Answer:
[tex]C_{eq}=1.97\ \mu F[/tex]
Explanation:
Given that,
Capacitance 1, [tex]C_1=0.5\ \mu F[/tex]
Capacitance 2, [tex]C_2=11\ \mu F[/tex]
Capacitance 3, [tex]C_3=1.5\ \mu F[/tex]
C₁ and C₂ are connected in series. Their equivalent is given by :
[tex]\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}[/tex]
[tex]\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}[/tex]
[tex]C'=0.47\ \mu F[/tex]
Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :
[tex]C_{eq}=C'+C_3[/tex]
[tex]C_{eq}=0.47+1.5[/tex]
[tex]C_{eq}=1.97\ \mu F[/tex]
So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.
Coco serves a tennis ball at vs = 50 m/s and charges the net at vc = 10 m/s. The opponent,x = 25 m away on the other side of the court, returns the ball with a speed half that of the serve. How close does Coco get to the net (x/2 away) before she meets the return?
Answer:
Distance from the net when she meets the return: 14,3 m
Explanation:
First we need to know how much time it takes the ball to reach the net, the kinematycs general equation for position:
(1) [tex]x = x_{0} + V_{s} t + \frac{1}{2} a t^{2}[/tex]
Taking the net as the origin (x = 0), [tex]x_{0} = 25m[/tex], velocity will be nagative [tex]V_{s} = - 50m/s[/tex] and assuming there is no friction wit air acceleration would be 0, so:
(2) [tex]x = x_{0} + V_{s} t [/tex]
we want to know the time when it reaches the net so when x=0, replacing de values:
(3) [tex]0 = 25m - 50 m/s t_{net} [/tex]
So: [tex]t_{net} = 0,5 s [/tex]
The opponent will return the ball at [tex]V_{ret} = 25m/s[/tex], the equation for the return of the ball will be:
(4) [tex]y = y_{0} + V_{ret} t + \frac{1}{2} a t^{2}[/tex]
Note that here it start from the origin, [tex]y_{0} = 0[/tex], as in the other case acceleration equals 0, and here we have to consider that the time starts when the ball reaches the net ([tex]t_{net} [/tex]) so the time for this equiation will be [tex]t - t_{net} [/tex], this is only valid for [tex]t >= t_{net} [/tex]:
(5) [tex]y = V_{ret} (t - t_{net}) [/tex]
Coco starts running as soon as he serves so his equiation for position will be:
(6) [tex]z = z_{0} + V_{c} t + \frac{1}{2} a t^{2}[/tex]
As in the first case it starts from 25m, [tex]z_{0} = 25m[/tex], acceleration equals 0 and velocity is negative [tex]V_{c} = - 10m/s[/tex]:
(7) [tex]z = z_{0} + V_{c} t[/tex]
To get the time when they meet we have that [tex]z = y[/tex], so from equiations (5) and (7):
[tex]V_{ret} (t - t_{net}) = z_{0} + V_{c} t[/tex]
[tex]t = \frac{z_{0} + V_{ret}* t_{net}}{V_{ret}-V_{c}}[/tex]
Replacing the values:
[tex]t = 1,071 s[/tex]
Replacing t in either (5) or (7):
[tex]z = y = 14,3 m[/tex]
This is the distance to the net when she meets the return
An ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 0.100 m . The spring constant is 25.0 N/m . When the ice cube is released, how far will it travel up the slope before reversing direction? Express your answer numerically in meters to three significant figures. d = nothing m
Answer:
0.6 m
Explanation:
When a spring is compressed it stores potential energy. This energy is:
Ep = 1/2 * k * x^2
Being x the distance it compressed/stretched.
When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.
The potential energy of the ice cube is:
Ep = m * g * h
This is vertical height and is related to the distance up the slope by:
sin(a) = h/d
h = sin(a) * d
Replacing:
Ep = m * g * sin(a) * d
Equating both potential energies:
1/2 * k * x^2 = m * g * sin(a) * d
d = (1/2 * k * x^2) / (m * g * sin(a))
d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m
Using the conservation of mechanical energy principle, we relate the initial potential energy in the compressed spring to the final potential energy due to the height gained by the ice cube sliding up a slope. In doing so, we calculate how far the ice cube travels up the slope before reversing direction to be approximately 0.607 meters.
Explanation:The question pertains to the concept of energy conservation involving the energies of spring compression (potential energy) and gravity (also potential energy). Here the energy initially stored in the compressed spring is used to propel the ice cube up the slope until its potential energy (due to height gained) fully consumes the initial energy from the spring.
The conservation of mechanical energy principle states the initial total energy is equal to the final total energy. Initially we have spring potential energy and finally it's all converted to gravitational potential energy. So, 1/2*k*x^2=m*g*h, where k is the spring constant, x is the spring compression, m is mass of the ice block, g is acceleration due to gravity, and h is height gained by the ice cube.
We solved for h to get h = (1/2*25*(0.100)^2) / (50*9.8) = 0.255 m. This is the vertical height gained, not the distance along the slope. To get the slope distance traveled by the cube (d), we divide the height by sin(25 degrees) to get: d = 0.255/sin(25) = 0.607 m.
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A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 27 m/s^2. If he reaches the ground with a speed of 14 m/s, how long was he in the air (in seconds) ?
