Answer:
the mass drop by 6.5cm before coming to rest.
Explanation:
Given that:
the mass of the block M= 1.40 kg
angle of inclination θ = 30°
spring constant K = 40.0 N/m
mass of the suspended block m = 60.0 g = 0.06 kg
initial downward speed = 1.40 m/s
The objective is to determine how far does it drop before coming to rest?
Let assume it drops at y from a certain point in the vertical direction;
Then :
Workdone by gravity on the mass of the block is:
[tex]w_1g = 1.4*9.81*y*sin30[/tex] = - 6.867y
Workdone by gravity on the mass of the suspended block is:
[tex]w_2g = 0.06*9.81[/tex] = 0.5886
The workdone by the spring = -1/2ky²
= -0.5 × 40y²
= -20 y²
The net workdone is = -20 y² - 6.867y + 0.5886y
According to the work energy theorem
Net work done = Δ K.E
-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²
-20 y² - 6.867y + 0.5886 = 0.5 × 0.1176
-20 y² - 6.867y + 0.5886 = 0.0588
-20 y² - 6.867y + 0.5886 - 0.0588
-20 y² - 6.867y + 0.5298 = 0
multiplying through by (-)
20 y² + 6.867y - 0.5298 = 0
Using the quadratic formula:
[tex]\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
where;
a = 20 ; b = 6.867 c= - 0.5298
[tex]\dfrac{-(6.867) \pm \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}[/tex]
[tex]= \dfrac{-(6.867) + \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)} \ \ \ OR \ \ \ \dfrac{-(6.867) - \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}[/tex]= 0.0649 OR −0.408
We go by the positive integer
y = 0.0649 m
y = 6.5 cm
Therefore; the mass drop by 6.5cm before coming to rest.
What is the main difference between work power and energy
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 0.13 kg ball is moving at 6.6 m/s when it is hit by a bat, causing it to reverse direction and having a speed of 10.3 m/s, What is the change in the magnitude of the momentum of the ball
Answer:
Change in momentum = 2.197 kgm/s
Explanation:
Momentum = MV
Initial momentum = MU
Final momentum = MV
Computation:
⇒ Change in momentum = MV - MU
⇒ Change in momentum = M (V - U)
⇒ Change in momentum = 0.13(-10.3 - 6.6)
⇒ Change in momentum = 0.13(16.9)
⇒ Change in momentum = 2.197 kgm/s
Jack and Jill went up the hill to fetch a pail of water. Jack, who’s mass is 75 kg, 1.5 times heavier than Jill’s mass, fell down and broke his crown after climbing a 15 m high hill. Jillcame tumbling after covering the same distance as Jack in 1/3rd of the time.Required:a. Who did the most work climbing up the hill? b. Who applied the most power?
Answer:
a) Jack does more work uphill
b) Numerically, we can see that Jill applied the most power downhill
Explanation:
Jack's mass = 75 kg
Jill's mass = [tex]1.5x = 75[/tex]
Jill's mass = [tex]x = \frac{75}{1.5}[/tex] = 50 kg
distance up hill = 15 m
a) work done by Jack uphill = mgh
where g = acceleration due to gravity= 9.81 m/s^2
work = 75 x 9.81 x 15 = 11036.25 J
similarly,
Jill's work uphill = 50 x 9.81 x 15 = 7357.5 J
this shows that Jack does more work climbing up the hill
b) assuming Jack's time downhill to be t,
then Jill's time = [tex]\frac{t}{3}[/tex]
we recall that power is the rate in which work id done, i.e
P = [tex]\frac{work}{time}[/tex]
For Jack, power = [tex]\frac{11036.25}{t}[/tex]
For Jill, power = [tex]\frac{3*7357.5}{t}[/tex] = [tex]\frac{22072.5}{t}[/tex]
Numerically, we can see that Jill applied the most power downhill
g A small car travels up the hill with a speed of v = 0.2 s (m/s) where s is the distance measured from point A in meters. Determine the magnitude of the car’s acceleration when it reaches s = 50 (m) where the road’s radius of curvature is r
Answer:
The magnitude of the acceleration of the car when [tex]s = 50\,m[/tex] is [tex]2\,\frac{m}{s^{2}}[/tex].
