For the instant represented, car A has an acceleration in the direction of its motion, and car B has a speed of 45 mi/hr which is increasing. If the acceleration of B as observed from A is zero for this instant, determine the magnitude of the acceleration of A and the rate at which the speed of B is changing.
Answers
Answer:
[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]
[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]
Explanation:
Firstly, there is supposed to be a diagram attached in order to complete this question;
I have attached the diagram below in order to solve this question.
From the data given;
The radius of the car R = 600 ft
Velocity of the car B, [tex]V_B = 45 mi / hr[/tex]
We are to determine the magnitude of the acceleration of A and the rate at which the speed of B is changing.
To start with the magnitude of acceleration A;
We all know that
1 mile = 5280 ft and an hour = 3600 seconds
Thus for ; 1 mile/hr ; we have :
5280 ft/ 3600 seconds
= 22/15 ft/sec
However;
for the velocity of the car B = 45 mi/hr; to ft/sec, we have:
= (45 × 22/15) ft/sec
= 66 ft/sec
A free body diagram is attached in the second diagram showing how we resolve the vector form
Now; to determine the magnitude of the acceleration of A; we have:
[tex]^ \to {a_A} = a_A sin 45^0 ^{\to} + a_A cos 45^0 \ j ^{\to} \\ \\ ^\to {a_B} = -(a_t)_B \ i ^ \to + (a_c )_B cos 45 ^0 \ j ^{\to}[/tex]
Where;
[tex](a_c)_B[/tex] = radial acceleration of B
[tex](a_t)_B[/tex] = tangential acceleration of B
From observation in the diagram; The acceleration of B is 0 from A
So;
[tex]a_B ^\to - a_A ^\to = a_{B/A} ^ \to[/tex]
[tex](-(a_t)_B - a_A sin 45^0 ) ^\to i+ ((a_t)_B-a_A \ cos \ 45^0) ^ \to j = 0[/tex]
[tex](a_c)_B = \dfrac{V_B^2}{R}[/tex]
[tex](a_c)_B = \dfrac{(66)^2}{600}[/tex]
[tex](a_c)_B = 7.26 ft/s^2[/tex]
Equating the coefficient of i and j now; we have :
[tex](a_t)_B = -a_A \ sin 45^0 --- (1)\\ \\ (a_c)_B = a_A cos \ 45^0 --- (2)\\ \\[/tex]
From equation (2)
replace [tex](a_c)_B[/tex] with 7.26 ft/s^2; we have
[tex]7.26 \ ft/s^2 = a_A cos \ 45^0 \\ \\ a_A = \dfrac{7.26 \ ft/s^2}{co s \ 45^0}[/tex]
[tex]\mathbf{a_A = 10.267 \ ft/s^2}[/tex]
Similarly;
From equation (1)
[tex](a_t)_B = -a_A \ sin 45^0[/tex]
replace [tex]a_A[/tex] with 10.267 ft/s^2
[tex](a_t)_B = -10.267 \ ft/s^2 * \ sin 45^0[/tex]
[tex]\mathbf{V_B = (a_t)_B =-7.26 \ ft/s^2}[/tex]
Related Questions
Difference between regular and irregular object.
Answers
Hope this helps....
Good luck on your assignment.....
Calculate the amount of kinetic energy the car stores if it has a mass of 1200 kg and speed of 15 m/s
Answers
Answer:
KE = 135,000 j or 135 KJ
Explanation:
KE=0.5mv^2
KE=0.5*1200*15^2
KE = 135,000 joules or 135 Kilo Joules
a cannon is fired with an initial horizontal velocity of 20m/s and an initial velocity of 25m/s. After 3s in the air, the cannon hits its target. How far away(in meters) was the cannon from its target
Answers
Answer:
60 m
Explanation:
After 3 seconds of travel at 20 m/s, the projectile is 3·20 = 60 meters horizontally from the cannon.
__
The vertical height after 3 seconds is 0.9 m, so the straight-line distance from cannon to target is √(60^2 +0.9^2) ≈ 60.007 meters.
A cart of mass 350 g is placed on a frictionless horizontal air track. A spring having a spring constant of 7.5 N/m is attached between the cart and the left end of the track. The cart is displaced 3.8 cm from its equilibrium position. (a) Find the period at which it oscillates. s (b) Find its maximum speed. m/s (c) Find its speed when it is located 2.0 cm from its equilibrium position.