Answer:
3.83 s
Explanation:
Initially for 50 m A falls with acceleration equal to g ie 9.8 m /s².In this journey his initial velocity u = 0 , a = 9.8 , v = ? , t = ?
h = ut + 1/ 2 g t²
50 = .5 x 9.8x t²
t = 3.19 s
v = u +gt
= o + 9.8 x 3.19
= 31.26 m /s
For motion under deceleration
initial speed u = 31.26 m/s
Final speed v = 14 m/s
deceleration a = - 27 m/s²
v = u - at
14 = 31.26 - 27 t
t = 0.64 s
So total time in the air
= 3.19 + .64 = 3.83 s
Runner A is initially 3.91 mi west of a flagpole and is running with a constant velocity of 6.04 mi/h due east. Runner B is initially 2.95 mi east of the flagpole and is running with a constant velocity of 5.00 mi/h due west. How far are the runners from the flagpole when they meet? (West is positive and east is negative).
Answer:
The runners are 0.159 mi to the west of the flagpole.
Explanation:
Let´s place the origin of the frame of reference at the point where the flagpole is located.
The initial position of runner A is 3.91 mi and his velocity is -6.04 mi/h
The initial position of runner B is -2.95 mi and his velocity is 5.00 mi/h
When both runners meet, their position is the same. The equation of the position of each runner is:
x = x0 + v · t
Where
x = position at time t
x0 = initial position
v = velocity
t = time
Then, at the meeting point:
x runner A = x runner B
3.91 mi - 6.04 mi/h · t = -2.95 mi + 5.00m/h · t
Solving for t:
3.91 mi + 2.95 mi = 5.00m/h · t + 6.04 mi/h · t
6.86 mi = 11.04 mi/h · t
t = 0.621 h
Now, we use this time to find the meeting point. We can use the equation of any runner. Let´s use the position of runner A:
x = 3.91 mi - 6.04 mi/h · 0.621 h = 0.159 mi
Since the position is positive, the runners met 0.159 mi to the west of the flagpole.
An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.250 m. (a) How long does the collision last? (b) What is his deceleration?
Answer:
a) 0.067 seconds
b) 112.5 m/s²
Explanation:
t = Time taken
u = Initial velocity = 7.5 m/s
v = Final velocity = 0
s = Displacement = 0.25 m
a = Acceleration
Eqaution of motion
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-7.5^2}{2\times 0.25}\\\Rightarrow a=-112.5\ m/s^2[/tex]
b) The deceleration of the football player is 112.5 m/s²
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-7.5}{-112.5}\\\Rightarrow t=0.067\ s[/tex]
a) The collision lasts for 0.067 seconds
Mary's glasses have +4.4D converging lenses. This gives her a near point of 20cm. What is the location of her near point when she is not wearing her glasses?
Express your answer with the appropriate units.
Answer:
The position of near point is 1.68 m.
Explanation:
Given that,
Power = + 4.4 D
Object distance = 20 cm
We need to calculate the focal length
Using formula of power
[tex]P =\dfrac{1}{f}[/tex]
[tex]f=\dfrac{1}{P}[/tex]
[tex]f=\dfrac{1}{4.4}[/tex]
[tex]f=0.227\ m[/tex]
We need to calculate the image distance
Using formula of lens
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}[/tex]
Put the value into the formula
[tex]\dfrac{1}{v}=\dfrac{1}{0.227}-\dfrac{1}{0.20}[/tex]
[tex]\dfrac{1}{v}=-\dfrac{135}{227}[/tex]
[tex]v=-1.68\ m[/tex]
Hence, The position of near point is 1.68 m.
Mary's near point without glasses is 22.73 cm, calculated by using the lens equation and the given lens power of her glasses with a knowledge of optics.
Explanation:The question involves determining the near point location when the student, Mary, is not using her glasses. Mary has glasses with +4.4D converging lenses, and with the glasses, her near point is at 20 cm. To find the near point without glasses, we must use the lens equation:
1/f = 1/di + 1/do
Where f is the focal length of the lens, di is the image distance, and do is the object distance. For her glasses:
The lens power P is +4.4D.
The focal length f is the inverse of power, f = 1/P.
Hence, f = 1/4.4 = 0.2273 meters (or 22.73 cm).
Since the near point with glasses is the closest distance at which she can see clearly, that means the image distance di (for her near point with glasses) is at infinity because the lenses correct her vision to normal. So:
1/f = 0 + 1/do
1/22.73 cm = 1/do
Therefore, the near point without glasses, do, is also 22.73 cm, since for the near point with glasses di is infinite - hence the person can see objects clearly at the near point with the glasses when the image is formed at infinity without the glasses.
A proton accelarates from rest in a unifrom electric field of 630 N/C. At some later time, it's speed is 1.3 * 10^6 m/s. a) Find the magnitude of the acceleration of the proton. b) How long does it take the proton to reach this speed? c) How far has it moved in that interval?
Answer:
a) 6.028*10^10 m/s^2
b)2.156*10^-5 s
c)14.01 m
Explanation:
Hello!
I will not consider relativistic efects since the velocity of the proton is 1% of the velocity of ligth.