Explanation:
The acceleration can be obtained by using the following differential equation:
[tex]a = v \cdot \frac{dv}{ds}[/tex]
Where [tex]a[/tex], [tex]v[/tex] and [tex]s[/tex] are the acceleration, speed and distance masured in meters per square second, meters per second and meters, respectively.
Given that [tex]v = 0.2\cdot s[/tex], its first derivative is:
[tex]\frac{dv}{ds} = 0.2[/tex]
The following expression is obtained by replacing each term:
[tex]a = 0.2\cdot 0.2\cdot s[/tex]
[tex]a = 0.04\cdot s[/tex]
The magnitude of the acceleration of the car when [tex]s = 50\,m[/tex] is:
[tex]a = 0.04\cdot (50\,m)[/tex]
[tex]a = 2\,\frac{m}{s^{2}}[/tex]
how much weight can a man lift in the jupiter if he can lift 100kg on the earth.calculate
Answer:
2479 NewtonSolution,
Mass=100 kg
Acceleration due to gravity(g)=24.79 m/s^2
Now,.
[tex]weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton[/tex]
hope this helps ..
Good luck on your assignment..
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.
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]
A 0.40-kg particle moves under the influence of a single conservative force. At point A, where the particle has a speed of 10 m/s, the potential energy associated with the conservative force is 40 J. As the particle moves from A to B, the force does 25 J of work on the particle. What is the value of the potential energy at point B
Answer:
The value of the potential energy of the particle at point B is 85 joules.
Explanation:
According to the Principle of Energy Conservation, the energy cannot be created nor destroyed, only transformed. The particle at point A has kinetic and potential energy and receives a work due to an external conservative force (Work-Energy Theorem), whose sum is equal to potential energy at point B. Mathematically speaking, the expression that describes the phenomenon is:
[tex]K_{A} + U_{A} + W_{A \rightarrow B} = U_{B}[/tex]
Where:
[tex]K_{A}[/tex] - Kinetic energy at point A, measured in joules.
[tex]U_{A}[/tex] - Potential energy at point A, measured in joules.
[tex]W_{A \rightarrow B}[/tex] - Work due to conservative force from A to B, measured in joules.
[tex]U_{B}[/tex] - Potential energy at point B, measured in joules.
The initial kinetic energy of the particle is:
[tex]K_{A} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Velocity, measured in meters per second.
If [tex]m = 0.4\,kg[/tex] and [tex]v = 10\,\frac{m}{s}[/tex], then:
[tex]K_{A} = \frac{1}{2}\cdot (0.4\,kg)\cdot \left(10\,\frac{m}{s} \right)^{2}[/tex]
[tex]K_{A} = 20\,J[/tex]
Finally, the value of the potential energy at point B is:
[tex]U_{B} = 20\,J + 40\,J + 25\,J[/tex]
[tex]U_{B} = 85\,J[/tex]
The value of the potential energy of the particle at point B is 85 joules.
The potential energy of the particle at point B is 85 J.
Given to us:
Mass of the particle, [tex]m=0.40\ kg[/tex]
velocity of the particle, [tex]v= 10\ m/s[/tex]
potential energy of the particle, [tex]PE= 40\ J[/tex]
Workdone from pt. A to B, [tex]WD_{(A\rightarrow B)} = 25\ J[/tex]
Calculating the kinetic energy of the particle,
[tex]\begin{aligned}KE&= \frac{1}{2}mv^2 \\\\&=\frac{1}{2}\times0.40\times (10)^2\\\\&=20 J\\\end{aligned}[/tex]
According to the Principle of Energy Conservation,
The energy cannot be created nor be destroyed, it can only be transformed from one form to another.Therefore,
Also,
Total Energy at point A ,
[tex]\begin{aligned}(TE)_A &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_A+ PE_A+UE_A+ WD_{(0\rightarrow A)}\\&=20+40+0+0\\&=60\ J\end{aligned}[/tex]
Total Energy at point B,
[tex]\begin{aligned}(TE)_B &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\end{aligned}[/tex]
As the total energy is conserved from point A to B and also an external work is done on the particle. we can write the above equation as,
[tex]\begin{aligned} TE_B&=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\\&=(KE_B+ PE_B+UE_B)+ WD_{(A\rightarrow B)}\\&= TE_A+ WD_{(A\rightarrow B)}\\&=60+25\\&=85\ J\end{aligned}[/tex]
Therefore, the total energy for the particle at point B is 85 J but as the particle is not moving neither work is done at point B, the total energy of the particle is the potential energy of the particle.
