Answer:
10.15m/s
Explanation:
The change in the frequency of sound (or any other wave) when the source of the sound and the receiver or observer of the sound move towards (or away from) each other is explained by the Doppler effect which is given by the following equation:
f₁ = [(v ± v₁) / (v ± v₂)] f ----------------------(i)
Where;
f₁ = frequency received by the observer or receiver
v = speed of sound in air
v₁ = velocity of the observer
v₂ = velocity of the source
f = original frequency of the sound
From the question, the observer is the bicyclist and the source is the car driver. Therefore;
f₁ = frequency received by the observer (bicyclist) = 357Hz
v = speed of sound in air = 330m/s
v₁ = velocity of the observer(bicyclist) = (1 / 3) v₂ = 0.33v₂
v₂ = velocity of the source (driver)
f = original frequency of the sound = 365Hz
Note: The speed of the observer is positive if he moves towards the source and negative if he moves away from the source. Also, the speed of the source is positive if it moves away from the listener and negative otherwise.
From the question, the cyclist and the driver are moving in the same direction. But then, we do not know which one is at the front. Therefore, two scenarios are possible.
i. The bicyclist is at the front. In this case, v₁ and v₂ are negative.
Substitute these values into equation (i) as follows;
357 = [(330 - 0.33v₂) / (330 - v₂)] * 365
(357 / 365) = [(330 - 0.33v₂) / (330 - v₂)]
0.98 = [(330 - 0.33v₂) / (330 - v₂)]
0.98 (330 - v₂) = (330 - 0.33v₂)
323.4 - 0.98v₂ = 330 - 0.33v₂
323.4 - 330 = (0.98 - 0.33)v₂
-6.6 = 0.65v₂
v₂ = -10.15
The value of v₂ is not supposed to be negative since we already plugged in the right value polarity into the equation.
iI. The bicyclist is behind. In this case, v₁ and v₂ are positive.
Substitute these values into equation (i) as follows;
357 = [(330 + 0.33v₂) / (330 + v₂)] * 365
(357 / 365) = [(330 + 0.33v₂) / (330 + v₂)]
0.98 = [(330 + 0.33v₂) / (330 + v₂)]
0.98 (330 + v₂) = (330 + 0.33v₂)
323.4 + 0.98v₂ = 330 + 0.33v₂
323.4 - 330 = (0.33 - 0.98)v₂
-6.6 = -0.65v₂
v₂ = 10.15
The value of v₂ is positive and that is a valid solution.
Therefore, the speed of the car is 10.15m/s
The diameter of a copper wire that is being used in a high voltage power line is 8.252 mm. What is the resistance in Ohms of a power line that is 361 miles long if it is 20oC
Answer: 182.45Ω
Explanation:
Hi to answer this question we have to apply the next formula:
Resistance = ρL /A
Where:
ρ = resistivity of copper at 20° ( 1.68 x 10^-8 Ωm)
L = lenght
A = transversal area = π (diameter /2)^2 = π (8.252x10^-03 /2)^2 =
5.35 x10^-05
Since
1 mile = 1,609 meters
361 miles = 580,849 meters (length)
1 mm = 0.001 meter
8.252 mm = 8.252x10^-03 m (diameter)
Resistance = [ 1.68 x 10^-8 (580,849)]/(5.35 x10^-05) = 182.45Ω
Feel free to ask for more if needed or if you did not understand something.
A wire carries a current of 4 A travelling to the left (-x direction). It is placed in a constant magnetic field of magnitude 0.05 T, pointing upward ( z direction). a. If 25 cm of the wire is in the magnetic field, what is the force on the current
Answer:
0.05 N
Explanation:
Data provided in the question
The Wire carries a current of 4A to the left direction
The constant magnetic field of magnitude = 0.05 T
Pointing upward i.e Z direction
The wire is in the magnetic field = 25 cm
Based on the above information, the force on the current is
[tex]= Current \times constant\ magnetic\ field\ of\ magnitude \times magnetic\ field[/tex]
[tex]= 4 \times 0.05 \times 0.25[/tex]
= 0.05 N
The direction will be the negative Y direction
Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10 m. The bridge is supported by two vertical stone pillars, one 2.0 m from the left end of the bridge and the other 2.0 m from the right end of the bridge. If a 200 kg knight stands on the bridge 4.0 m from the left end, what force is applied by the left support
Answer:
F = 2123.33N
Explanation:
In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.