In order to find the acceleration we need to calculate first the force, this is done by multiplying the electric field times the charge of the proton (1e=1.6*10^-19)
[tex]ma=F=630\times1.6\times10^{-19}N[/tex]
Since the mass of the proton is 1.6726219 × 10^-27 kilograms
The acceleration it suffers due to the electric field is:
[tex]a = 6.028 \times10^{10}m/s^{2}[/tex]
Since the proton accelerates from rest, the velocity as a function of time is given by:
[tex]v = at[/tex]
So
[tex]t=\frac{1.3*10^{6}m/s}{6.028 \times10^{10}m/s^{2}}=2.156\times10^{-5}s[/tex]
Finally, the length traveled by the proton in that interval is given by:
[tex]x(t=2.156\times10^{-5}s)=\frac{1}{2} 6.028 \times10^{10}m/s^{2}\times(2.156\times10^{-5}s)^{2}=14.01 m[/tex]
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm from the normal height. What is the atmospheric pressure? (The density of mercury is 13.59 g/cm3.)
Final answer:
To calculate the atmospheric pressure, we can use the formula: Pressure = Density × Acceleration due to gravity × Height. Given the density of mercury, the height difference, and the acceleration due to gravity, we can calculate the atmospheric pressure.
Explanation:
The normal atmospheric pressure is 1.013 × 10^5 Pa. In this case, the height of a mercury barometer drops by 27.1 mm due to the approaching storm. To calculate the atmospheric pressure, we can use the formula:
Pressure = Density × Acceleration due to gravity × Height
Given that the density of mercury is 13.59 g/cm3, the height difference is 27.1 mm, and the acceleration due to gravity is approximately 9.8 m/s2, we can calculate the atmospheric pressure:
Pressure = (13.59 g/cm3)(9.8 m/s2)(27.1 mm)
An aluminum alloy rod has a length of 6.3243 cm at 16.00°C and a length of 6.3568 cm at the boiling point of water. (a) What is the length of the rod at the freezing point of water? (b) What is the temperature if the length of the rod is 6.3689 cm?
Answer:
Part a)
[tex]L_o = 6.3181 cm[/tex]
Part b)
[tex]T = 131.3 ^oC[/tex]
Explanation:
Let the length of the rod at 0 degree Celsius is given as Lo
now we have
[tex]L = L_o( 1 + \alpha \Delta T)[/tex]
now we know that
[tex]L_o[/tex] = Length of rod at zero degree C
Part a)
[tex]6.3243 = L_o(1 + \alpha (16 - 0))[/tex]
[tex]6.3568 = L_o(1 + \alpha (100 - 0))[/tex]
now we have
[tex]\frac{6.3568}{6.3243} = \frac{1 + 100 \alpha}{1 + 16 \alpha}[/tex]
[tex]1.005(1 + 16 \alpha) = 1 + 100 \alpha[/tex]
[tex]83.918\alpha = 5.138\times 10^{-3}[/tex]
[tex]\alpha = 6.12 \times 10^{-5}[/tex]
now we have
[tex]L_o = 6.3181 cm[/tex]
Part b)
length of the rod is 6.3689 cm
now we have
[tex]L = L_o(1 + \alpha\Delta T)[/tex]
[tex]6.3689 = 6.3181(1 + \alpha \Delta T)[/tex]
[tex]6.3689 = 6.3181(1 + (6.12 \times 10^{-5})(T - 0))[/tex]
[tex]1.008 = 1 + (6.12 \times 10^{-5})T[/tex]
[tex]T = 131.3 ^oC[/tex]
Two small particles of mass m1 and mass m2 attract each other with a force that varies inversely with the cube of their separation. At time t0,m1 has a velocity of magnitude v0, directed towards m2, which is at rest a distance d away. At time t1, the particles collide. Calculate L, the distance traveled by particle 1 during the time interval t1 − t0. Express your answer using some or all of the following variables: m1, m2, t0, t1, v0, and d.
Answer:
[tex]r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}[/tex]
Explanation:
Given,
mass of the first particle = [tex]m_1[/tex]velocity of the first particle = [tex]v_o[/tex]mass of the second particle = [tex]m_2[/tex]velocity of the second particle = [tex]v_2 = 0[/tex]Time interval = [tex](t_1\ -\ t_o)[/tex]Let [tex]v_{cm}[/tex] be the initial velocity of the center of mass of the system of particle at time [tex]t_o[/tex]
[tex]\therefore v_{cm}\ =\ \dfrac{m_1v_1\ +\ m_2v_2}{m_1\ +\ m_2}\\\Rightarrow v_{cm}\ =\ \dfrac{m_1v_0}{m_1\ +\ m_2}[/tex]
Assuming that the first particle is at origin, distance of the second particle from the origin is 'd'
[tex]x_1\ =\ 0[/tex][tex]x_2\ =\ d[/tex]Center of mass of the system of particles
[tex]x_{cm}\ =\ \dfrac{m_1x_1\ +\ m_2x_2}{m_1\ +\ m_2}\\\Rightarrow x_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\\[/tex]
Hence, at time [tex]t_0[/tex], the center of mass of the system is at [tex]x_0\ =\ \dfrac{m_2d}{m_1\ +\ m_2}[/tex] at an initial speed of [tex]v_{cm}[/tex]
Both the particles are assumed to be the point masses, therefore at the time [tex]t_1[/tex] the center of mass is at the position of the second particle which should be equal to the total distance traveled by the first particle because the second particle is at rest.
