Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
You're driving a vehicle of mass 850 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 80 m. The maximum horizontal component of the force that the road can exert on the tires is only 0.22 times the vertical component of the force of the road on the tires (in this case the vertical component of the force of the road on the tires is , the weight of the car, where as usual 9.8 N/kg, the magnitude of the gravitational field near the surface of the Earth). The factor 0.22 is called the "coefficient of friction" (usually written "", Greek "mu") and is large for surfaces with high friction, small for surfaces with low friction.(a) What is the fastest speed you can drive and still make it around the turn? Invent symbols for the various quantities and solve algebraically before plugging in numbers.
maximum speed = ____ m/s
(b) Which of the following statements are true about this situation?
The net force is nonzero and points away from the center of the kissing circle.The momentum points toward the center of the kissing circle.The net force is nonzero and points toward the center of the kissing circle.The rate of change of the momentum is nonzero and points toward the center of the kissing circle.The centrifugal force balances the force of the road, so the net force is zero.The rate of change of the momentum is nonzero and points away from the center of the kissing circle.
(c) Look at your algebraic analysis and answer the following question. Suppose your vehicle had a mass 3 times as big (5250 kg). Now what is the fastest speed you can drive and still make it around the turn?
maximum speed = ____ m/s
(d) Look at your algebraic analysis and answer the following question. Suppose you have the original 1750 kg vehicle but the turn has a radius twice as large (166 m). What is the fastest speed you can drive and still make it around the turn?
maximum speed = ____m/s
Answer:
(a) v = 13.13 m/s
(b) The centrifugal force balances the force of the road, so net force is zero.
(c) v = 13.13 m/s
(d) v = 18.92 m/s
Explanation:
(a)
To make it around the turn without skidding the frictional force on cat must balance the centrifugal force. Therefore:
Frictional Force = Centrifugal Force
μR = mv²/r
where,
R = Normal Reaction = Weight of Car = mg
Therefore,
μmg = mv²/r
μg = v²/r
v = √μgr
where,
v = maximum possible velocity of car = ?
μ = coefficient of friction = 0.22
g = 9.8 m/s²
r = radius of curvature = 80 m
Therefore,
v = √[(0.22)(9.8 m/s²)(80 m)
v = 13.13 m/s
(b)
In order for the car to move without skidding around the turn, all the forces in horizontal direction must be equal. Hence, the centrifugal force and the frictional force (force of the road) must balance each other. So the true statement is:
The centrifugal force balances the force of the road, so net force is zero.
(c)
v = √μgr
Since the formula for speed is independent of mass. Therefore, the speed will remain same.
v = 13.13 m/s
(d)
v = √μgr
v = √[(0.22)(9.8 m/s²)(166 m)
v = 18.92 m/s
A 20 g "bouncy ball" is dropped from a height of 1.8 m. It rebounds from the ground with 80% of the speed it had just before it hit the ground. Assume that during the bounce the ground causes a constant force on the ball for 75 ms. What is the force applied to the ball by the ground in N?
The following are not correct: 0.513 N, 0.317 N, 0.121 N. Please show your work so I can understand!
Answer:
F = 0.314 N
Explanation:
In order to calculate the applied force to the ball by the ground, you first calculate the speed of the ball just before it hits the ground. You use the following formula:
[tex]v^2=v_o^2+2gy[/tex] (1)
y: height from the ball starts its motion = 1.8 m
vo: initial velocity = 0 m/s
g: gravitational acceleration = 9.8 m/s^2
v: final velocity of the ball = ?
You replace the values of the parameters in the equation (1):
[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(1.8m)}=5.93\frac{m}{s}[/tex]
Next, you take into account that the force exerted by the ground on the ball is given by the change, on time, of the linear momentum of the ball, that is:
[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}=m\frac{v_2-v_1}{\Delta t}[/tex] (2)
m: mass of the ball = 20g = 20*10^-3 kg
v1: velocity of the ball just before it hits the ground = 5.93m/s
v2: velocity of the ball after it impacts the ground (80% of v1):
0.8(5.93m/s) = 4.75 m/s
Δt: time interval o which the ground applies the force on the ball = 75*10^-3 s
You replace the values of the parameters in the equation (2):
[tex]F=(20*10^{-3}kg)\frac{4.75m/s-5.93m/s}{75*10^{-3}s}=-0.314N[/tex]
The minus sign means that the force is applied against the initial direction of the motion of the ball.