[tex]\Sigma \tau=0[/tex]
Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.
Then, you obtain the following formula:
[tex]-\tau_l+\tau_p+\tau_{cm}=0[/tex] (1)
τl: torque produced by the left support
τp: torque produced by the person
τcm: torque produced by the center of mass of the wooden
The torque is given by:
[tex]\tau=Fd[/tex] (2)
F: force applied
d: distance to the pivot of the torque, in this case, distance to the right support.
You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:
[tex]-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N[/tex]
Where d1, d2 and d3 are distance to the right support.
You solve the equation for F and replace the values of the other parameters:
[tex]F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N[/tex]
The force applied by the left support is 2123.33 N
A car travels north at 30 m/s for one half hour. It then travels south at 40 m/s for 15 minutes. The total distance the car has traveled and its displacement are: Group of answer choices 36 km; 36 km N. 90 km; 18 km N. 90 km; 36 km N. 36 km; 36 km S. 18 km; 18 km S.
Answer:
xtotal = 90km
displacement = 18km N
Explanation:
To find the total distance traveled by the car, you first calculate the distance traveled by the car when it travels to north. You use the following formula:
[tex]x=vt[/tex] (1)
x: distance
v: speed of the car = 30 m/s
t: time = one half hour
In order to calculate the distance you convert the time from hours to seconds:
[tex]t=0.5\ h*\frac{3600s}{1\ h}=1800s[/tex]
Then, you replace the values of t and v in the equation (1):
[tex]x=(30m/s)(1800s)=54000m[/tex] (2)
Next, you calculate the distance traveled by the car when it travels to south:
[tex]x'=v't'\\\\v'=40\frac{m}{s}\\\\t'=15\ min[/tex]
You convert the time from minutes to seconds:
[tex]t'=15\ min*\frac{60s}{1min}=900s[/tex]
[tex]x'=(40m/s)(900s)=36000m[/tex]
Finally, you sum both distances x and x':
[tex]x_{total}=x+x'=54000m+36000m=90000m=90km[/tex]
The total distance traveled by the car is 90km
The total displacement is the final distance of the car respect to the starting point of the motion. This is calculated by subtracting x' to x:
[tex]d=x-x'=54000m-36000m=18000m=18km[/tex]
The total displacement of the car is 18km to the north from its starting point of motion.
A ray in glass (n = 1.51) reaches a boundary with air at 49.2 deg. Does it reflect internally or refract into the air? Enter 0 for reflect, and 1 for refract.
Answer:
0 - Then, the ray is totally reflected
Explanation:
The ray reaches the boundary between the two mediums at 49.2°.
If the ray is totally reflected it is necessary that the crictical angle is lower that the incidet angle.
You use the following to calculate the critical angle:
[tex]\theta_c=sin^{-1}(\frac{n_2}{n_1})[/tex] (1)
n2: index of refraction of the second medium (air) = 1.00
n1: index of refraction of the first medium (glass) = 1.51
You replace the values of the parameters in the equation (1):
[tex]\theta_c=sin^{-1}(\frac{1.00}{1.51})=41.47\°[/tex]
The critical angle is 41.47°, which is lower than the incident angle 49.2°.
Then, the ray is totally reflected.
0
A box of mass 45 kg is at rest on a horizontal frictionless surface. A constant horizontal force F then acts on the box and accelerates it to the right. It is observed that it takes the box 7.5 seconds to travel 40 meters. What is the magnitude of the force
Answer:
F = 64.0 N
the magnitude of the force is 64 N
Explanation:
Given;
Mass m = 45kg
time t = 7.5 s
Displacement d = 40 m
Force = mass × acceleration
F = ma ........1
From equation of motion;
d = vt + 0.5at^2
Since initial velocity v = 0 (initially at rest)
d = 0.5at^2
Making a the subject of formula;
a = d/(0.5t^2)
Substituting the given values;
a = 40/(0.5×7.5^2)
a = 1.42 m/s^2
From equation 1;
F = ma
Substituting the given values;
Force F = 45 kg × 1.42 m/s^2
F = 64.0 N
the magnitude of the force is 64 N
A coastline on Earth experienced a high tide at 6 a.m. At what approximate time will the next high tide occur, and why?
Answer:
about 6:25 p.m.
Explanation:
There are two complete cycles of the tides in each lunar day of about 24 hours and 50 minutes. Thus peak tides are about half a lunar day apart, or 12 hours and 25 minutes.
