Answer:
β = 30º
Explanation:
By definition, the vector flux across a surface, can be found integrating the dot product of the vector field (electric field E in this case) and the differential surface dA, across the entire surface.If the surface is placed perpendicular to the electric field, and this field is uniform, the total flux across the surface can be expressed as follows:Φ0 = E*A*cos 0º = E*A.
If the magnitude of the electric flux, is reduced to half of its original value, we can write the following equality:Φ0/2 = E*A*cos θ⇒ = Φ0 * cos θ (1)
where θ, is the angle between the electric field E and the vector
perpendicular to the plane traversed by E.
Rearranging terms in (1) we can solve for θ, as follows:
⇒ cos θ = 1/2 ⇒ θ = arc cos (1/2) = 60º
As this is the angle between the electric field, and the surface vector,
which is by definition, perpendicular to the plane, the angle between
the electric field and the plane can be found as follows:
β = 90º - θ = 90º - 60º = 30º
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
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
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.
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 .
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
. A very fine sample is placed 0.15 cm from the objective of a microscope. The focal length of the objective is 0.14 cm and of the eyepiece is 1.0 cm. The near-point distance of the person using the microscope is 25.0 cm. What is the final magnification of the microscope?
Answer:
Final magnification = -375
Explanation:
The total magnification of a compound microscope is expressed mathematically as the product of the magnifying power of each of the lenses that are combined in the compound microscope.
Final magnification = (magnifying power of the objective lens) × (magnifying power of the eyepiece lens) = m₁ × m₂
Magnifying power of a lens is defined as the ratio of the least distance of distinct vision to the focal length of the lens.
For the eyepiece, the least distance of distinct vision is the distance from the object to the near point = 25 + 1 + 0.15 = 26.15 cm
Focal length = 1 cm
Magnifying power of the eyepiece length = (26.15/1) = 26.15
For the objective lens,
The focal length = 0.14 cm
Least distance of distinct vision = -(1 + 1.00 + 0.15 - 0.14) = -2.01 cm (the negative sign means the image seen is upside down)
Magnifying power of the objective lens = (-2.01/0.14) = -14.357
Final magnification = -14.357 × 26.15 = -375
Hope this Helps!!!
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!
A cheetah goes from 0m/s to 25m/s in 2.5 s. What is the cheetah's rate of acceleration?
Answer:
10 m/s²
Explanation:
Acceleration: This the rate of change of velocity. The unit of acceleration is m/s²
From the question,
a = (v-u)/t.................... Equation 1
Where a = acceleration of the cheetah, v = final velocity of the cheetah, u = initial velocity of the cheetah, t = time.
Given: u = 0 m/s, v = 25 m/s, t = 2.5 s.
Substitute these values into equation 1
a = (25-0)/2.5
a = 25/2.5
a = 10 m/s²
Hence the acceleration of the cheetah = 10 m/s²
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]
Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10 m. The bridge is supported by two vertical stone pillars, one 2.0 m from the left end of the bridge and the other 2.0 m from the right end of the bridge. If a 200 kg knight stands on the bridge 4.0 m from the left end, what force is applied by the left support
Answer:
F = 2123.33N
Explanation:
In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.
[tex]\Sigma \tau=0[/tex]
Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.
Then, you obtain the following formula:
[tex]-\tau_l+\tau_p+\tau_{cm}=0[/tex] (1)
τl: torque produced by the left support
τp: torque produced by the person
τcm: torque produced by the center of mass of the wooden
The torque is given by:
[tex]\tau=Fd[/tex] (2)
F: force applied
d: distance to the pivot of the torque, in this case, distance to the right support.
You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:
[tex]-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N[/tex]
Where d1, d2 and d3 are distance to the right support.
You solve the equation for F and replace the values of the other parameters:
[tex]F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N[/tex]
The force applied by the left support is 2123.33 N
A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a constant acceleration of 2.10 m/s2 and the car an acceleration of 3.40 m/s2. The automobile overtakes the truck after the truck has moved 60.0 m.Requried:a. How much time does it take the car to overtake the truck? b. How far was the car behind the truck initially? c. What is the speed of each when they are abreast? d. On a single graph, sketch the position of each vehicle as a function of time. Take x = 0 at the initial location of the truck.
Answer:
A = 7.56s
B = 37.16m
C = for car 25.704m/s and for truck = 15.876m/s
Explanation:
Hello,
This is an example of relative motion between two moving bodies and equation of motion are usually modified to solve problems involving relative motion.
In this question, we'll be making use of
x - x₁ = v₁t + ½at²
The above equation is a modification of
x = vt + ½at²
Where x = distance
v = velocity of the body
a = acceleration of the body
t = time
x - x₁ = v₁t + ½at²
Assuming the starting point of the bodies x₁ = 0 which is at rest, the car sped past the truck at 60m.
x - x₁ = v₁t + ½at²
x₁ = 0
v₁ = 0
x = 60
a = 2.10
t = ?
60 - 0 = 0 × t = ½ × 2.10t²
Solve for t
60 = 1.05t²
t² = 60 / 1.05
t² = 57.14
t = √(57.14)
t = 7.56s
The time it took the car to over take the truck is 7.56s
b).
From our previous equation,
x - x₁ = v₁t + ½at²
Same variable except
a = 3.40m/s² and t = 7.56s
60 - x₁ = 0 + ½ × 3.40 × 7.56²
60 - x₁ = 97.16
x₁ = 60 - 97.16
x₁ = -37.16m
The car was behind the truck by 37.16m or the truck was ahead of the car by 37.16m.
c).