Hence, the potential energy of the particle at point B is 85 J.
To know more visit:
https://brainly.com/question/122902
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.
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
1. Consider the ball example in the introduction when a ball is dropped from 3 meters. After the ball bounces, it raises to a height of 2 meters. The mass of the ball is 0.5 kg a. Calculate the speed of the ball right before the bounce. b. How much energy was converted into heat after the ball bounced off the ground
Answer:
(a) 7.67 m/s.
(b) 4.9 J
Explanation:
(a) From the law of conservation of energy,
P.E = K.E
mgh = 1/2(mv²)
therefore,
v = √(2gh)....................... Equation 1
Where v = speed of the ball before bounce, g = acceleration due to gravity, h = height from which the ball was dropped.
Given: h = 3 m, g = 9.8 m/s²
Substitute into equation 1
v = √(2×9.8×3)
v = √(58.8)
v = 7.67 m/s.
(b) Energy of the ball before the bounce = mgh = 0.5×9.8×3 = 14.7 J
Energy of the ball after the bounce = mgh' = 0.5(9.8)(2) = 9.8 J
Amount of energy converted to heat = 14.7-9.8 = 4.9 J
A surface is bombarded by particles, each of mass small 'm', which have velocity
normal to the surface. On average, n particles strike unit area of the surface each second
and rebound elastically. What is the pressure on the surface?
A. nmv
B. 2nmv
C. nmv²
D. 1/2nmv²
Answer:
B. 2nmv
Explanation:
Pressure is force over area.
P = F / A
Force is mass times acceleration.
F = ma
Acceleration is change in velocity over change in time.
a = Δv / Δt
Therefore:
F = m Δv / Δt
P = m Δv / (A Δt)
The total mass is nm.
The change in velocity is Δv = v − (-v) = 2v.
A = 1 and Δt = 1.
Plugging in:
P = (nm) (2v) / (1 × 1)
P = 2nmv
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
what happen to the volume of liquid displaced
when the density of liquid is changed
explain ?
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.
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
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
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
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
What is the particle arrangement in a liquid
Answer:
the particle arrangement in liquid are close together with no regular arrangement
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.
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 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?
The figure shows an arrangement of four charged particles, with θ = 20.0° and d1 = 3.00 cm, which is the distance from the origin to a charge q1. Charge q1 is unknown, but q2= +7.00×10‒19 C and q3 = q4 = ‒2.00×10‒19 C. If there is no nett electrostatic force on q1 due to the other charges (the nett electrostatic force on q1 is zero), calculate the distance from the origin to q2, given by d2, in cm. Assume that all forces apart from the electrostatic forces in the system are negligible
Answer:
[tex]d_2=3.16cm[/tex]
Explanation:
So, in order to solve this problem, we must start by building a diagram of the problem itself. (See attached picture) And together with the diagram, we must build a free body diagram, which will include the forces that are being applied on the given charged particle together with their directions.
