
T_y = T cos 30
Tₓ = T sin 30
Let's write the equilibrium condition for the system.
y-axis.
T_y -W = 0
T cos 30 = W
[tex]T = \frac{W}{cos 30}[/tex]
x-axis.
R - Tₓ = 0
R = T sin 30
We substitute
[tex]R = \frac{W}{cos 30} \ sin 30 \\R = W \ tan 30[/tex]
Let's calculate.
R = 600 tan 30
R = 346.4 N
This force is directed from the wall towards the ball, by Newton's third law the force of the ball is of equal magnitude and opposite direction, that is, directed towards the wall.
In conclusion with the equilibrium condition we can find the result for the force of the ball against the wall is:
The force of the ball directed towards the wall is 346.4 N
Learn more about the equilibrium condition here: brainly.com/question/18117041
A ball is kicked horizontally at 4.6 m/s off of a cliff 13.4 m high. How far from the cliff will it land.
7.53 m
Explanation:We are given:
Initial Horizontal Velocity of the Ball = 4.6 m/s
Initial Vertical Velocity of the Ball = 0 m/s
Height from which ball is kicked = 13.4 m
Time taken by the ball to reach the ground:
The ball has an initial vertical velocity of 0 m/s
it also has a downward acceleration of 10 m/s² due to gravity
Solving for the time taken:
s = ut + 1/2(at²) [second equation of motion]
replacing the values
13.4 = (0)(t) + 1/2 (10)(t²)
13.4 = 5t²
t² = 13.4/5 [dividing both sides by 5]
t² = 2.68
t = 1.637 seconds [taking the square root of both sides]
Horizontal distance covered by the ball:
Since there are no horizontal opposing forces on the ball,
the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground
We calculated that the ball will hit the ground in 1.637 seconds
Distance covered:
s = ut + 1/2 (at²) [seconds equation of motion]
s = ut [since a = 0m/s² in the horizontal plane]
replacing the values
s = 4.6 * 1.637
s = 7.53 m
Hence, the ball landed 7.53 m from the cliff
A 4.0 kg mess kit sliding on a frictionless surface explodes into two 2.0 kg parts: 3.0 m/s, due north, and 5.0 m/s, 30° north of east.What is the original speed of the mess kit?
Answer:
3.49m/s
Explanation:
We have mess of kit = 4kg
Then pieces are 2kg
3m/s due north
5m/s, 30⁰ due south
MiVi = m1v1 + m2v2
We break velocities to have x and y components
MiVi = m1v1(cosθ+sin θ1j) + m2v2(v2cos θ2i+sin2 θj)
θ1 = 90⁰north
θ2 = 30⁰ north east
Vi = 2/4x3m/s(cos90i + sin90j)+2/4x5m/s(cos30i + sin30j)
= 2.16i + 2.75j
Vi = √Vx²+Vy²
Vi = √ (2.16)²+(2.75)²
Vi = 3.49m/s
This is the original speed of the kit
2. Which bicyclist was traveling the fastest at the end of the race?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, to determine the bicyclist that traveled the fastest at the end of the race, the speed of the bicyclists at the end of the race will determine this (not the bicyclist that came first nor there overall speed). The speed of the bicyclist at the end of the race can be determined by using the formula below
s = d ÷ t
Where s is the speed of each bicyclist at the end of the race
d is the specific distance covered by the bicyclist at the end of the race
t is the time taken for the bicyclist to complete that distance
It should be noted that to get an accurate result, the distance covered at the end of the race must be the same for all the bicyclists.
3 Magnetised mineral ore was found in
A Africa
B Australia
C America
D Magnesia
E None of the above
Answer:
D Magnesia
Explanation:
Ore particles Fe3O4 (magnetite) were found in the region called Magnesia.
Magnesia was an antic city in Asia, named after the inhabitants of Greek Magnesia.
Magnetite is used as an ore, abrasive, in paint production, electrophotography, as a micronutrient fertilizer, and as a high-density concrete aggregate.
A 350 g mass on a 45-cm-long string is released at an angle of 4.5° from vertical. It has a damping constant of 0.010 kg/s. After 25 s, how many oscillations has it completed?