12 hours and 25 minutes after 6 a.m. is 6:25 p.m.
Ignoring details associated with friction, extra forces exerted by arm and leg muscles, and other factors, we can consider a pole vault as the conversion of an athlete's running kinetic energy to gravitational potential energy. If an athlete is to lift his body 3.8 m during a vault, what speed (in m/s) must he have when he plants his pole
Answer:
v = 8.63 m/s
Explanation:
Neglecting the frictional forces, the law of conservation of energy can be applied to this situation as follows:
Potential Energy Gained By Athlete = Kinetic Energy Lost By Athlete
mgh = (0.5)mv²
gh = (0.5)v²
v = √2gh
where,
v = speed that the athlete must have when he plants the pole = ?
g = acceleration due to gravity = 9.8 m/s²
h = Height to be achieved by the athlete = 3.8 m
Therefore,
v = √(2)(9.8 m/s²)(3.8 m)
v = 8.63 m/s
That 85 kg paratrooper from the 50's was moving at constant speed of 56 m/s because the air was applying a frictional drag force to him that matched his weight. If he fell this way for 40 m, how much heat was generated by this frictional drag force in J
Answer:
46648 J
Explanation:
mass m= 85 Kg
velocity v = 56 m/s
distance covered s =40 m
According to Question,
frictional drag force to him that matched his weight
[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]
Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J
W=Q= F_d×s
=833×56 = 46648 J
Q3. A car is initially heading east at 55.0 km h when it turns a corner and heads N 55° Eat
65.0 km h -1. Determine the change in the velocity of the car during this time.
Answer:
37.3 km/h at N 2.70° E
Explanation:
Find the horizontal and vertical components of the initial velocity (u) and final velocity (v). Remember that N 55° E means 55° east of north.
uₓ = 55.0 km/h
uᵧ = 0 km/h
vₓ = 65.0 km/h × sin 55° = 53.24 km/h
vᵧ = 65.0 km/h × cos 55° = 37.28 km/h
Find the difference.
vₓ − uₓ = -1.76 km/h
vᵧ − uᵧ = 37.28 km/h
The magnitude is:
|v − u| = √((-1.76 km/h)² + (37.28 km/h)²)
|v − u| = 37.3 km/h
The direction is:
θ = tan⁻¹(37.28 / -1.76)
θ = 92.70°
θ = N 2.70° E
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the floor. A man weighing 700 N could climb up to 7.0 m before slipping. What is the coefficient of static friction between the floor and the ladder
Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37
WHO WANTS BRAINLIEST THEN ANSWER THIS QUESTION
look at my previous last question they relate
so
the car slows down to 50 mph
stae the new speed of the car relative to the lorry
if the velocity of the car reduces from 70km/h to 50km/h then the speed of the car will be equal to the speed of the lorry...
thus the relative velocity will be 0
How to give the spheres exactly equal but opposite charges?
A. Make sure spheres are not in contact with each other. Rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Bring the rod near, but not touching, one of the spheres. Do the same with piece of silk and another sphere. Remove the galss rod and the piece of silk.
B. Make sure spheres are not in contact with each other. Rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Touch one of the spheres with the glass rod then touch another one with the piece of silk.
C. Put the two metal spheres in contact with each other. Then rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Touch one of the spheres with the glass rod then separate the spheres. Touch another sphere with the piece of silk.
D. Put the two metal spheres in contact with each other. Then rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Bring the rod near, but not touching, one of the spheres. Separate the spheres and remove the rod.
Answer:
B) Make sure spheres are not in contact with each other. Rub the glass rod on the silk to charge the rod positively (hence to charge piece of silk negatively). Touch one of the spheres with the glass rod then touch another one with the piece of silk.
Explanation:
In static electricity, frictional rubbing can excite electrons, and make them jump from one material to another. Rubbing the glass rod and the silk together makes electron flow from the glass rod to the silk piece. On separation, the rod and silk will be seen to have equal amount of charge but, the glass rod will be positively charged from losing electron, and the silk piece will be negatively charged from gaining extra electrons. When they are each made to touch the metal spheres that are not in contact, the spheres will acquire equal amount of charges opposite the charge on the material that was used to charge them.
Answer:
Friction
Explanation:
During the deceleration of an ascending elevator, the normal force on the feet of a passenger is _____ her weight. During the deceleration of a descending elevator, the normal force on the feet of a passenger is _____ her weight.