Let [tex]r_{cm}[/tex] be the distance traveled by the center of mass of the system of particles in the time interval [tex](t_1\ +\ t_0)[/tex]
From the kinematics,
[tex]s\ =\ x_0\ +\ vt\\\Rightarrow r_{cm}\ =\ x_{cm}\ +\ v_{cm}{t_1\ -\ t_0}\\\Rightarrow r_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\ +\ \left ( \dfrac{m_1v_0}{m_1\ +\ m_2}\ \right )\times (t_1\ -\ t_0)\\\Rightarow r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}[/tex]
Hence, this is the required distance traveled by the first mass to collide with the second mass which is at rest.
Calculate The water pressure at the bottom of the Marianas Trench is approxi mately 1,100 kPa. With how much force would the water pressure at the bottom of the Marianas Trench push on a fish with a surface area of 0.50 m^2?
Answer:
Force, F = 550000 N
Explanation:
Given that,
Pressure at the bottom of the Marianas Trench, [tex]P=1100\ kPa=11\times 10^5\ Pa[/tex]
Surface area, [tex]A=0.5\ m^2[/tex]
We need to find the force with which the water pressure at the bottom of the Marianas Trench push on a fish. Mathematically, the pressure is given by :
[tex]P=\dfrac{F}{A}[/tex]
[tex]F=P\times A[/tex]
[tex]F=11\times 10^5\ Pa\times 0.5\ m^2[/tex]
F = 550000 N
So, the force with which the water pressure at the bottom of the Marianas Trench push on a fish is 550000 N. Hence, this is the required solution.
The water pressure at the bottom of the Marianas Trench would exert a force of 550,000 Newtons on a fish with a surface area of 0.50 m^2.
Explanation:The water pressure exerted on a fish at the bottom of the Marianas Trench can be determined by using the formula for pressure (P), which is force (F) divided by area (A). Given the pressure at the bottom of the trench is approximately 1,100 kPa and the surface area of the fish is 0.50 m2, the force can be calculated using the equation F = P x A. By inserting the given values, we get F = 1,100,000 Pa x 0.50 m2 = 550,000 N. Therefore, the water pressure would push on the fish with a force of 550,000 Newton.
A car moves along an x axis through a distance of 980 m, starting at rest (at x = 0) and ending at rest (at x = 980 m). Through the first 1/4 of that distance, its acceleration is + 4.20 m/s^2. Through the next 3/4 of that distance, its acceleration is - 1.40 m/s2. What are (a) its travel time through the 980 m and (b) its maximum speed?
Answer:
travel time is 32.4 s
maximum speed is 45.36 m/s
Explanation:
given data
distance = 980 m
acceleration = 4.20 m/s² for first 1/4 of that distance
acceleration = -1.40 m/s² for next 3/4 of that distance
to find out
travel time through the 980 m and maximum speed
solution
we know for first 1/4 of that distance is = [tex]\frac{980}{4}[/tex] = 245 m
so equation of motion
s = ut + 0.5 ×at² .............1
here u is initial speed = 0 and a is acceleration an t is time
s = ut + 0.5 ×at²
245 = 0+ 0.5 ×4.20 (t)²
t = 10.80 s
so
maximum speed at 1/4 of that distance
use equation of motion
v² - u² = 2as
put here value
v² - 0 = 2(4.20)× (245)
v = 45.36 m/s
so maximum speed is 45.36 m/s
and
for 3/4 distance
use equation of motion
v = u + at
here u is here 45.36 and a is acceleration and t is time and v final speed is 0
0 = 45.36 + (-1.40) × t
t = 32.4 s
so travel time is 32.4 s
A tired worker pushes with a horizontal force of 440 N on a 170 kg crate initially resting on a thick pile carpet. The coefficients of static and kinetic friction are 0.5 and 0.2, respectively. The acceleration of gravity is 9.81 m/s^2 . Find the frictional force exerted by the carpet on the crate. Answer in units of N.
Answer:
440 N
Explanation:
Force applied, F = 440 N
mass of crate, m = 170 kg
μs = 0.5, μk = 0.2
g = 9.81 m/s^2
The normal reaction acting on the crate, N = m g = 170 x 9.81 = 1667.7 N
The maximum value of static friction force acting on the crate
[tex]f_{s}=\mu _{s}N=0.5 \times 1667.7 = 833.85 N[/tex]
The maximum value of static friction force is more than the applied force so the crate does not move and teh applied force becomes friction force.
thus, the friction force acting on the crate is 440 N.
The frictional force exerted by the carpet on the crate is equal to the force of static friction.
Explanation:To find the frictional force exerted by the carpet on the crate, we need to calculate the force of static friction first. The maximum force of static friction is given by fs(max) = μsN, where μs is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of the crate, which is given by N = mg. In this case, the normal force N = (170 kg)(9.81 m/s^2) = 1666.7 N.
Therefore, the maximum force of static friction is fs(max) = (0.5)(1666.7 N) = 833.35 N. Since the worker is pushing with a force of 440 N, which is less than the maximum force of static friction, the crate remains at rest and the frictional force exerted by the carpet on the crate is equal to the force of static friction, which is 833.35 N.
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A race car accelerates uniformly from 12 m/s to 52 m/s in 2 seconds. Determine the acceleration of the car.