Answer:
The value is [tex]n = 18.5 \ oscillations[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 350 \ g = 0.350 \ kg[/tex]
The length is [tex]L = 45 \ cm = 0.45 \ m[/tex]
The angle is [tex]\theta = 4.5^o[/tex]
The damping constant is [tex]b = 0.010 \ kg/s[/tex]
The time taken is [tex]t = 25 \ s[/tex]
Generally the angular frequency of this damped oscillation is mathematically evaluated as
[tex]w = \sqrt{ \frac{ g}{L} + \frac{b^2}{4m^2} }[/tex]
=> [tex]w = \sqrt{ \frac{9.80 }{ 0.45} + \frac{0.010 ^2}{4* 0.350^2} }[/tex]
=> [tex]w = 4.667 \ s^{-1}[/tex]
Generally the period of the oscillation is mathematically represented as
[tex]T = \frac{2 \pi }{w}[/tex]
=> [tex]T = \frac{2 * 3.142 }{ 4.667 }[/tex]
=> [tex]T = 1.35 \ s[/tex]
Generally the number of oscillation is mathematically represented as
[tex]n = \frac{t}{T}[/tex]
=> [tex]n = \frac{25}{1.35}[/tex]
=> [tex]n = 18.5 \ oscillations[/tex]
Is a parked car potential or kinetic ?
Answer:
Potential energy is the energy that is stored in an object. ... When you park your car at the top of a hill, your car has potential energy because the gravity is pulling your car to move downward; if your car's parking brake fails, your vehicle may roll down the hill because of the force of gravity.
3. The force exerted by gravity on kg = 98N
Answer:
960.4N
Explanation:
Given parameter:
Mass of the body = 98kg
Unknown:
Force exerted by gravity = ?
Solution:
Force exerted by gravity on the body is the weight
Weight = force x acceleration due to gravity
Acceleration due to gravity = 9.8m/s²
Weight = 98kg x 9.8m/s²
Weight = 960.4N
When particles get close to the surface, they interact with atoms in
the
(Finish the sentence)
A ball is kicked horizontally at 4.6 m/s off of a cliff 12.4 m high. How far from the cliff will it land?
A) 2.5 m
B) 1.6 m
C) 7.4 m
D) 4.1 m
Answer:
42
Explanation:
42
Ned is learning about cells in science class. He drew a picture to model a cell. What can he show with his cell drawing?
A)
how big a cell is
B)
how fast cells grow
c)
what cells do to move
D)
what the cell parts are
Answer:I need the answer pls
Explanation:
I don't have one
Answer:
The answer is D, what the cell parts are.
Explanation:
In the days before scuba gear, some divers descended to underwater depths in diving bells, which are basically just upside-down containers whose open ends face down. The bell allows the person inside to breathe the air trapped inside it, observe underwater objects and marine life, and work under the water.
Required:
a. If the bell is submerged to a depth of 30m below sea level, what is the water pressure at the air-water interface inside the bell?
b. If the air pressure inside the bell before submersion into the water was 1 atm (101.825 kPa), what air pressure does the person experience at that depth?
Answer:
a) P = 4.03 10⁵ Pa, b) P = 4.03 10⁵ Pa
Explanation:
a) The pressure as a function of the depth of a fluid is
P =[tex]P_{atm}[/tex] + ρ g y
where Patm is the atmospheric pressure, the sea density of about 1025 kg / m³
let's calculate
P = 1.01825 10⁵ + 1025 9.8 30
P = 4.03 10⁵ Pa
b) When the hood is submerged, the water exerts a perpendicular force on the entire surface, in the equilibrium position, the air is compressed by this force until the pressure it exerts is equal to the external pressure (open at the lower), therefore the air pressure is
P = 4.03 105 Pa
"[tex]3.958\times 10^5 \ Pa[/tex]" would be the pressure outside the bell .as well as the air pressure inside it. A further explanation is below.
According to the question,
Depth,
[tex]d = 20 \ m[/tex]Pressure,
[tex]P_o = 1 \ atm[/tex][tex]= 1.018\times 10^{5} \ Pa[/tex]
Now,
→ The water pressure outside the bell will be:
= [tex]P_o +p\times g\times d[/tex]
By putting the values, we get
= [tex]1.018\times 10^5+1000\times 9.8\times 30[/tex]
= [tex]3.958\times 10^5 \ Pa[/tex]
And inside the air pressure will be same as the water pressure.
Thus the response above is right.
Learn more:
https://brainly.com/question/22786261
Ms. Reitman's scooter starts from rest and the final velocity is
14 m/s in 7 seconds. What is the average acceleration?
Answer:
2 m/s 2
Explanation: Just is....
Complete this statement: When a substance undergoes fusion, it:
a. condenses
b. freezes
c. evaporates
d. sublimes
e vaporizes
When a substance undergoes fusion, it melts.