Answer: Smaller than ; larger than
Explanation:
When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of
N= mg +ma, (g is gravity and a is acceleration)
here ma is negative so the N= mg-ma
Hence, it feels smaller than its original weight.
When the elevator is moving downward , then the force acting will be positive in nature
N= mg+ma,
here ma will be positive so it feels larger the original weight of passenger.
Nitrogen at an initial state of 300 K, 150 kPa, and 0.45 m3 is compressed slowly in an isothermal process to a final pressure of 800 kPa. Determine the work done during this process
Answer:
W = -113.0 kJ.
Explanation:
The work done during the isothermal process is the following:
[tex]W_{a-b} = p_{a}V_{a}ln(\frac{p_{a}}{p_{b}})[/tex]
Where:
[tex]p_{a}[/tex] = 150 kPa
[tex]V_{a}[/tex] = 0.45 m³
[tex]p_{b}[/tex] = 800 kPa
[tex]W_{a-b} = 150 kPa*0.45 m^{3}ln(\frac{150 kPa}{800 kPa}) = -113.0 \cdot 10^{3} J[/tex]
Therefore, the work done during this process is -113.0 kJ.
I hope it helps you!
A disk-shaped merry-go-round of radius 2.93 m and mass 135 kg rotates freely with an angular speed of 0.621 rev/s . A 62.4 kg person running tangential to the rim of the merry-go-round at 3.21 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim.What is the final angular speed of the merry-go-round? rad/s
Answer:
2.55 rad/sec
Explanation:
Use conservation of angular momentum
The merry go round has
I = ½mr²
I = ½ * 135 * 2.93²
I = 579 kgm²
the person has
I = mr²
I = 62.4 * 2.93²
I = 536 kgm²
Converting 0.621 rev/sec to rad/sec we have 0.621 * 2π
0.621 * 2 * 3.14 rad/sec
3.9 rad/sec
for the person v/r = w
w = 3.21 / 2.93
w = 1.10 rad/sec
So
579 * 3.9 + 536 * 1.1 = (579+536)*w
2258 + 589.6 = 1115w
2847.6 = 1115w
solve for w
w = 2847.6 / 1115
w = 2.55 rad/sec
Thus, the final angular speed of the merry go round is 2.55 rad/sec
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:
The correct answer will be "[tex]-\frac{6C_{6}}{x^{7}}[/tex]". The further explanation is given below.
Explanation:
The potential energy will be,
⇒ [tex]U(x)= -\frac{C_{6}}{x^6}[/tex]
The expression of force will be,
⇒ [tex]F=-\frac{dU(x)}{dx}[/tex]
⇒ [tex]=-(C_{6}(-6)x^{-7})[/tex]
⇒ [tex]=-\frac{6C_{6}}{x^{7}}[/tex]
Force seems to be appealing because the expression has been negative. It therefore means that the force or substance is acting laterally in on itself.
An electron is traveling at 2.0E5 m/s parallel to a uniform electric field of 9.11e-3 N/C strength. The electron is traveling to the right. The electric field lines point to the right. What is the speed of the electron (in m/s) after traveling 1.8 meters to the right?
Answer:
v = 1.85*10^5 m/s
Explanation:
In order to calculate the speed of the electron after it has traveled 1.8m, you first take into account that the electric field generates a desceleration on the electron, because the direction of the electron and electric field are the same.
You use the Newton second law, to calculate the deceleration of the electron:
[tex]F_e=qE=ma[/tex] (1)
q: charge of the electron = 1.6*10^-19C
m: mass of the electron = 9.1*10^-31kg
E: magnitude of the electric field = 9.11*10^-3N/C
a: deceleration = ?
You solve the equation (1) for a, and replace the values of the other parameters:
[tex]a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(9.11*10^{-3}N/C)}{9.1*10^{-31}kg}\\\\a=1.6*10^9\frac{m}{s^2}[/tex]
Next, you use the following formula to calculate the final speed of the electron:
[tex]v^2=v_o^2-2ax[/tex] (2)
v: final speed of the electron = ?