Answer:
The acceleration of the car is [tex]20 m/s^{2}[/tex]
Solution:
According to the question:
Initial velocity of the car, u = 12 m/s
Final velocity of the car, v = 52 m/s
Time interval for the change in velocity, [tex]\DeltaT = 2 s[/tex]
The rate at which the velocity of an object varies is referred to as its acceleration:
[tex]a = \frac{v - u}{\Delta T}[/tex]
[tex]a = \frac{52 - 12}{2} = 20 m/s^{2}[/tex]
A ball is thrown vertically up with a velocity of 65 ft/sec at the edge of a 680-ft cliff. Calculate the height h to which the ball rises and the total time t after release for the ball to reach the bottom of the cliff. Neglect air resistance and take the downward acceleration to be 32.2 ft/sec2.
Answer:
a) [tex]y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2} =640 ft+(65ft/s)(2.018s)+(\frac{1}{2})(-32.2ft/s^{2} )(2.018s)^{2}[/tex]
[tex]y=705.6ft[/tex]
b) [tex]t=8.63 s[/tex]
Explanation:
We start the exercise knowing that a ball is thrown up with an initial velocity of 65 ft/s with an initial height of 680 ft.
To calculate the maximum heigh, we know that at the top of the motion the ball stop going up and start going down because of the gravity
First of all, we need to calculate the time that takes the ball to reach the maximum point.
a) [tex]v_{y}=v_{oy}+gt[/tex]
[tex]t=\frac{-v_{oy} }{g}=\frac{-65ft/s}{-32.2ft/s^{2} } =2.018s[/tex]
Knowing that time, we can calculate the height to which the ball rises:
[tex]y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2} =640 ft+(65ft/s)(2.018s)+(\frac{1}{2})(-32.2ft/s^{2} )(2.018s)^{2}[/tex]
[tex]y=705.6ft[/tex]
b) Now, to know the time that the ball reach the bottom of the cliff, we know that the final height is y=0ft
[tex]y=y_{0}+v_{oy} t+\frac{1}{2} gt^{2}[/tex]
[tex]0=640ft+(65ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}[/tex]
This is a classic quadratic equation, that can be solve using the quadratic formula
[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]
a=-16.1
b=65
c=640
Solving for t, we have that
[tex]t=-4.6015 s[/tex] or [tex]t=8.6387s[/tex]
Since the time can not be negative:
[tex]t=8.6387s[/tex]
Given a particle that follows the acceleration a(t) 10 (-t + 2 ) (t-5)+ 100 m/s2, find: a. Find the displacement at 2 seconds. Assume from rest and a starting point of 3m. b. Find the velocity at 4 seconds. Assume an initially at rest. c. Find the time at which maximum displacement occurs (use calculus because it is way easier, not kinematics). d. Find the value of maximum velocity over the interval 0
Answer:
a. x= 83.03 m at t= 2 s
b. v=346.7 m/s at t= 4s
c. t=10.5s : maximum displacement occurs
d. t= 7s : maximum velocity
Explanation:
Definitions
acceleration :a(t) = dv/dt :Derived from velocity with respect to time
Velocity : V(t)=dx/dt : Derived from Displacement: with respect to time
Displacement: X(t)
Developing of problem
we have a(t) =10 (-t + 2 ) (t-5)+ 100
a(t) =(-10 t + 20) (t-5)+ 100 = -10 t²+50t+ 20t-100+ 100=--10 t²+70t
a(t)=--10 t²+70t Equation (1)
a(t) = dv/dt
-10 t²+70t=dv/dt
dv=(-10 t²+70t)dt
We apply integrals to both sides of the equation
∫dv=∫(-10 t²+70t)dt
[tex]v=\frac{-10t^{3} }{3} +\frac{70t^{2} }{2} +C_{1}[/tex]
[tex]v=\frac{-10t}{3} +35t^{2} +C_{1}[/tex]
at time t=0, v=0, then, C₁=0
[tex]v=\frac{-10t^{3}}{3} +35t^{2}}[/tex] Equation (2)
[tex]v=\frac{dx}{dt}[/tex]
dx= vdt
We apply integrals to both sides of the equation and we replace v(t) of the equation (2)
[tex]\int\limits {\, dx =\int\limits{\frac{(-10t^{3} }{3}+35t^{2} ) } \, dt[/tex]
[tex]x=\frac{-5t^{4} }{6} +\frac{35t^{3} }{3} +C_{2}[/tex]
at time t=0, x=3, then,C₂=3
[tex]x=\frac{-5t^{4} }{6} +\frac{35t^{3} }{3} +3}[/tex] Equation (3)
a)Displacement at t=2s
We replace t=2s in the Equation (3)
[tex]x=\frac{-5(2)^{4} }{6} +\frac{35(2)^{3} }{3} +3}[/tex]
x= 83.03 m at t= 2 s
b) Velocity at t=4s
We replace t=4s in the Equation (2)
[tex]v=\frac{-10(4)^{3}}{3} +35(4)^{2}}[/tex]
v=346.7 m/s at t= 4s
c)Time at which maximum displacement occurs
at maximum displacement :v= [tex]\frac{dx}{dt} =0[/tex]
in the Equation (2) :
[tex]v=\frac{-10t^{3}}{3} +35t^{2}}[/tex] =0
[tex]35t^{2} } =\frac{10t}{3}[/tex] ,we divide by t on both sides of the equation
[tex]t=\frac{35*3}{10}[/tex]
t=10.5s
d. maximum velocity over the interval
at maximum velocity :a= [tex]\frac{dv}{dt} =0[/tex]
in the Equation (1)
-10 t²+70t = 0 we multiply the equation by -1 and factor
10t ( t-7) =0
t= 7 s
A rock is rolled in the sand. It starts at 5.0 m/s, moves in
astraight line for a distance of 3.0 m, and then stops. What is
themagnitude of the average acfeleration?