Fusion is a physical process. The process of fusion of substances takes place during the conversion of a solid substance into a liquid.
For this process to take place, heat is being applied to the solid substance, and the particles of the solid loose, then move freely thereby reducing the intermolecular forces in the solid substance before changing into liquid.
Learn more about physical changes here:
https://brainly.com/question/25014732?referrer=searchResults
A Ferris wheel has a diameter of 60 m and a period of rotation of 70 s. A passenger weighs 500 N. What is her apparent weight at the highest point of the Ferris wheel?
A. 298 N
B. 350 N
C. 487 N
D. 513 N
Answer:
A
Explanation:
Hopefully this helps.
You were chosen to co-pilot a mission to Mars. After successfully reaching a stable orbit, a piece of space-junk hits the Shuttle and you are sent out on a space walk with out a rope and only with a large roll of duct tape to repair the damage. You lose your grip during the walk and start to float away from the shuttle and realize you don’t have a safety line to grab. Should Mr. Wright call your parents to tell them you floated out into space or is it possible you can get back to the ship? Explain your answer. If possible, include a force diagram in your explanation. Hint: Think about newton’s laws.
Answer:
im just so focused on the fact that im going to mars :O
A sealed cubical container 28.0 cm on a side contains three times Avogadro's number of molecules at a temperature of 24.0°C. Find the force exerted by the gas on one of the walls of the container.
Answer:
3.32 atm
__________________________________________________________
We are given:
side of the cubical container = 28 cm
number of molecules in the container = 3 * Nₐ
[where Nₐ is the Avogadro's number]
Temperature = 24°C OR 297 K
We need to find the pressure exerted by the gas on the walls of the container
__________________________________________________________
Some Calculations:
Volume of the container
we are given the side of the cubical container = 28 cm
Volume of the cubical container = side³
Volume = 28³
Volume = 21952 cm³
We know that 1 cm³ = 1 mL
So,
Volume = 21952 mL
We also know that 1 L = 1000 mL
Volume = 21.952 L
Number of moles of Gas
We know that:
number of moles = number of molecules / Avogadro's number
number of moles = 3 * Nₐ / Nₐ [number of molecules = 3 * Nₐ]
number of moles = 3 moles
__________________________________________________________
Pressure Exerted by the Gas:
Using the ideal gas equation:
PV = nRT
Since the volume is in L, and Temperature is in K. R is equal to
0.082 L atm /mol K and the pressure will be in atm
P(21.952) = 3*(0.082)*(297)
P = 3.32 atm
Hence, the gas will exert a pressure of 3.32 atm on the walls of the container
While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius 12.0 cm. If the angular speed of the front sprocket is 0.600 rev/s, what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be 5.00 m/s?
Answer:
2.9 cm
Explanation:
Assuming that the rear wheel has a radius of 0.330 m
Given that
r(a) = 12 cm -> 0.12 m
w(a) = 0.6 rev/s -> 3.77 rad/s
v = 5 m/s
r(w) = 0.330 m
The speed on any point on the rim at the sprocket in the front is
v(a) = w(a).r(a) = 3.77 * 0.12 = 0.4524 m/s
Also,
v(a) = speed at any point on the chain
v(b) = speed at any point on the rim of the rear sprocket
v(a) = v(b)
where v(b) = w(b).r(b)
Recall that the speed at any point on the rear wheel is v, where
v = w(b).r(w)
5 = w(b) * 0.330
w(b) = 5/0.330
w(b) = 15.15 rad/s
On substituting this in the equation, we have
v(b) = w(b).r(b).
Remember also, that v(a) = v(b), so
0.4524 = 15.15 * r(b)
r(b) = 0.4524 / 15.15
r(b) = 0.029 m -> 2.9 cm
Therefore, the radius of the rear sprocket needed is 2.9 cm
A submarine sends out a sonar signal (sound waves) in a direction directly downward it take 2.3 s for the sound waves to travel from the submarine to the ocean bottom and back to the submarine how high (approx) up from the ocean floor is the submarine?speed of sound in water is 1490 m/s
Explanation:
Using the formula;
2x = vt
x is the distance up from the ocean floor the submarine is
v is the speed of sound in water
t is the time
Given
t = 2.3s
v = 1490m/s
Required
how high (approx) up from the ocean floor is the submarine x
From the formula;
x = vt/2
x = 1490(2.3)/2
x = 745(2.3)
x = 1,713.5m
Hence the submarine is 1713.5m high up from the ocean floor