The initial speed of the car ?
v = u + at
u = 0m/s
a = 3.40m/s²
v = 0 + 3.40 × 7.56
v = 25.704m/s
The initial speed of the truck = ?
v = u + at
u = 0m/s
a = 2.10m/s
v = 0 + 2.10 × 7.56
v = 15.876m/s
d).
Due to some circumstance, kindly check attached document for a sketch of the graph
What is the particle arrangement in a liquid
Answer:
the particle arrangement in liquid are close together with no regular arrangement
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.
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..
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
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 0.13 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
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
At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 65.9 V/m. Find the magnetic field strength B, the total energy density u, and the power flow per unit area, all at the same instant and location.
Answer:
B = 2.19*10^-7 T
u = 1.92*10^-18 J/m^3
P = (4pi r^2)cεo E^2
Explanation:
In order to find the magnetic field strength of the electromagnetic wave you use the following formula:
[tex]B=\frac{E}{c}[/tex] (1)
B: magnitude of the magnetic field
E: magnitude of the electric field = 65.9V/m
c: speed of light = 3*10^8m/s
[tex]B=\frac{65.9V/m}{3*10^8m/s}=2.19*10^{-7}T[/tex]
The magnitude of the magnetic field 2.19*10^-7 T
The energy density of the electromagnetic wave is:
[tex]u=\frac{1}{2}\epsilon_oE^2[/tex] (2)
εo: dielectric permittivity = 8.85*10^-12C^2/Nm^2
[tex]u=\frac{1}{2}(8.85*10^{-12}C^2/Nm^2)(65.9V/m)^2=1.92*10^{-8}\frac{J}{m^3}[/tex]
The energy density of the electromagnetic wave is 1.92*10^-8J/m^3
The power is given by:
[tex]P=IA=c\epsilon_oE^2(4\pi r^2)[/tex]
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
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 29.4 s. The distance between two successive crests is 31.4 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed
Answer:
(a) 5.88 s
(b) 0.17 Hz
(c) 31.4 m
(d) 5.338 m/s
Explanation:
From the question,
(a) Period = time between successive crest = t/n............ Equation 1
where t = time, n = number of wave crest
Given: t = 29.4, n = 5
therefore,
Period (T) = 29.4/5 = 5.88 s.
(b) Frequency = 1/period
Frequency = 1/5.88
Frequency = 0.17 Hz.
(c) wavelength = distance between two successive crest = 31.4 m
(d) using,
v = λf............... Equation 2
Where v = speed, f = frequency, λ = wavelength
given: f = 0.17 Hz, λ = 31.4 m
Substitute into equation 2
v = 0.17(31.4)
v = 5.338 m/s
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.
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
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
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
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:.
g 0 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.390 m along the surface before stopping. Part A What was the initial speed of the bullet? Express your answer with the appropriate units. v = nothing
Answer:
vb = 298 m/s
Explanation:
Given:-
- The mass of the bullet, mb = 5.00 g
- The mass of the wooden block, mw = 1.2 kg
- The coefficient of kinetic friction between block and horizontal surface, [tex]u_k = 0.2[/tex]
- The bullet and block combined moves a distance after impact, s = 0.390
Solution:-
- We will first consider the motion of the block and bullet combined after the bullet is embedded into the block.
- We are told that the wooden block is rested on a horizontal floor with coefficient of kinetic friction ( uk ).
- When the bullet is embedded into the block. Both combined will move with a velocity ( V ). Both will eventually loose all of their kinetic energy by doing work against the friction.
- We will apply the work - energy principle to the system ( block + bullet ) as follows:
[tex]W = dE_k[/tex]
Where,
W: Work done against friction
dEk: The change in kinetic energy of the system
- The work-done against friction is the product of frictional force ( Ff ) and the displacement over which the system travels ( s ).
- The frictional force ( Ff ) is proportional to the contact force ( N ) between the system and the surface.
- Apply static balance on the system in the direction normal to the surface as follows:
[tex]N - ( m_b + m_w )*g = 0\\\\N = ( m_b + m_w )*g[/tex]
- The frictional force is defined as:
[tex]F_f = u_k*N\\\\F_f = u_k* ( m_b + m_w ) * g[/tex]
- The work done against friction ( W ) is defined as:
[tex]W = F_f*s\\\\W = u_k*(m_b + m_w )*g*s[/tex]
Where,
g: the gravitational acceleration constant = 9.81 m/s^2
- Now use the energy balance to determine the velocity of the system after impact ( V ):
[tex]u_k*( m_b + m_w )*g*s = 0.5*( m_b + m_w )*V^2\\\\V = \sqrt{2*u_k*g*s } \\\\V = \sqrt{2*0.2*9.81*0.39 }\\\\V = 1.23707 m/s[/tex]
- Now we will again consider an independent system of bullet and the wooden block resting on the horizontal surface.
- The bullet is fired with velocity ( vb ) towards the wooden block. The system can be considered to be isolated and all other fictitious effects can be ignored. This validates the use of conservation of linear momentum for this system.
- The conservation of linear momentum denotes:
[tex]P_i = P_f[/tex]
Where,
Pi: the inital momentum of the system
Pf: the final momentum of the system
- Initially the block was at rest and bullet had a velocity ( vb ) and after striking the bullet is embedded into the block and moves with a velocity ( V ).
[tex]m_b*v_b = ( m_b + m_w )*V\\\\v_b = \frac{ ( m_b + m_w )}{m_b}*V\\\\v_b = \frac{ ( 0.005 + 1.2 )}{0.005}*(1.23707)\\\\v_b = 298 m/s[/tex]
Answer: The initial speed of the bullet is equivalent to the speed of the bullet just before the impact as vb = 298 m/s. This is under the assumption that forces like ( air resistance or gravitational or impulse) have negligible effect.
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.
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
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
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
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