(a) The time the baseball spends in the air is 0.92 s.
(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
Time spent in air by the baseball
h = vt - ¹/₂gt²
-2.1 = (4.05 x sin 34)t - ¹/₂(9.8)(t²)
-2.1 = 2.26t - 4.9t²
4.9t² - 2.26t - 2.1 = 0
t = 0.92 s
Horizontal distance traveled by the baseballR = Vx(t)
R = (4.05 x cos 34)(0.92)
R = 3.1 m
Thus, the time the baseball spends in the air is 0.92 s.
The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.
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A small object with mass 3.95 kg moves counterclockwise with constant speed 1.65 rad/s in a circle of radius 2.95 m centered at the origin. It starts at the point with position vector 2.95î m. Then it undergoes an angular displacement of 9.10 rad.
(a) What is its new position vector? _____ m
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis? ______ quadrant at _______°
(c) What is its velocity? _____ m/s
(d) In what direction is it moving? _____ ° from the +x direction
(e) What is its acceleration? _____ m/s2
(f) Make a sketch of its position, velocity, and acceleration vectors.
a)New position vector in vector form= r = 0.233404 î + 2.94056j m
b) Lies in second quadrant at 161.391°
c)Velocity =4.8675 m/s
d)It is moving in a direction making 161.391° with positive x-direction.
e)Acceleration will be centripetal acceleration (8.031 m/s²).
Given:
Mass of the object m = 3.95 kg
ω=1.65 rad/s
Radius of the circle = 2.95 m
a)
new position vector in vector form
=R cos1.65 î + R sin 1.65 j
= 2.95 cos1.65 î +2.95 sin1.65 j
= 2.95 x 0.07912 î + 2.95 x 0.9968 j
r = 0.233404 î + 2.94056j
b)
Angular Displacement = θ₀ = 9.10 rad
9.10 radian = 180/π× 9/10 degree
= 521.391°
=521.391°- 360°
=161.391°
This will lie in second quadrant.
Angle made with positive x-axis
=161.391°
c)
Velocity
v = ω R
= 1.65 x 2.95
=4.8675 m/s
d)
It is moving in a direction making 161.391° with positive x-direction.
e)
Acceleration will be centripetal acceleration.
= v²/R
=(4.8675)² / 2.95
=23.6925562 / 2.95
=8.031 m/s²
f) Position, Velocity and Acceleration graph:
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a) New position vector is r = 0.233404 î + 2.94056j m
b) Lies in second quadrant at 161.391°
c) Velocity =4.8675 m/s
d) It is moving in a direction making 161.391° with positive x-direction.
e) Acceleration will be centripetal acceleration (8.031 m/s²).
Given:
Mass of the object m = 3.95 kg
ω=1.65 rad/s
The radius of the circle = 2.95 m
a) new position vector in vector form
r =R cos1.65 î + R sin 1.65 j
r = 2.95 cos1.65 î +2.95 sin1.65 j
r = 2.95 x 0.07912 î + 2.95 x 0.9968 j
r = 0.233404 î + 2.94056j
b) Angular Displacement = θ₀ = 9.10 rad
9.10 radian = 180/π× 9/10 degree
= 521.391°
= 521.391°- 360°
=161.391°
This will lie in the second quadrant.
Angle made with the positive x-axis =161.391°
c) Velocity
v = ω R
v = 1.65 x 2.95
v = 4.8675 m/s
d) It is moving in a direction making 161.391° with positive x-direction.
e) Acceleration will be centripetal acceleration.= v²/R
=(4.8675)² / 2.95
=23.6925562 / 2.95
=8.031 m/s²
f) Position, Velocity, and Acceleration graph:
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How do I connect with my higher self?
Answer:
Create space
Watch your breath
Watch your thoughts
Be gentle with yourself
Affirm what you want
Explanation:
Object A has the same mass as Object B but Object A is traveling faster. What can we say about the momentum of Object A compared to that of Object B?