The applied force by the ground on the bouncy ball is 0.314 N
Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 11.3 m/s. Olaf's mass is 75.0 kg. (a) If Olaf catches the ball, with what speed v_f do Olaf and the ball move afterward
Answer:
v = 0.059 m/s
Explanation:
To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:
[tex]mv_{1i}+Mv_{2i}=(m+M)v[/tex] (1)
m: mass of the ball = 0.400kg
M: mass of Olaf = 75.0 kg
v1i: initial velocity of the ball = 11.3m/s
v2i: initial velocity of Olaf = 0m/s
v: final velocity of Olaf and the ball
You solve the equation (1) for v and replace the values of all variables:
[tex]v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}[/tex]
Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s
Two carts undergo an inelastic collision where they stick together. Cart A has an initial velocity v0, and the second cart B is initially at rest. After the collision, it is observed that the ratio of the final kinetic energy system to its initial kinetic energy is KfK0= 1/6. Determine the ratio of the carts' masses, mBmA. (Assume the track is frictionless.)
Answer:
Explanation:
Initial kinetic energy of the system = 1/2 mA v0²
If Vf be the final velocity of both the carts
applying conservation of momentum
final velocity
Vf = mAvo / ( mA +mB)
kinetic energy ( final ) = 1/2 (mA +mB)mA²vo² / ( mA +mB)²
= mA²vo² / 2( mA +mB)
Given 1/2 mA v0² / mA²vo² / 2( mA +mB) = 6
mA v0² x ( mA +mB) / mA²vo² = 6
( mA +mB) / mA = 6
mA + mB = 6 mA
5 mA = mB
mB / mA = 5 .
first law of equilibrium
Answer:
For an object to be an equilibrium it must be experiencing no acceleration.
Explanation:
Hope it helps.
At an accident scene on a level road, investigators measure a car’s skid mark (mass of car is M) to be of length d. It was a rainy day and the coefficient of friction was estimated to be μk.
A) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.B) Why does the car's mass not matter?1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation.2) Since the work done by friction does not depend on mass.3) Since the change in kinetic energy and the work done by friction do not depend on mass.
Answer:
1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation
Explanation:
The kinetic friction works against the kinetic energy of the car and the car stops when these two equalises .
friction force = μk x R , μk is coefficient of kinetic friction and R is reaction from the ground.
= μk x mg
work done by friction
= force x displacement
= μk x mg x d
kinetic energy of car at the time of accident = 1/2 m v²
kinetic energy = work done by friction
1/2 m v² = μk x mg x d
d = v² / (2 μk x g)
v² = 2dμk g
v = √(2dμk g)
Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation
I need someone that knows physics. I have a test in 10 hrs and Im not good at it. Can anyone help me?
Answer:
I can help! What level of physics is it and what are your main topics?
Two cars start moving from the same point. One travels south at 60 miyh and the other travels west at 25 miyh. At what rate is the distance between the cars increasing two hours later?
Answer:
65 m/h
Explanation:
Let the distance of the car moving south be y.
Let the distance of the car moving west be x.
Let the distance between the two cars be a.
These three distances can be represented as a right angled triangle. So we can say:
[tex]a^2 = x^2 + y ^2[/tex]
Let us differentiate with respect to time, since the distances are changing with respect to time:
[tex]2a\frac{da}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \\\\=>a\frac{da}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}[/tex]__________(1)
da/dt = rate of change of distance between two cars
The speed of the car moving south (dy/dt) is 60 m/h and the speed of the car moving west (dx/dt) is 25 m/h.
Therefore:
dy/dt = 60 m/h and dx/dt = 25 m/h
After two hours, the distance of the two cars will be:
y = 2 * 60 = 120 miles
x = 2 * 25 = 50 miles
Therefore:
[tex]a^2 = 50^2 + 120^2\\\\a^2 = 2500 + 14400 = 16900\\\\a = \sqrt{16900}\\ \\a = 130 miles[/tex]
From (1):
130(da/dt) = 50(25) + 120(60)
130(da/dt) = 1250 + 7200 = 8450
da/dt = 8450/130 = 65 m/h
Therefore, after two hours, the distance between the two cars is changing at a rate of 65 m/h.