vo: initial speed of the electron = 2.0*10^5 m/s
x: distance traveled by the electron = 1.8m
You solve the equation (2) for v and replace the values of the other parameters:
[tex]v=\sqrt{v_o^2-2ax}=\sqrt{(2.0*10^5m/s)^2-2(1.6*10^9m/s^2)(1.8m)}\\\\v=1.85*10^5\frac{m}{s}[/tex]
The speed of the electron after it has traveled 1.8m is 1.85*10^5 m/s
An air bubble underwater has the same pressure as that of the surrounding water. As the air bubble rises toward the surface (and its temperature remains constant), the volume of the air bubble.
a) increase or decrease, depending on the rate it rises.
b) decreases.
c) increases.
d) remains constant
Answer:
D Remains constant
Explanation:
A toy rocket, launched from the ground, rises vertically with an acceleration of 20 m/s2 for 6.0 s until its motor stops. Disregarding any air resistance, what maximum height above the ground will the rocket achieve?
Answer:
h = 1094.69m
The maximum height above the ground the rocket will achieve is 1094.69m.
Explanation:
The maximum height h is;
h = height covered during acceleration plus height covered when the motor stops.
h = h1 + h2 .......1
height covered during acceleration h1 can be derived using the equation of motion;
h1 = ut + 0.5at^2
Initial speed u = 0
h1 = 0.5at^2
acceleration a = 20 m/s^2
Time t = 6.0 s
h1 = 0.5×(20 × 6^2)
h1 = 0.5(20×36)
h1 = 360 m
height covered when the motor stops h2 can be derived using equation of motion;
h2 = ut + 0.5at^2 .......2
Where;
a = g = acceleration due to gravity = -9.8 m/s^2
The speed when the motor stops u;
u = at = 20 m/s^2 × 6.0 s = 120 m/s
Time t2 can be derived from;
v = u - gt
v = 0 (at maximum height velocity is zero)
u = gt
t = u/g
t = 120m/s / 9.8m/s^2
t = 12.24 seconds.
Substituting the values into equation 2;
h2 = 120(12.24) - 0.5(9.8×12.24^2)
h2 = 734.69376 m
h2 = 734.69 m
From equation 1;
h = h1 + h2 . substituting the values;
h = 360m + 734.69m
h = 1094.69m
The maximum height above the ground the rocket will achieve is 1094.69m.
The block on this incline weighs 100 kg and is connected by a cable and pulley to a weight of 10 kg. If the coefficient of friction between the block and incline is o.3, the block will:
Answer:
a. 94.54 N
b. 0.356 m/s^2
Explanation:
Given:-
- The mass of the inclined block, M = 100 kg
- The mass of the vertically hanging block, m = 10 kg
- The angle of inclination, θ = 20°
- The coefficient of friction of inclined surface, u = 0.3
Find:-
a) The magnitude of tension in the cable
b) The acceleration of the system
Solution:-
- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.
- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.
- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:
[tex]T - m*g = m*a\\\\[/tex] ... Eq 1
Where,
a: The acceleration of the system
- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.
- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.
- We will employ the static equilibrium of the inclined block in the normal direction and we have:
[tex]N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )[/tex]
- The frictional force ( F ) is proportional to the contact force ( N ) as follows:
[tex]F = u*N\\\\F = u*M*g *cos ( Q )[/tex]
- Now we will apply the Newton's second law of motion parallel to the plane as follows:
[tex]M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\[/tex] .. Eq2
- Add the two equation, Eq 1 and Eq 2:
[tex]M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}[/tex]
- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).
- Using equation 1, we determine the tension ( T ) in the cable as follows:
[tex]T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N[/tex]
A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v = 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
Answer:
The answer is 3.48 seconds
Explanation:
The kinematic equation
y= y0+V0*t+1/2*a*(t*t)
-50=0+(0)t+1/2(-9.8)*(t*t)
t=3.194 seconds
During ribbons ball,
x=x0+ Vt+1/2*a*(t*t)
x= 0+(15)*(3.194)+1/2*(0)* (3.194*3.194)
x= 47.9157m
So, distance (D) = 100-47.9157= 52.084m
52.084m=0+15(t)+1/2*(0)(t*t)
t=52.084/15=3.472286= 3.48seconds
1- A 30 gram bullet travels at 300 m/s. How much kinetic energy does it have?
Answer:
1.35 kJ
Explanation:
KE = ½mv² = ½ × 0.030 kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ
Given:-
Mass (m) of the bullet = 30 gramsVelocity of the bullet (v) = 300 m/sTo Find: Kinetic energy of the bullet.