Answer:
magnitude of the average acceleration is 4.16 m/s²
Explanation:
given data
initial velocity u = 5 m/s
distance s = 3 m
final velocity v = 0
to find out
magnitude of the average acceleration
solution
we will apply here linear motion equation that is
v²-u² = 2×a×s ..............1
put here all these value
v is final velocity and u is initial velocity and s is distance
0²-5² = 2×a×3
a = -4.16
so magnitude of the average acceleration is 4.16 m/s²
Answer: The magnitude of the average acceleration is 4.16m/s^2
(a) How strong is the attractive force between a glass rod with a 0.700 μC charge and a silk cloth with a –0.600 μC charge, which are 12.0 cm apart, using the approximation that they act like point charges? (b) Discuss how the answer to this problem might be affected if the charges are distributed over some area and do not act like point charges.
Answer:
a) [tex]F=-3.1465 N[/tex]
b) Greater attractive force
Explanation:
Assuming they are point charges, we can use coulomb's law to know the magnitude of the force between these objects:
[tex]F=\frac{kq_1q_2}{d^2}[/tex]
Here, K is the Coulomb constant, [tex]q_1[/tex] and [tex]q_2[/tex] are the point charges and d is the distance bewteen the charges.
a) So, we have:
[tex]F=\frac{(8.99*10^9\frac{Nm^2}{C^2})(7*10^{-6}C)(-6*10^{-6}C)}{12*10^{-2}m}\\F=-3.1465 N[/tex]
b) There will be a greater amount of charge on the side closest to the object with opposite charge, therefore this will increase the charge at that point, increasing the attractive force.
The attractive force between a 0.700 µC charged glass rod and a -0.600 µC charged silk cloth 12.0 cm apart is 0.2625 N, as calculated by Coulomb's law. If charges are distributed over an area instead of acting as point charges, the force calculation might differ, depending on the geometry and distribution of the charges.
Explanation:To calculate the attractive force between a glass rod with a 0.700 µC charge and a silk cloth with a -0.600 µC charge that are 12.0 cm apart, we can use Coulomb's law. Coulomb's law is given by the formula F = k * |q1*q2| / r^2, where F is the force between the charges, q1 and q2 are the values of the charges, r is the distance between the charges, and k is Coulomb's constant (approximately 8.988 × 10^9 Nm^2/C^2). Plugging in the values, we get F = (8.988 × 10^9 Nm^2/C^2) * |0.700 × 10^-6 C * -0.600 × 10^-6 C| / (0.12 m)^2, which results in a force of approximately 0.2625 N.
Discussing how the answer to this problem might be affected if the charges are distributed over some area rather than acting like point charges: The resulting force might differ from the point charge approximation. This is because the distribution of charge affects the electric field produced by the charges, potentially reducing the force if the charges are spread out over a larger area compared to being concentrated at a point. The effect typically depends on the actual distribution and geometry of the charges.
A team is building a ballistic ball launcher. The target is 18 ft above the ground, and it needs to catch the ball at the top of its trajectory. Your launcher throws balls from 0.50 m above the ground and must be located 6 m from the target. At what speed must the launcher toss the food in m/s? At what angle above the horizontal must the launcher toss the food? Explain.
Answer:
[tex]\theta = 58.98[/tex]°
Explanation:
given data:
h = 18 ft = 5.48 m
from figure
[tex]h_{max} = 5.48 - 0.50 = 4.98 m[/tex]
[tex]h_{max} = \frac{ v_y^2}{2g}[/tex]
[tex]v_y =\sqrt{2gh_{max}[/tex]
[tex]v_y = \sqrt{2*9.8* 4.98} m/s[/tex]
[tex]v_y = 9.88 m/s[/tex]
[tex]t = \frac{v_y}{g} =\frac{9.88}{9.8} = 1.01 s[/tex]
[tex]v_x = \frac{d}{t} = \frac{6}{1.01} = 5.49 m/s
[tex]v = \sqrt{v_x^2+v_y^2} = \sqrt{5.95^2+9.88^2}[/tex]
v = 11.53 m/s
[tex]tan\theta = \frac[v_x}{v_y}[/tex]
[tex]\theta = tan^{-1} \frac{9.88}{5.94}[/tex]
[tex]\theta = 58.98[/tex]°
The tub of a washing machine goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it is turning at 4.00 rev/s. At this point, the lid of the washing machine is opened, and a safety switch turns it off. The tub then smoothly slows to rest in 11.0 s. Through how many revolutions does the tub rotate while it is in motion?
Answer:
36 rev
Explanation:
See it in the pic
In a 770 kW hydroelectric plant, 300 m^3 of water passes through the turbine each minute. Assuming complete conversion of the water's initial gravitational potential energy to electrical energy, what distance does the water fall? Assume two significant figures.
Answer:
16 m
Explanation:
given,
power of hydraulic plant = 770 kW
volume of water pass through the turbine = 300 m³
density of water = 1000 kg/m³
m =ρ × V
mass of water pass each minute = 300 × 1000 = 3 × 10⁵
assume height of the fall be h
potential head of the water = mgh
[tex]\dfrac{mgh}{60}= 770 \times 10^3[/tex]
[tex]3\times 10^5 \times 9.81\times h= 770 \times 10^3\times 60[/tex]
[tex]h = \dfrac{770 \times 10^3\times 60}{3\times 10^5 \times 9.81}[/tex]
h = 15.69 m ≈ 16 m
the distance of the water fall is equal to 16 m.