In this case we only care about the x-direction of the force, since the y-forces cancel each other. So if we do a sum of forces on the x-direction, we get the following:
[tex]\sum{F_{x}}=0[/tex]
so:
[Tex]-F_{12}+F_{13x}+F_{14x}=0[/tex]
Since [tex]F_{13x}=F_{14x}[/tex] we can simplify the equation as:
[tex]-F_{12}+2F_{13x}=0[/tex]
we can now solve this for [tex]F_{12}[/tex] so we get:
[tex]F_{12}=2F_{13x}[/tex]
Now we can substitute with the electrostatic force formula, so we get:
[tex]k_{e}\frac{q_{1}q_{2}}{r_{12}^{2}}=2k_{e}\frac{q_{1}q_{3}}{r_{13}^{2}}cos \theta[/tex]
We can cancel [tex]k_{e}[/tex] and [tex]q_{1}[/tex]
so the simplified equation is:
[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{r_{13}^{2}}cos \theta[/tex]
From the given diagram we know that:
[tex]cos \theta = \frac{d_{1}}{r_{13}}[/tex]
so when solving for [tex]r_{13}[/tex] we get:
[tex]r_{13}=\frac{d_{1}}{cos\theta}[/tex]
and if we square both sides of the equation, we get:
[tex]r_{13}^{2}=\frac{d_{1}^{2}}{cos^{2}\theta}[/tex]
and we can substitute this into our equation:
[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{d_{1}^{2}}cos^{3} \theta[/tex]
so we can now solve this for [tex]r_{12}[/tex] so we get:
[tex]r_{12}=\sqrt{\frac{d_{1}^{2}q_{2}}{2q_{3}cos^{3}\theta}}[/tex]
which can be rewritten as:
[tex]r_{12}=d_{1}\sqrt{\frac{q_{2}}{2q_{3}cos^{3}\theta}}[/tex]
and now we can substitute values.
[tex]r_{12}=(3cm)\sqrt{\frac{7x10^{-19}C}{2(2x10^{-19}C)cos^{3}(20^{o})}}[/tex]
which solves to:
[tex]r_{12}=6.16cm[/tex]
now, we must find [tex]d_{2}[/tex] by using the following equation:
[tex]r_{12}=d_{1}+d_{2}[/tex]
when solving for [tex]d_{2}[/tex] we get:
[tex]d_{2}=r_{12}-d_{1}[/tex]
when substituting we get:
[tex]d_{2}=6.16cm-3cm[/tex]
so:
[tex]d_{2}=3.16cm[/tex]
A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 5,200 rev/min.
Required:
a. Find the kinetic energy stored in the flywheel.
b. If the flywheel is to supply energy to the car as would a 15.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.
Answer:
a
[tex]KE = 7.17 *10^{7} \ J[/tex]
b
[tex]t = 6411.09 \ s[/tex]
Explanation:
From the question we are told that
The radius of the flywheel is [tex]r = 1.50 \ m[/tex]
The mass of the flywheel is [tex]m = 430 \ kg[/tex]
The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]
The power supplied by the motor is [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]
Generally the moment of inertia of the flywheel is mathematically represented as
[tex]I = \frac{1}{2} mr^2[/tex]
substituting values
[tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]
[tex]I = 483.75 \ kgm^2[/tex]
The kinetic energy that is been stored is
[tex]KE = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]
[tex]KE = 7.17 *10^{7} \ J[/tex]
Generally power is mathematically represented as
[tex]P = \frac{KE}{t}[/tex]
=> [tex]t = \frac{KE}{P}[/tex]
substituting the value
[tex]t = \frac{7.17 *10^{7}}{11190}[/tex]
[tex]t = 6411.09 \ s[/tex]
4. A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strikes the ground at a horizontal distance of 160 m from the base of the building 5.0 s after being thrown. Determine the speed with which the rock was thrown.
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s
Answer:
55.42 m/s
Explanation:
Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by
u_x = d/t
Where
d = 160 m is the distance covered
t = 5.0 s is the time taken
Substituting, we get
u_x =160/5 = 32 m/s.
Along the vertical direction, the rock is in free-fall - so its motion is a uniform accelerated motion with constant acceleration g = -9.8 m/s^2 (downward). Therefore, the vertical distance covered is given by the
[tex]S=u_yt+\frac{1}{2}at^2[/tex]
where
S = -100 m is the vertical displacement
u_y is the initial vertical velocity
Replacing t = 5.0 s and solving the equation for u_y, we find
-100 = u_y(5) + (-9.81)(5)^2/2
u_y = 45.25 m/s
Therefore, the speed with which the rock was thrown u
[tex]u= \sqrt{u_x^2+u_y^2} \\=\sqrt{32^2+45.25^2}\\ = 55.42 m/s[/tex]
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?