Answers
Answer:
(a) T = 1.35 s
(b) vmax = 0.17 m/s
(c) v = 0.056 m/s
Explanation:
(a) In order to calculate the period of oscillation you use the following formula for the period in a simple harmonic motion:
[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex] (1)
m: mass of the cart = 350 g = 0.350kg
k: spring constant = 7.5 N/m
[tex]T=2\pi \sqrt{\frac{0.350kg}{7.5N/m}}=1.35s[/tex]
The period of oscillation of the car is 1.35s
(b) The maximum speed of the car is given by the following formula:
[tex]v_{max}=\omega A[/tex] (2)
w: angular frequency
A: amplitude of the motion = 3.8 cm = 0.038m
You calculate the angular frequency:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{1.35s}=4.65\frac{rad}{s}[/tex]
Then, you use the result of w in the equation (2):
[tex]v_{max}=(4.65rad/s)(0.038m)=0.17\frac{m}{s}[/tex]
The maximum speed if 0.17m/s
(c) To find the speed when the car is at x=2.0cm you first calculate the time t by using the following formula:
[tex]x=Acos(\omega t)\\\\t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\t=\frac{1}{4.65rad/s}cos^{-1}(\frac{0.02}{0.038})=0.069s[/tex]
The speed is the value of the following function for t = 0.069s
[tex]|v|=|\omega A sin(\omega t)|\\\\|v|=(4.65rad/s)(0.038m)sin(4.65rad/s (0.069s))=0.056\frac{m}{s}[/tex]
The speed of the car is 0.056m/s
A shell (a large bullet) is shot with an initial speed of 20 m/s, 60 degrees above the horizontal. At the top of the trajectory, the bullet explodes into two fragments of equal mass. One fragment has a speed of zero just after the explosion and simply drops straight down. How far from the gun does the other fragment land, assuming that the ground is level and that the air drag is negligible.
Answers
Answer:
17.656 m
Explanation:
Initial speed u = 20 m/s
angle of projection α = 60°
at the top of the trajectory, one fragment has a speed of zero and drops to the ground.
we should note that the top of the trajectory will coincide with halfway the horizontal range of the the projectile travel. This is because the projectile follows an upward arc up till it reaches its maximum height from the ground, before descending down by following a similar arc downwards.
To find the range of the projectile, we use the equation
R = [tex]\frac{u^{2}sin2\alpha }{g}[/tex]
where g = acceleration due to gravity = 9.81 m/s^2
Sin 2α = 2 x (sin α) x (cos α)
when α = 60°,
Sin 2α = 2 x sin 60° x cos 60° = 2 x 0.866 x 0.5
Sin 2α = 0.866
therefore,
R = [tex]\frac{20^{2}*0.866 }{9.81}[/tex] = 35.31 m
since the other fraction with zero velocity drops a top of trajectory, distance between the two fragments assuming level ground and zero air drag, will be 35.31/2 = 17.656 m
A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 60.0 cm, and the density of iron is 7.87 g∕c m cubed . Find the inner diameter in cm. Express to 3 sig figs.
Answers
Answer:
The inner diameter is 57.3 cm
Explanation:
The inner diameter of the hollow spherical iron shell can be found using the weight of the sphere ([tex]W_{s}[/tex]) and the weight of the water displaced ([tex]W_{w}[/tex]):
[tex] W_{s} = W_{w} [/tex]
[tex] m_{s}*g = m_{w}*g [/tex]
[tex] D_{s}*V_{s} = D_{w}*V_{w} [/tex]
Where D is the density and V is the volume
[tex] D_{s}*\frac{4}{3}\pi*(\frac{d_{o}^{3} - d_{i}^{3}}{2^{3}}) = \frac{4}{3}\pi*(\frac{d_{o}}{2})^{3} [/tex]
Where [tex]d_{o}[/tex] is the outer diameter and [tex]d_{i}[/tex] is the inner diameter
[tex] D_{s}*(d_{o}^{3} - d_{i}^{3}) = d_{o}^{3} [/tex]
[tex] D_{s}*d_{i}^{3} = d_{o}^{3}(D_{s} - 1) [/tex]
[tex] 7.87*d_{i}^{3} = 60.0^{3}(7.87 - 1) [/tex]
[tex] d_{i} = 57.3 cm [/tex]
Therefore, the inner diameter is 57.3 cm.
I hope it helps you!