We know,
Eₖ = ½mv²
where,
Eₖ = Kinetic energy,m = Mass &v = Velocity.thus,
Eₖ = ½(30 g)(300 m/s)²
= (15 g)(90000 m²/s²)
= 1350000 g m²/s²
= 1350 kg m²/s²
= 1350 J
= 1.35 kJ (Ans.)
To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 57.0° above the horizontal. It explodes on the mountainside 36.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?
Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t ² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m
The initial velocity of a car moving along a straight path was 0 meters/second. After 10 seconds, it reached a velocity of 25 meters/second. Calculate the car’s average acceleration during this time period.
Answer:
The answer is 2.5 m/s^2.
Explanation:
Average acceleration is (final velocity - Initial velocity) ÷ time.
a car travels 12 miles due north and then 12 miles due west going from town A to town B. The magnitude of the car's displacement is --- miles
Answer:
The magnitude of the displacement of the car = 16.97 miles (North-West of A)
Explanation:
Attached to this answer is a diagram to give you a visual on what is going on i the question
Let the magnitude of the car's displacement be 'd'
The triangle formed is a right angled triangle, using the Pythagoras theorem:
d² = 12² + 12² (Hyp² = Opp² + Adj²)
d² = 144 +144 = 288
d =√ 288 = 16.97 miles
Therefore the magnitude of the displacement of the car = 16.97 miles (North-West of A)
When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1850 kg car traveling to the right at 1.40 m/s collides with a 1450 kg car going to the left at 1.10 m/s. Measurements show that the heavier car's speed just after the collision was 0.250 m/s in its original direction.
A) What is the speed of the lighter car just after collision?
B) Calculate the change in the combined kinetic energy of the two-car system during this collision.
Answer:
Explanation:
Given that,
Mass of the heavier car m_1 = 1750 kg
Mass of the lighter car m_2 = 1350 kg
The speed of the lighter car just after collision can be represented as follows
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}[/tex]
[tex]v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s[/tex]
b) the change in the combined kinetic energy of the two-car system during this collision
[tex]\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))[/tex]
substitute the value in the equation above
[tex]=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J[/tex]
Hence, the change in combine kinetic energy is -2534.78J
An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 28.9 m/s . It then flies a further distance of 45300 m , and afterwards, its velocity is 47.5 m/s . Find the airplane's acceleration.
Answer:
a = 0.016 m/s^2
the airplane's acceleration is 0.016 m/s^2
Explanation:
Applying the equation of motion;
v^2 = u^2 + 2as ........1
Where;
v = final velocity = 47.5 m/s
u = initial velocity = 28.9 m/s
a = acceleration
s = displacement = 45300m
From equation 1, making a the subject of formula;
v^2 = u^2 + 2as
2as = v^2 - u^2
a = (v^2 - u^2)/2s
Substituting the given values;
a = (47.5^2 - 28.9^2)/(2×45300)
a = 0.015684768211 m/s^2
a = 0.016 m/s^2
the airplane's acceleration is 0.016 m/s^2
A boxcar at a rail yard is set into motion at the top of a hump. The car rolls down quietly and without friction onto a straight, level track where it couples with a flatcar of smaller mass, originally at rest, so that the two cars then roll together without friction. Consider the two cars as a system from the moment of release of the boxcar until both are rolling together.
(a) is the mechanical energy of the system conserved?
(b) is the momentum of this system conserved?
2) If the centripetal force on an object in uniform circular motion is increased, what is the effect on (a) the frequency of rotation f (with r constant) and (b) f and r when both are free to vary
Answer:
a) Frequency will increase
b)The f and r will vary similarly
Explanation:
from equation F=mω²r
ω=2πf
⇒Force, F is directly proportional to f and r
Centripetal force of an object F = [tex]mv^2/r[/tex] or [tex]mrW^2[/tex]
What is Centripetal force?A centripetal force is defined as the force that causes a body to follow a curved path, their direction always being orthogonal to the motion of the body and towards a fixed point of instantaneous center of curvature of the path.
This word comes from the Latin words centrum for "center" and petere, meaning "to seek" which may be considered the center-seeking force. The component of this force that is perpendicular to the velocity is the part that results in the centripetal force.
For above given information,
W = 2pif
so
If Force increases, frequency also increases and when both f and r are free to vary, frequency also do vary directly to both of them
Thus, Centripetal force of an object F = [tex]mv^2/r[/tex] or [tex]mrW^2[/tex]
Learn more about Centripetal force, here:
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