A tank of volume 0.25 ft is designed to contain 50 standard cubic feet of air. The temperature is 80° F.Calculate the pressure inside the tank. One standard cubic foot of air occupies foot at standard 'temperature and pressure (T 59° F and p= 2116 lb/ft2). one cubic
Answer:
The inside Pressure of the tank is [tex]4499.12 lb/ft^{2}[/tex]
Solution:
As per the question:
Volume of tank, [tex]V = 0.25 ft^{3}[/tex]
The capacity of tank, [tex]V' = 50ft^{3}[/tex]
Temperature, T' = [tex]80^{\circ}F[/tex] = 299.8 K
Temperature, T = [tex]59^{\circ}F[/tex] = 288.2 K
Now, from the eqn:
PV = nRT (1)
Volume of the gas in the container is constant.
V = V'
Similarly,
P'V' = n'RT' (2)
Also,
The amount of gas is double of the first case in the cylinder then:
n' = 2n
[tex]\]frac{n'}{n} = 2[/tex]
where
n and n' are the no. of moles
Now, from eqn (1) and (2):
[tex]\frac{PV}{P'V'} = \frac{nRT}{n'RT'}[/tex]
[tex]P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}[/tex]
Water drips from the nozzle of a shower onto the floor 181 cm below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops?
Answer:
Explanation:
If t be the time of fall of drops from a height of 181 s ,
181 = 1/2 g t²
t = 6 s approx
There are 4 drops in vertical line , one touching the floor, one at 181 m height on the verge of falling down , and two drops in mid air.
The time interval between dripping of two consecutive drops
= 6 / 3
2 s
time duration of fall of second drop = 2 x 2 = 4 s
Position of drop below nozzle
= 1/2 g x 4²
= 78.4 m
b )
Time duration of fall of third drop
= 2 s
Position of drop below nozzle
= 1/2 9.8 x 2²
= 19.6 m
A surveillance satellite has a camera which detects 550 nm light and is equipped with a 35 cm diameter lens. If the satellite is at an altitude of 160 km, what is the minimum spacing between two objects on the ground that the camera can barely resolve (in m)?
Answer:
0.3067 m
Explanation:
Resolving power of a lens is given by the expression
R.P = 1.22λ / D
where λ is wavelength of light used, D is diameter of the lens
Substituting the data given
R.P in radian =
[tex]\frac{1.22\times550\times10^{-9}}{35\times10^{-2}}[/tex]
R P = 19.17 X 10⁻⁷ radian
If d be the minimum spacing between two objects on the ground that is resolvable by lens
[tex]\frac{d}{160\times10^3} =19.17\times10^{-7}[/tex]
d = 0.3067 m
A stone is tied to a string and whirled around in a circle at
aconstant speed. Is the string more likely to break when the
circleis horizontal or when it is vertical? Account for your
answer,assuming the constant speed is the same in each case.
Answer:
Vertical circle.
Explanation:
According to the question:
When the stone tied string moves in horizontal circle, then the tension in the string is provided by the centripetal force.
When the stone tied string moves in a vertical circle, then the tension is provided by the centripetal force as well as its weight.
Thus the probability of breaking in the vertical circle is more.
Refer to the fig shown:
In case of horizontal circle:
The centripetal force for a circle of radius R is given by:
[tex]F_{c} = \frac{mv_{P}^{2}}{R} = mg[/tex]
Now, at point P:
[tex]mg - T = \frac{mv_{P}^{2}}{R} [/tex]
Since, T = 0
[tex]mg - 0 = \frac{mv_{P}^{2}}{R} [/tex]
[tex]v_{P}^{2} = Rg[/tex] (1)
Now, at point Q:
[tex]T - mg = \frac{mv_{Q}^{2}}{R} [/tex] (2)
Also, by using the law of conservation of energy, total mechanical energy at point P and Q will be conserved:
[tex]\frac{1}{2}mv_{P}^{2} + mg(2R) = \frac{1}{2}mv_{Q}^{2} + mg(0)[/tex]
Using eqn (1):
[tex]\frac{1}{2}gR + 2gR = \frac{1}{2}v_{Q}^{2}[/tex]
[tex]v_{Q}^{2} = 5gR[/tex] (3)
Now, using eqn (2) and (3):
[tex]T = \frac{m5gR}{R} + mg = 6mg[/tex]
Thus tension at point Q is greater than the force at point P.
A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Find the velocity of the object when it is 23.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.) _______ m/s (b) Find the total distance the object travels during the fall. ________ m
Answer:
(a). The velocity of the object is -2.496 m/s.
(b). The total distance of the object travels during the fall is 23.80 m.