When using a mercury barometer , the vapor pressure of mercury is usually assumed to be zero. At room temperature mercury's vapor pressure is about 0.0015 mm-Hg. At sea level, the height hhh of mercury in a barometer is about 760 mm.Required:a. If the vapor pressure of mercury is neglected, is the true atmospheric pressure greater or less than the value read from the barometer? b. What is the percent error? c. What is the percent error if you use a water barometer and ignore water's saturated vapor pressure at STP?
Answers
Answer:
Explanation:
(a)
The true atmospheric pressure will has more value than the reading in the barometer. If Parm is the atmospheric
pressure in the tube then the resulting vapour pressure is
Patm - pgh = Prapor
The final reading ion the barometer is
pgh = Palm - Proper
Hence, the true atmospheric pressure is greater.
you can find the answer in this book
physics principles with Applications, Global Edition Problem 67P: Chapter: CH 13 Problem:67p
A projectile is launched in the horizontal direction. It travels 2.050 m horizontally while it falls 0.450 m vertically, and it then strikes the floor. How long is the projectile in the air
Answers
Answer:
0.303s
Explanation:
horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m
Using equation of linear motion
Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative
-0.45 = 0 - 0.5 × 9.81×t²
0.45 / (0.5 × 9.81) = t²
t = √0.0917 = 0.303 s
What is the main difference between work power and energy
Answers
Answer:
Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.
Energy can also be defined as the ability to do work.
A large crate of mass m is place on the flatbed of a truck but not tied down. As the truck accelerates forward with acceleration a, the crate remains at rest relative to the truck. What force causes the crate to accelerate?
Answers
Answer:
Friction
Explanation:
There are tiny bumps and grooves on every object, which make them rough and more difficult to rub against each other. Even though the crate remains at rest at first, the frictional force causes it to stay in place and accelerate with the truck. Hope this helps!
A butcher grinds 5 and 3/4 lb of meat then sells it for 2 and 2/3 pounds to the customer what is the maximum amount me that the butcher can sell to the next customer
Answers
Answer:
The maximum amount of meat that the butcher can sell is [tex]3\frac{1}{12}\:lb[/tex]
Explanation:
The maximum amount can be found by taking the difference of mixed numbers.
[tex]5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\[/tex]
Best Regards!
what is the most likely elevation of point x?
A. 150 ft
B. 200 ft
C. 125 ft
D. 250 ft
Answers
Because, sea level is 0 and the elevation gets higher the closer you get towards the center. The x was closest to the 100.
explanation: sea level is at 0 and the elevation gets higher the closer you get towards the center. X was the closest one to 100
Strontium decays by beta decay part of the nuclear equation is shown below fill in the blank with a number? 90/38Sr -> 0/-1e 90/blankY
Answers
Answer : The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:
[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]
Explanation :
Beta decay : It is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.
The released beta particle is also known as electron.
The beta decay reaction is:
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]
The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:
[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]
Answer:
the blank is 39
Explanation: a p e x
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.
Answers
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 skydiver stepped out of an airplane at an altitude of 1000m fell freely for 5.00s opened her parachute and slowed to 7.00m/s in a negligible time what was the total elapsed time from leaving the airplane to landing on the ground
Answers
Answer:
t = 17.68s
Explanation:
In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:
[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex] (1)
y: height for a time t
yo: initial height = 1000m
vo: initial velocity = 0m/s
g: gravitational acceleration = 9.8m/s^2
t: time = 5.00 s
You replace the values of the parameters to get the values of the new height of the skydiver:
[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]
Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.
You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0
[tex]0=877.5-7.00t-4.9t^2[/tex] (2)
The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:
[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]
You use the positive value of t1 because it has physical meaning.
Finally, you sum the times of both parts of the trajectory:
total time = 5.00s + 12.68s = 17.68s
The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s
The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s) of the orbit and shaded segments are the same?
Answers
Answer: not sure
Explanation:
A 2-kg block is released from rest at the top of a 20-mlong frictionless ramp that is 4 m high. At the same time, an identical block is released next to the ramp so that it drops straight down the same 4 m. What are the values for each of the following for the blocks just before they reach ground level.