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
Which three terms are needed to describe the energy a BASE jumper has as
she falls toward the ground?
O A. Potential
B. Electromagnetic
C. Gravitational
D. Kinetic
Answer:
I’m saying kinetic gravitational and electromagnetic and I will comment on this if I got it right
Explanation:.
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
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
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 26.5 m/s (about 59 mph ) around the turn, what is the race car's centripetal (radial) acceleration
Answer:
The centripetal acceleration of the car will be 12.32 m/s² .
Explanation:
Given that
radius ,R= 57 m
Velocity , V=26.5 m/s
We know that centripetal acceleration given as follows
[tex]a_c=\dfrac{V^2}{R}[/tex]
Now by putting the values in the above equation we get
[tex]a_c=\dfrac{26.5^2}{57}=12.32\ m/s^2[/tex]
Therefore the centripetal acceleration of the car will be 12.32 m/s² .
Calculate the electric potential due to a dipole whose dipole moment is 5.2×10−30 C⋅m at a point r = 2.8×10−9 m away. Suppose that r≫ℓ, where ℓ is the distance between the charges in the dipole.
Answer:
V = 8.01*10^-12 V
Explanation:
In order to calculate the electric potential produced by the dipole you use the following formula:
[tex]V=k\frac{p}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
p: dipole moment = 5.2×10−30 C⋅m
r: distance to the dipole = 2.8*10^-9m
You replace the values of the parameters:
[tex]V=(8.98*10^9Nm^2/C^2)\frac{5.2*10^{-30}Cm}{(2.8*10^{-9}m)}\\\\V=8.01*10^{-12}V[/tex]
The electric potential of the dipole is 8.01*10^-12V
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.
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!
Rock can melt at a depth of about
below Earth's surface.
O A. 50 m
B. 500 km
C. 100 km
O D. 2000 km
Rock can melt at a depth of about below Earth's surface 100 km
What role does surface physics play in the world?On the other hand, interface physics offers a wide variety of spectroscopic & microscopic techniques to characterize that deposition & structure creation process upon that sub-nanometer size and, therefore, to pave the way for effective manufacturing techniques.
How does Surface Chemistry work?It is the study of both the chemical processes that take place at the meeting point of two surfaces, such as solid-liquid, solid-gas, sturdy, liquid-gas, etc. Surface engineering refers to a few surface chemistry applications.
To know more about surface visit:
https://brainly.com/question/31464585
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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.
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
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
The flowers of the bunchberry plant open with astonishing force and speed, causing the pollen grains to be ejected out of the flower in a mere 0.30 ms at an acceleration of 2.5 × 104 m. s2 If the acceleration is constant, what impulse is delivered to a pollen grain with a mass of 1.0 × 10−7g?
Answer:
I = 7.5*10^-10 kg m/s
Explanation:
In order to calculate the impulse you first take into account the following formula:
[tex]I=m\Delta v=m(v-v_o)[/tex] (1)
m: mass of the pollen grain = 1.0*10^-7g = 1.0*10^-10 kg
v: final speed of the pollen grain = ?
vo: initial speed of the pollen grain = 0 m/s
Next, you calculate the final speed of the pollen grain by using the information about the acceleration and time. You use the following formula:
[tex]v=v_o+a t[/tex] (2)
a: acceleration = 2.5*10^4 m/s^2
t: time = 0.30ms = 0.30*10^-3 s
[tex]v=0m/s+(2.5*10^4m/s^2)(0.30*10^{-3}s)=7.5\frac{m}{s}[/tex]
Next, you replcae this value of v in the equation (1) and calculate the impulse:
[tex]I=m(v-v_o)=(1.0*10^{-10}kg)(7.5m/s-0m/s)=7.5*10^{-10}kg.\frac{m}{s}[/tex]
The impulse delivered to the pollen grain is 7.5*10^-10 kg m/s