Explanation:
Given that,
Time = 1.95 s
Distance = 23.5 m
(a). We need to calculate the velocity
Using equation of motion
[tex]s = ut+\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2[/tex]
[tex]u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}[/tex]
[tex]u=-2.496\ m/s[/tex]
(b). We need to calculate the total distance the object travels during the fall
Using equation of motion
[tex]v = u+gt[/tex]
Put the value in the equation
[tex]-2.496=0-9.8\times t[/tex]
[tex]t =\dfrac{2.496}{9.8}[/tex]
[tex]t=0.254\ sec[/tex]
The total time is
[tex]t'=t+1.95[/tex]
[tex]t'=0.254+1.95[/tex]
[tex]t'=2.204\ sec[/tex]
We need to calculate the distance
Using equation of motion
[tex]s = ut+\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]s=0+\dfrac{1}{2}\times9.8\times(2.204)^2[/tex]
[tex]s=23.80\ m[/tex]
Hence, (a). The velocity of the object is -2.496 m/s.
(b). The total distance of the object travels during the fall is 23.80 m.
If one motor has three times as much power as another, then the smaller power motor: A. can do the same work in the same time. B. can do the same work in three times the time. C. can never do the same work as the larger motor. D. can do the same work in one-third the time.
Answer:B
Explanation:
Given
First motor has three times as much as power as another
Let the power of smaller motor be P
Therefore bigger motor has power 3 P
and we know that energy [tex]E=power\times time[/tex]
Larger motor [tex]E=3P\times t=3Pt[/tex]
Smaller motor [tex]E=P\times 3t=3Pt[/tex]
Therefore smaller motor will do same work if it takes three times the time taken by Larger motor
You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg is placed against a light spring that is compressed 0.280 m. The spring has force constant 48.0 N/m . The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is μk = 0.300. When the box has traveled 0.280 m and the spring has reached its equilibrium length, the box loses contact with the spring. (a) What is the speed of the box at the instant when it leaves the spring? (b) What is the maximum speed of the box during its motion?
Final answer:
The speed of the box as it leaves the spring can be calculated using conservation of energy, accounting for the initial potential energy in the spring and work done by friction. The maximum speed of the box is achieved when all the spring's potential energy is converted into kinetic energy, before friction does any work.
Explanation:
To solve for the speed of the box at the instant it leaves the spring and its maximum speed during the motion, we can apply the principle of conservation of energy and the work-energy theorem.
Part (a): Speed of Box When It Leaves the Spring
The initial potential energy stored in the compressed spring is equal to the kinetic energy of the box plus the work done by friction as the box slides the distance where the spring is compressed:
PE_spring = KE_box + Work_friction
Using the formula for potential energy PE_spring = (1/2)[tex]kx^2[/tex], where k is the spring constant and x is the compression distance, and the work done by friction Work_friction = μ_kmgx, where μ_k is the coefficient of kinetic friction, m is the mass, and g is the acceleration due to gravity, we can find the kinetic energy which will then be used to calculate the speed using KE = (1/2)[tex]mv^2[/tex].
To find the actual speed, we rearrange to v = √(2(PE_spring - Work_friction)/m).
Part (b): Maximum Speed of the Box
The maximum speed of the box occurs at the point where all the potential energy in the spring has been converted to kinetic energy and before any work has been done by friction. Therefore, v_max = √(2PE_spring/m).
A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line that carries steam at 1 MPa and 300 ºC. Now the valve is opened and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, at which point the valve is closed. Determine the final temperature of the steam in the tank.
Answer:
Explanation:
The steam was earlier at 300 ° C. and pressure of 1 MPa. When the gas is allowed to expand against vacuum , work done by the gas is nil because there is no external pressure against which it has to work . Therefore there will not be any change in its internal energy. Since the tank is insulated therefore there is no possibility of external heat to increase its internal energy.
Hence the temperature of gas will remain unaffected. It will remain stagnant at 300
Interstellar space (far from any stars) contains atomic hydrogen (H) with a density of 1 atom/cm3 and at a temperature of about 2.7 K. Determine (a) the pressure in interstellar space, (b) root-mean square speed of the atoms and (c) The kinetic energy stored in 1 km3 of space.
Explanation:
Given that,
Number density [tex]n= 1\ atom/cm^{3} =10^{6}\ atom/m^3[/tex]
Temperature = 2.7 K
(a). We need to calculate the pressure in interstellar space
Using ideal gas equation
[tex]PV=nRT[/tex]
[tex]P=\dfrac{nRT}{V}[/tex]
[tex]P=\dfrac{10^{6}\times8.314\times2.7}{6.023\times10^{23}}[/tex]
[tex]P=3.727\times10^{-17}\ Pa[/tex]
[tex]P=36.78\times10^{-23}\ atm[/tex]
The pressure in interstellar space is [tex]36.78\times10^{-23}\ atm[/tex]
(b). We need to calculate the root-mean square speed of the atom
Using formula of rms
[tex]v_{rms}=\sqrt{\dfrac{3RT}{Nm}}[/tex]
Put the value into the formula
[tex]v_{rms}=\sqrt{\dfrac{3\times8.314\times2.7}{1.007\times10^{-3}}}[/tex]
[tex]v_{rms}=258.6\ m/s[/tex]
The root-mean square speed of the atom is 258.6 m/s.
(c). We need to calculate the kinetic energy
Average kinetic energy of atom
[tex]E=\dfrac{3}{2}kT[/tex]
Where, k = Boltzmann constant
Put the value into the formula
[tex]E=\dfrac{3}{2}\times1.38\times10^{-23}\times2.7[/tex]
[tex]E=5.58\times10^{-23}\ J[/tex]
The kinetic energy stored in 1 km³ of space is [tex]5.58\times10^{-23}\ J[/tex].
Hence, This is the required solution.