Required:
a. Gravitational potential energy Block a_____ J Block b _____ J
b. Kinetic energy Block a _____ J Block b _____
c. Speed Block a _____ J Block b _____ J
d. Momentum Block a _____ J Block b _____ J
Answers
Answer:
A.) 78.4 J for both
B.) 78.4 J for both
C.) 8.85 m/s for both
D.) 17.7 kgm/s
Explanation:
Given information:
Mass m = 2 kg
Distance d = 20 m
High h = 4 m
A.) Gravitational potential energy can be calculated by using the formula
P.E = mgh
P.E = 2 × 9.8 × 4
P.E = 78.4 J
Since the two objects are identical, the gravitational potential energy of the block for both a and b will be 78.4 J
B.) According to conservative energy,
Maximum P.E = Maximum K.E.
Therefore, the kinetic energy of the two blocks will be 78.4 J
C.) Since K.E = 1/2mv^2 = mgh
V = √(2gh)
Solve for velocity V by substituting g and h into the formula
V = √(2 × 9.8 × 4)
V = √78.4
V = 8.85 m/s
The velocities of both block will be 8.85 m/s
D.) Momentum is the product of mass and velocity. That is,
Momentum = MV
Substitute for m and V into the formula
Momentum = 2 × 8.85 = 17.7 kgm/s
Both block will have the same value since the ramp Is frictionless.
In the 25 ftft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 W/m^2 at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.)
Required:
Find the average radiation pressure (in pascals and in atmospheres) on
a. A totally absorbing section of the floor.
b. A totally reflecting section of the floor.
c. Find the average momentum density (momentum per unit volume) in the light at the floor.
Answers
Answer:
a) 8.33 x [tex]10^{-6}[/tex] Pa or 8.22 x [tex]10^{-11}[/tex] atm
b) 1.66 x [tex]10^{-5}[/tex] Pa or 1.63 x [tex]10^{-10}[/tex] atm
c) 2.77 x [tex]10^{-14}[/tex] kg/m^2-s
Explanation:
Intensity of light = 2500 W/m^2
area = 25 ft^2
a) average radiation pressure on a totally absorbing section of the floor[tex]Pav = \frac{I}{c}[/tex]
where I is the intensity of the light
c is the speed of light = [tex]3*10^{8} m/s[/tex]
[tex]Pav = \frac{2500}{3*10^{8} }[/tex] = 8.33 x [tex]10^{-6}[/tex] Pa
1 pa = [tex]9.87*10^{-6}[/tex]
8.33 x [tex]10^{-6}[/tex] Pa = 8.22 x [tex]10^{-11}[/tex] atm
b) average radiation for a totally radiating section of the floor
[tex]Pav = \frac{2I}{c}[/tex]
this means that the pressure for a totally radiating section is twice the average pressure of the totally absorbing section
therefore,
Pav = 2 x 8.33 x [tex]10^{-6}[/tex] = 1.66 x [tex]10^{-5}[/tex] Pa
or
Pav in atm = 2 x 8.22 x [tex]10^{-11}[/tex] = 1.63 x [tex]10^{-10}[/tex] atm
c) average momentum per unit volume is
[tex]m = \frac{I}{c^{2} }[/tex]
[tex]m = \frac{2500}{(3*10^{8}) ^{2} }[/tex] = 2.77 x [tex]10^{-14}[/tex] kg/m^2-s
Giving quadrilateral a(2,-1 ) b ( 1,3) c(6,5) d(7,1) you want to prove that it is a parallelogram by showing opposite sides are congruent . what formula would you use ? show that sb is congruent to cd
Answers
Answer:
AB = CD = √17
Explanation:
The distance formula is used to find the length of a line segment between two points. Here, we want to show the distance AB is the same as the distance CD.
d = √((x1 -x1)² +(y2 -y1)²)
__
AB: d = √((1 -2)² +(3 -(-1))²) = √((-1)² +4²) = √17
CD: d = √((7-6)² +(1-5)²) = √(1² +(-4)²) = √17 . . . . same as AB
Segment AB is congruent to segment CD.
what happen to the volume of liquid displaced
when the density of liquid is changed
explain ?
Answers
Answer:
Density depends on the temperature and the gap between particles of the liquid. In most of cases temperature is inversely proportional to density means if the temperature increases then the density decreases and the space between particles of that liquid is also inversely proportional to the density means if the intraparticle space increases then the density decreases.