A balloon with a radius of 16 cm has an electric charge of 4.25 10 –9 C.
Determine the electric field strength at a distance of 40.0 cm from the balloon’s centre.
Answer:
239 N/C
Explanation:
Electric field strength at distance R from a charge Q is given by the expression
E = k Q / R² where Q is charge , R is distance of charge from the point . k is a constant .
R = 40 cm , Q = 4.25 x 10⁻⁹
Putting the given values
E = 9 x 10⁹ x 4.25 x 10⁻⁹ / ( 40 x 10⁻²)²
= 239 N/C .
A 1KW electric heater is switched on for ten minutes
How much heat does it produce?
Explanation:
P=W/T ==> 1000w = Q/600 ==> Q=600000j
If a 1 - kilowatt electric heater is switched on for ten minutes then the heat produced by the electric heater would be 600 - kilo Joules .
What is thermal energy ?It can be defined as the form of the energy in which heat is transferred from one body to another body due to their molecular movements, thermal energy is also known as heat energy .
As given in the problem , we have to find out the heat produced by the 1 - kilo watts electric heater if it is switched on for ten minutes ,
The heat produced by the electric heater = Power × time
= 1000 × 600 Joules
= 600 kilo - Joules
Thus , the heat produced by the electric heater would be 600 - kilo Joules .
To learn more about thermal energy here , refer to the link ;
brainly.com/question/3022807
#SPJ2
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed
Complete Question
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?
A It should stay the same
B It should be quadrupled.
C It should be quintupled
D It should be doubled.
E It should be tripled
Answer:
Option D is the correct option
Explanation:
Generally electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon_o}[/tex]
Where [tex]\sigma[/tex] is the charge per unit area (Charge density )
From the question we are told that [tex]\sigma[/tex] is doubled hence the
[tex]E = \frac{2 \sigma }{\epsilon_o}[/tex]
Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density
At a time when mining asteroids has become feasible, astronauts have connected a line between their 3220-kg space tug and a 6240-kg asteroid. They pull on the asteroid with a force of 362 N. Initially the tug and the asteroid are at rest, 311 m apart. How much time does it take for the ship and the asteroid to meet
-- F = m a ... ==> a = F/m
-- The tension in the rope is 362 N. That same force acts on the asteroid and on the tug, pulling them together.
-- The asteroid's acceleration is 362N / 6240 kg = 0.058 m/s², headed for a point on the rope somewhere between the asteroid and the tug.
-- The tug's acceleration is 362 N / 3220 kg = 0.112 m/s², also headed for a point on the rope somewhere between the tug and the asteroid.
-- So now we have a gap between them, initially 311 m long, closing with a speed that starts at zero and accelerates at 0.170 m/s² .
-- D = (1/2) a T²
311 m = (1/2) (0.170 m/s²) (T²)
T² = 311 m / 0.085 m/s²
T = √(311/0.085) seconds
T = 60.41 seconds
The answer I get is so durn near 60 seconds (1 minute) that it suggests two things to me: ==> That's where the weird numbers of 362N and 311m came from, and ==> there's a good chance that my answer is correct.
Note: It's important to me that you know that 5 points for this one is really cheap and chintzy, and the reason I decided to try it was only to see whether I could.
A 330-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,110 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (Use 3.156 107 for the number of seconds in a year.)
Answer:
t = 402 years
Explanation:
To find the number of year that electrons take in crossing the complete transmission line, you first calculate the drift speed of the electrons. Then, you use the following formula for the current in a wire:
[tex]I=nqv_dA[/tex] (1)
n: number of mobile charge carrier per volume = 8.50*10^28 e/m^3
q: charge of the electron = 1.6*10^-19 C
vd: drift velocity of electron in the metal = ?
A: cross sectional area of the wire = π r^2 = π (0.02m/2)^2 = 3.1415*10^-4 m^2
I: current in the wire = 1110 A
You solve the equation (1) for vd:
[tex]v_d=\frac{I}{nqA}=\frac{110A}{(8.50*10^{28}m^{-3})(1.6*10^{-19}C)(3.1415*10^{-4}m^2)}\\\\v_d=2.59*10^{-4}m/s[/tex]
Next, you calculate the time by using the information about the length of the line transmission:
[tex]x=v_dt\\\\x=330km=330000m\\\\t=\frac{x}{v_d}=\frac{330000m}{2.59*10^{-4}m/s}=1,270,184,865s\\\\1,270,184,865s*\frac{1\ year}{3,156,107}=402.45\ years[/tex]
hence, the electrons will take aproximately 402 years in crossing the line of transmission
A 110-kg football player running at 8.00 m/s catches a 0.410-kg football that is traveling at 25.0 m/s. Assuming the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: the final velocity if the ball and player are going in the same directio
Answer:[tex]8.062\ m/s[/tex]
Explanation:
Given
masss of football player [tex]M=110\ kg[/tex]
Velocity of football player [tex]u_1=8\ m/s[/tex]
mass of football [tex]m=0.41\ kg[/tex]
velocity of football [tex]u_2=25\ m/s[/tex]
Final velocity will be given by applying conservation of linear momentum
After catching the ball Player and ball moves with same velocity
[tex]\Rightarrow Mu_1+mu_2=(M+m)v[/tex]
[tex]\Rightarrow 110\times 8+0.41\times 25=(110+0.41)v[/tex]
[tex]\Rightarrow 880+10.25=110.41\times v[/tex]
[tex]\Rightarrow v=\frac{890.25}{110.41}=8.063\ m/s[/tex]
So, final velocity will be [tex]8.062\ m/s[/tex]
first law of equilibrium
Answer:
for an object to be in equilibrium, it must be experiencing no acceleration. Both the net force and the net torque must be zero.
Hope I helped
Answer:
An object in static equilibrium has zero net force acting upon it.
The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes. This can be written as
F = F1+ F2+ F3+ F4+. . . = 0
A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target holder and the steel ball she uses has a mass of 0.0120 kg. She finds that the target ball travels a distance of 1.40 m after it is struck. Assume g = 9.80 m/s2. What is the kinetic energy (in joules) of the target ball just after it is struck?
Answer:
11.8 Joules
Explanation:
Given:-
- The height of the target ball, si = 0.860 m
- The mass of target and steel ball, m = 0.012 kg
- The target ball travels a distance ( x ) after being struck = 1.40 m
Find:-
What is the kinetic energy (in joules) of the target ball just after it is struck?
Solution:-
- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.
- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:
[tex]vf^2 = vi^2 - 2*g*x[/tex]
Where,
vf: The final velocity of target ball at maximum height = 0
g: The gravitational acceleration constant = 9.8 m/s^2
- Plug in the required parameters and evaluate the ( vi ) as follows:
[tex]0^2 = vi^2 - 2*( 9.80 )*( 1.40 )\\\\vi^2 = 27.44\\\\vi = \sqrt{27.44} = 5.24 m/s[/tex]
- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:
[tex]E_k = 0.5*m*(vi)^2\\\\E_k = 0.5*0.86*27.44\\\\E_k = 11.8 J[/tex]
Answer: The kinetic energy of the target ball just after it is struck is 11.8 Joules.
two blocks with masses 2 kg and 4 kg are pushed from rest by the same amount of fore for a distance of 100 m on a frictionless floor. the final kinetic energy of the 2 kg block after the 100 m distance is
Answer:
the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J
Explanation:
mass of block A = 2 kg
mass of block B = 4 kg
distance the blocks were pushed = 100 m
NB: Blocks were pushed the same distance at the same equal time period. And the ground is without friction.
Work done in moving the 2 kg mass along the 100 m distance is,
work = force x distance moved
force exerted by the 2 kg mass = 2 x 9.81 m/s^2(acceleration due to gravity)
force = 19.62 N
therefore,
work done = 19.62 x 100 = 1962 Joules of work.
According to energy conservation principles, the kinetic energy impacted of the 2 kg mass through this distance will be equal to the work done in moving the 2 kg mass through this distance.
Therefore, the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J
Which of these boxes will not accelerate!
30 Newtons
40 Newtons
50 kg
15 Newton
B.
10 kg
30 Newtons
C.
30 Newtons
80 kg
20 Newtons
20 Newtons
20 Newtons
D.