A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two fragments, each 1.0 kg, move in directions that make 60o angle above and below the positive x-axis and their speeds are 60 m/s each. What is the velocity of the 4.0-kg fragment
Answers
Answer:
-15 m/s
Explanation:
The computation of the velocity of the 4.0 kg fragment is shown below:
For this question, we use the correlation of the momentum along with horizontal x axis
Given that
Weight of stationary shell = 6 kg
Other two fragments each = 1.0 kg
Angle = 60
Speed = 60 m/s
Based on the above information, the velocity = v is
[tex]1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0[/tex]
[tex]\frac{60}{2} + \frac{60}{2} - 4\ v = 0[/tex]
[tex]v = \frac{60}{4}[/tex]
= -15 m/s
Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the ball, causing it to swing around the pole. What is the total initial acceleration of a tether ball on a 2.0 m rope whose angular velocity changes from 13 rad/s to 7.0 rad/s in 15 s
Answers
Answer:
a_total = 14.022 m/s²
Explanation:
The total acceleration of a uniform circular motion is given by the following formula:
[tex]a=\sqrt{a_c^2+a_T^2}[/tex] (1)
ac: centripetal acceleration
aT: tangential acceleration
Then, you first calculate the centripetal acceleration by using the following formula:
[tex]a_c=r\omega^2[/tex]
r: radius of the circular trajectory = 2.0m
w: final angular velocity of the ball = 7.0 rad/s
[tex]a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}[/tex]
Next, you calculate the tangential acceleration. aT is calculate by using:
[tex]a_T=r\alpha[/tex] (2)
α: angular acceleration
The angular acceleration is:
[tex]\alpha=\frac{\omega_o-\omega}{t}[/tex]
wo: initial angular velocity = 13 rad/s
t: time = 15 s
Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:
[tex]a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}[/tex]
Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:
[tex]a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}[/tex]
The total acceleration of the ball is 14.022 m/s²
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands
Answers
Complete question is;
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.
Answer:
Tangential speed of foot just before it lands is; v = 5.37m/s
Explanation:
Let U (potential energy) be zero on the ground.
So, initially, U = mgh
where, h = 0.98/2 = 0.49m (midpoint of the leg)
Now just before the leg hits the floor it would have kinetic energy as;
K = ½Iω²
where ω = v/r and I = ⅓mr²
So, K = ½(⅓mr²)(v/r)²
K = (1/6) × (mr²)/(v²/r²)
K = (1/6) × mv²
From principle of conservation of energy, we have;
Potential energy = Kinetic energy
Thus;
mgh = (1/6) × mv²
m will cancel out to give;
gh = (1/6)v²
Making v the subject, we have;
v = √6gh
v = √(6 × 9.81 × 0.49)
v = √28.8414
v = 5.37m/s
Imagine you are in a small boat on a small pond that has no inflow or outflow. If you take an anchor that was sitting on the floor of the boat and lower it over the side until it sits on the ground at the bottom of the pond, will the water level rise slightly, stay the same, or lower slightly?Two students, Ian and Owen, are discussing this. Ian says that the anchor will still displace just as much water when it is sitting on the bottom of the pond as it does when it is in the boat. After all, adding the anchor to the boat causes the water level in the lake to rise, and so would immersing the anchor in the pond. So Ian reasons that both displacements would be equal, and the lake level remains unchanged.
Answers
Answer;
The pond's water level will fall.
Explanation;
Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.
When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.
Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.
When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.
Hope this Helps!!!
Buoyancy is an upward force exerted by a fluid on a body partially or completely immersed in it
The pond water level will lower slightly
According to Archimedes principle, the up thrust on the boat by the water is given by the volume of the water displaced
When a boat floats, the weight of the boat and all its contents and passengers is equal to the displaced water, so that larger boats with more wider opening can displace more water and therefore, carry more loadWith regards to lowering the anchor from the boat into the pond, the weight of the anchor is no longer carried by the boat but by the bottom of the pond, therefore, the weight of the boat reduces, and the boat rises, while the volume initially occupied by the boat is taken up by the water available, therefore, the water level lowers slightly
Learn more here;
https://brainly.com/question/24529607
At least how many Calories does a mountain climber need in order to climb from sea level to the top of a 5.42 km tall peak assuming the muscles of the climber can convert chemical energy to mechanical energy with an efficiency of 16.0 percent. The total mass of the climber and the equipment is 78.4 kg. (Enter your answer as a number without units.)
Answers
Answer:
Ec = 6220.56 kcal
Explanation:
In order to calculate the amount of Calories needed by the climber, you first have to calculate the work done by the climber against the gravitational force.