75 kg
Answer:
(possibly) Box D
Explanation:
The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.
Your electric drill rotates initially at 5.35 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.331 rad/s2 for 4.81 s. What is the drill's angular displacement during that time interval?
Answer:
The angular displacement is [tex]\theta = 29.6 \ rad[/tex]
Explanation:
From the question we are told that
The initial angular speed is [tex]w = 5.35 \ rad/s[/tex]
The angular acceleration is [tex]\alpha = 0.331 rad /s^2[/tex]
The time take is [tex]t = 4.81 \ s[/tex]
Generally the angular displacement is mathematically represented as
[tex]\theta = w * t + \frac{1}{2} \alpha * t^2[/tex]
substituting values
[tex]\theta = 5.35 * 4.81 + \frac{1}{2} * 0.331 * (4.81)^2[/tex]
[tex]\theta = 29.6 \ rad[/tex]
A whistle of frequency 516 Hz moves in a circle of radius 64.3 cm at an angular speed of 17.9 rad/s. What are (a) the lowest and (b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle
Answer:
(a) 498.6 Hz
(b) 534.6 Hz
Explanation: Please see the attachments below
When you take your 1900-kg car out for a spin, you go around a corner of radius 56 m with a speed of 14 m/s. The coefficient of static friction between the car and the road is 0.88. Part A Assuming your car doesn't skid, what is the force exerted on it by static friction
Answer:
6,650 newtons
Explanation:
The computation of the force exerted on it by static friction is shown below:
Data provided in the question
Mass of car = m = 1,900 kg
speed = v = 14 m/s
radius = r = 56 m
Let us assume friction force be f
And, the Coefficient of friction = [tex]\mu[/tex]= 0.88
As we know that
[tex]f = \frac{mv^2}{r}[/tex]
[tex]= \frac{1,900 \times 14^2}{56}[/tex]
= 6,650 newtons
We simply applied the above formula so that the force exerted could come
g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positive constant. Part A What is the force that one atom exerts on the other? Express your answer in terms of C6 and x. Fx = nothing Request Answer Part B Is this force attractive or repulsive? Is this force attractive or repulsive? attractive repulsive
Answer:
[tex]F_x = -\frac{6 C_6}{2^7}[/tex]
Attractive
Explanation:
Data provided in the question
The potential energy of a pair of hydrogen atoms given by [tex]\frac{C_6}{X_6}[/tex]
Based on the given information, the force that one atom exerts on the other is
Potential energy μ = [tex]\frac{C_6}{X_6}[/tex]
Force exerted by one atom upon another
[tex]F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X} (-\frac{C_6}{X^6})[/tex]
or
[tex]F_x = \frac{\partial}{\partial X} (\frac{C_6}{X^6})[/tex]
or
[tex]F_x = -\frac{6 C_6}{2^7}[/tex]
As we can see that the [tex]C_6[/tex] comes in positive and constant which represents that the force is negative that means the force is attractive in nature
Jackson heads east at 25 km/h for 20 minutes before heading south at 45 km/h for 20 minutes. Hunter heads south at 45 km/h for 10 minutes before heading east at 40 km/h for 30 minutes. Find average velocity (magnitude and direction) of each person
Answer:
The average velocity of Jackson is 18.056 m/s South
The average velocity of Hunter is 10.65 m/s East
Explanation:
initial velocity of Jackson, u = 25 km/h east = 6.944 m/s east
time for this motion, [tex]t_i[/tex] = 20 minutes = 1200 seconds
⇒initial displacement of Jackson, [tex]x_i[/tex] = (6.944 m/s) x (1200 s) = 8332.8 m
Final velocity of Jackson, v = 45 km/h South = 12.5 m/s South
time at Jackson's final position, [tex]t_f[/tex] = 20 minutes + [tex]t_i[/tex] = 20 minutes + 20 minutes
time at Jackson's final position, [tex]t_f[/tex] = 40 minutes = 2400 s
⇒Final displacement of Jackson,[tex]x_f[/tex] = (12.5 m/s) x (2400 s) = 30,000m
Average velocity of Jackson;
[tex]= \frac{x_f-x_i}{t_f-t_i} \\\\= \frac{30,000-8332.8}{2400-1200} \\\\= 18.056 \ m/s \ South[/tex]
initial velocity of Hunter, u = 45 km/h South = 12.5 m/s South
time for this motion, [tex]t_i[/tex] = 10 minutes = 600 seconds
⇒initial displacement of Hunter, [tex]x_i[/tex] = (12.5 m/s) x (600 s) = 7500 m
Final velocity of Hunter, v = 40 km/h east = 11.11 m/s east
time at Hunter's final position, [tex]t_f[/tex] = 30 minutes + [tex]t_i[/tex] = 30 minutes + 10 minutes
time at Hunter's final position, [tex]t_f[/tex] = 40 minutes = 2400 s
⇒Final displacement of Hunter,[tex]x_f[/tex] = (11.11 m/s) x (2400 s) = 26,664m
Average velocity of Hunter;
[tex]= \frac{x_f-x_i}{t_f-t_o} \\\\= \frac{26,664-7500}{2400-600} \\\\= 10.65 \ m/s \ east[/tex]
You are at a stop light in your car, stuck behind a red light. Just before the light is supposed to change, a fire engine comes zooming up towards you traveling at a horrendous 85.0 km/h. If the siren has a rated frequency 665 Hz, what frequency of the sound do you hear
Answer:
The frequency of the sound you will hear is 713.85 Hz
Explanation:
Given;
speed of your car, [tex]v_s[/tex] = 85.0 km/h
frequency of the siren, f = 665 Hz
Speed of sound in air, v = 345 m/s
The frequency of the sound you hear, can be calculated as;
[tex]f' = f(\frac{v}{v-v_s})[/tex]
Convert the speed of the car to m/s
[tex]85 \ km/h =\frac{85 \ km}{h} (\frac{1000\ m}{1 \ km})(\frac{1 \ h}{3600 \ s} ) = 23.61 \ m/s[/tex]
[tex]f' = f(\frac{v}{v-v_s} )\\\\f' = 665(\frac{345}{345-23.61} )\\\\f' = 665 (1.07346)\\\\f' = 713.85 \ Hz[/tex]
Therefore, the frequency of the sound you will hear is 713.85 Hz
If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the inductance per unit length does not exceed 50 nH per meter? Express your answer using two significant figures.
Answer:
Inner radius = 2 mm
Explanation:
In a coaxial cable, series inductance per unit length is given by the formula;
L' = (µ/(2π))•ln(R/r)
Where R is outer radius and r is inner radius.
We are given;
L' = 50 nH/m = 50 × 10^(-9) H/m
R = 2.6mm = 2.6 × 10^(-3) m
Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m
Plugging in the relevant values, we have;
50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)
Rearranging, we have;
(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)
0.25 = ln((2.6 × 10^(-3))/r)
So,
e^(0.25) = (2.6 × 10^(-3))/r)
1.284 = (2.6 × 10^(-3))/r)
Cross multiply to give;
r = (2.6 × 10^(-3))/1.284)
r = 0.002 m or 2 mm
A hornet circles around a pop can at increasing speed while flying in a path with a 12-cm diameter. We can conclude that the hornet's wings must push on the air with force components that are Group of answer choices down and backwards. down, backwards, and outwards. down and inwards. down and outwards. straight down.
Answer:
down, backwards, and outwards.
Explanation:
For a hornet that is accelerating in flight, this means that there is a net forward motion at a relatively constant vertical height above the ground.
For this flight, the wings beat downwards to counter the weight of the hornet due to gravity, keeping it at that height above the floor.
For the hornet to accelerate forward, there has to be a net backwards force by the wing on the air. This backwards force accelerates tr forward due to the absence of an equal opposing force in the opposite direction save for a little drag.
The wings also beat with forces directed outwards to provide centripetal force to keep the hornet stable. The absence of this would cause it to spiral out of control.
1. Calculate the centripetal force exerted on a 900kg900kg car that rounds a 600m600m radius curve on horizontal ground at 25.0m/s25.0m/s. 2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
Explanation:
It is given that,
Mass of a car is 900 kg
Radius of curve is 600 m
Speed of the car in the curve is 25 m/s
We need to find the centripetal force exerted on a car. The formula used to find the centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{900\times (25)^2}{600}\\\\F=937.5\ N[/tex]
So, the centripetal force exerted on a car is 937.5 N.