You use the following formula:
[tex]W_c=Mgh[/tex] (1)
Wc: work done by the climber
g: gravitational constant = 9.8 m/s^2
M: mass of the climber = 78.4 kg
h: height reached by the climber = 5.42km = 5420 m
You replace in the equation (1):
[tex]W_c=(78.4kg)(9.8m/s^2)(5420m)=4,164,294.4\ J[/tex] (2)
Next, you use the fact that only 16.0% of the chemical energy is convert to mechanical energy. The energy calculated in the equation (2) is equivalent to the mechanical energy of the climber. Then, you have the following relation for the Calories needed:
[tex]0.16(E_c)=4,164,294.4J[/tex]
Ec: Calories
You solve for Ec and convert the result to Cal:
[tex]E_c=\frac{4,164,294.4}{016}=26,026,840J*\frac{1kcal}{4184J}\\\\E_c=6220.56\ kcal[/tex]
The amount of Calories needed by the climber was 6220.56 kcal
A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 50.2 s. The distance between two successive crests is 30.2 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed
Answers
Explanation:
(a) The period of a wave is the time required for one complete cycle. In this case, we have the time of five cycles. So:
[tex]T=\frac{t}{n}\\\\T=\frac{50.2s}{5}\\T=10.04s[/tex]
(b) The frequency of a wave is inversely proportional to its period:
[tex]f=\frac{1}{T}\\f=\frac{1}{10.04s}\\f=0.01Hz[/tex]
(c) The wavelength is the distance between two successive crests, so:
[tex]\lambda=30.2m[/tex]
(d) The speed of a wave is defined as:
[tex]v=f\lambda\\v=(0.1Hz)(30.2m)\\v=3.02\frac{m}{s}[/tex]
The temperature coefficient of resistivity for the metal gold is 0.0034 (C )1, and for tungsten it is 0.0045 (C )1. The resistance of a gold wire increases by 7.0% due to an increase in temperature. For the same increase in temperature, what is the percentage increase in the resistance of a tungsten wire
Answers
Answer:
% increase in resistance of tungsten = 9.27%
Explanation:
We are given:
Co-efficient of resistivity for the metal gold; α_g = 0.0034 /°C
Co-efficient of resistivity for tungsten;α_t = 0.0045 /°C
% Resistance change of gold wire due to temperature change = 7%
Now, let R1 and R2 be the resistance before and after the temperature change respectively.
Thus;
(R2 - R1)/R1) x 100 = 7
So,
(R2 - R1) = 0.07R1
R2 = R1 + 0.07R1
R2 = 1.07R1
The equation to get the change in temperature is given as;
R2 = R1(1 + αΔt)
So, for gold,
1.07R1 = R1(1 + 0.0034*Δt)
R1 will cancel out to give;
1.07 = 1 + 0.0034Δt
(1.07 - 1)/0.0034 = Δt
Δt = 20.59°C
For this same temperature for tungsten, let Rt1 and Rt2 be the resistance before and after the temperature change respectively and we have;
Rt2 = Rt1(1 + α_t*Δt)
So, Rt2/Rt1 = 1 + 0.0045*20.59
Rt2/Rt1 = 1.0927
From earlier, we saw that;
(R2 - R1)/R1) x 100 = change in resistance
Similarly,
(Rt2 - Rt1)/Rt1) x 100 = change in resistance
Simplifying it, we have;
[(Rt2/Rt1) - 1] × 100 = %change in resistance
Plugging in the value of 1.0927 for Rt2/Rt1, we have;
(1.0927 - 1) × 100 = %change in resistance
%change in resistance = 9.27%
At rest, a car's horn sounds at a frequency of 365 Hz. The horn is sounded while the car is moving down the street. A bicyclist moving in the same direction with one-third the car's speed hears a frequency of 357 Hz. What is the speed of the car?
Answers
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
What is the particle arrangement in a liquid
Answers
Answer:
the particle arrangement in liquid are close together with no regular arrangement
Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous value. If the speed of the water in the larger section of the pipe was what is its speed in this smaller section if the water behaves like an ideal incompressible fluid
Answers
Answer:
Explanation:
The speed of the water in the large section of the pipe is not stated
so i will assume 36m/s
(if its not the said speed, input the figure of your speed and you get it right)
Continuity equation is applicable for ideal, incompressible liquids
Q the flux of water that is Av with A the cross section area and v the velocity,
so,
[tex]A_1V_1=A_2V_2[/tex]
[tex]A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}[/tex]
the diameter decreases 86% so
[tex]d_2 = 0.86d_1[/tex]
[tex]v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}[/tex]
Thus, speed in smaller section is 48.6 m/s