Static friction prevents the car from slipping. It means that the magnitude of centripetal force is balanced by the frictional force. So, the frictional force of 937.5 N is acting on the car.
a research submarine what is the maximum depth it can go
Answer: 36, 200 feet deep according to information on google
Explanation:
A small submarine, the bathyscape Trieste, made it to 10,916 meters (35,813 feet) below sea level in the deepest point in the ocean, the Challenger Deep in the Marianas Trench, a few hundred miles east of the Philippines. This part of the ocean is 11,034 m (36,200 ft) deep, so it seems that a submarine can make it as deep as it's theoretically possible to go
A uniform rod of mass 2.30 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.30 kg is attached to one end and a second mass m2 = 3.50 kg is attached to the other end of the rod. Treat the two masses as point particles.
A) What is the moment of inertia of the system?B) If the rod rotates with an angular speed of 2.00 rad/s, how much kinetic energy does the system have?C) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?D) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.00 rad/s?
Answer:
Explanation:
Moment of inertia of the rod = 1/12 m L²
m is mass of the rod and L is its length
= 1/2 x 2.3 x 2 x 2
= 4.6 kg m²
Moment of inertia of masses attached with the rod
= m₁ d² + m₂ d²
m₁ and m₂ are masses attached , and d is their distance from the axis of rotation
= 5.3 x 1² + 3.5 x 1²
= 8.8 kg m²
Total moment of inertia = 13.4 kg m²
B )
Rotational kinetic energy = 1/2 I ω²
I is total moment of inertia and ω is angular velocity
= .5 x 13.4 x 2²
= 26.8 J .
C )
when mass of rod is negligible , moment of inertia will be due to masses only
Total moment of inertia of masses
= 8.8 kg m²
D )
kinetic energy of the system
= .5 x 8.8 x 2²
= 17.6 J .
(A) Total moment of inertia is 13.4 kgm²
(B) Total kinetic energy is 26.8J
(C) Moment of inertia is 8.8 kgm²
(D) Kinetic energy is 17.6J
Rotational motion:
(A) The moment of inertia of the rod is given by:
I = 1/12 mL²
where m is the mass of the rod
and L is the length
So,
I = (1/12) × 2.3 × 2²
I = 4.6 kgm²
Now, the moment of inertia of masses attached to the rod is given by:
I' = m₁ d² + m₂d²
where m₁ and m₂ are masses
and d is their distance from the axis of rotation
I' = 5.3 × 1² + 3.5 × 1²
I' = 8.8 kgm²
The total moment of inertia of the system is given by:
I(tot) = I + I'
I(tot) = 13.4 kgm²
(B) The rotational kinetic energy of an object with a moment of inertia I and angular velocity ω is given by:
KE = 1/2 I(tot)ω²
KE = 0.5 × 13.4 × 2²
KE = 26.8J
(C) If the mass of the rod is negligible, then the moment of inertia of the rod will be zero. So the total moment of inertia will be
I(tot) = I' = 8.8 kgm²
(D) the kinetic energy of the system when the mass of the rod is negligible and the angular speed is 2 rad/s is given by:
KE = 1/2 I'ω²
KE = 0.5 × 8.8 × 2²
KE = 17.6J
Learn more about rotational motion:
https://brainly.com/question/15120445?referrer=searchResults
A 18-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 30 N. Starting from rest, the sled attains a speed of 2.0 m/s in 8.5 m. Find the coefficient of kinetic friction between the runners of the sled and the snow. Num
Answer:
Coefficient of kinetic friction = 0.146
Explanation:
Given:
Mass of sled (m) = 18 kg
Horizontal force (F) = 30 N
FInal speed (v) = 2 m/s
Distance (s) = 8.5 m
Find:
Coefficient of kinetic friction.
Computation:
Initial speed (u) = 0 m/s
v² - u² = 2as
2(8.5)a = 2² - 0²
a = 0.2352 m/s²
Nweton's law of :
F (net) = ma
30N - μf = 18 (0.2352)
30 - 4.2336 = μ(mg)
25.7664 = μ(18)(9.8)
μ = 0.146
Coefficient of kinetic friction = 0.146