Answer:
Object A's momentum is larger
Explanation:
as the formula for momentum goes:
P = M * V
where P is momentum, M is mass, V is velocity.
so where Va (Object A's velocity) is larger than Vb (Object B's velocity) we get:
( i ) Va > Vb
as the masses of both objects are equal, we mark:
( ii) Ma = Mb = M
we multiply both sides of ( i ) by M to get:
( iii ) Va × M > Vb × M
and we finally get:
( iv ) Pa > Pb
Answer: Object A's momentum of larger than compared to that of object B.
Explanation:
A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car moves 30.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car.
(a) the speed v
______m/s
(b) the horizontal force exerted on the car (Enter the magnitude.)
_______N
The horizontal force applied is 160 N while the velocity is 2.03 m/s.
What is the speed of the car?The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres apart 53696.25N and the minimum number of horses required is 37 .
( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 )
(1)The contact area between the hemispheres is (π x 0.430²) = 0.5805m²
Pressure difference = (940 - 15) = 925millibars.
(925 x 100) = 92,500N/m^2.
(92500 x 0.5805) = 53696.25N. is the force required to part the hemispheres.
(2)[tex]\frac{53696.25N}{1450N}[/tex] = 37 horses for each side .
37 + 37 = 74 horses will be required.
Force is something which can change the motion of an object, stop it or move it, change its shape or size with it. There are two types of forces, contact forces and non-contact forces. Here, it is a contact force at first, then when the horses come it becomes non-contact force.
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200 g of water is heated and its temperature goes from 280 K to 300 K. What was the change in temperature for this process?
A. 280 K
B. 20 F
C. 20 K
D. 300 K
The change in temperature is 20 kelvin
What is Temperature?
Temperature can simply be described as how hot or how cold an object is at a particular period in time. The unit of temperature is Kelvin.
The formula for calculating change in temperature is
Final temperature - Initial temperature
Final temperature = 300 kelvin
Initial temperature= 280 kelvin
Change in temperature= 300-280
= 20 kelvin
Thus the change in temperature for this process is 20 kelvin
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The image shows a wheel that’s wound up and released. The wheel moves up and down as shown. Identify the position of the wheel when its potential energy is greatest.
The highest point of the wheel is the position of the wheel when its potential energy is greatest.
At what position of the wheel potential energy is greatest?The position of the wheel when its potential energy is greatest when it is at the highest point because potential energy depends on the height of an object. If the object is at more height then it has more potential energy and vice versa.
So we can conclude that the highest point of the wheel is the position of the wheel when its potential energy is greatest.
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A particle with a mass of 1.98×10^−4 kg carries a negative charge of -3.50×10^−8 C. The particle is given an initial horizontal velocity that is due north and has a magnitude of 3.78×10^4 m/s.
a) What is the magnitude of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?
b) What is the direction of the minimum magnetic field?
(a) The magnitude of the minimum magnetic field is 1.467 T.
(b) The direction of the minimum magnetic field will be eastwards.
Minimum magnetic fieldF = qvB
mg = qvB
B = mg/qv
where;
B is magnetic fieldv is velocitym is mass of the particleq is charge of the particleB = (1.98 x 10⁻⁴ x 9.8) / (3.5 x 10⁻⁸ x 3.78 x 10⁴)
B = 1.467 T
Direction of the magnetic fieldThe magnetic field will be directed towards the negative charge. Thus, the direction of the minimum magnetic field will be eastwards.
Thus, the magnitude of the minimum magnetic field is 1.467 T and the direction is eastwards.
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When using a stream table in a classroom setting what are three factors that can be controlled?
Answer:
These include the slope of the land, the nature of the land surface, the placement of dams, and the direction of topsoil disturbance as created by farming activities. Materials: Students should work in groups of 3 or 4, or as materials allow.Explanation:
Hope this helpsA 550-g squirrel with a surface area of 945 cm2 falls from a 4.0-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the squirrel can be approximated as a rectanglar prism with cross-sectional area of width 11.6 cm and length 23.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the ground. Give the squirrel's terminal velocity, not it's velocity as it hits the ground.)
m/s
What will be the velocity of a 55.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
m/s
The velocity of a 55.0-kg person hitting the ground, is mathematically given as
vt=39.5983m/s
What will be the velocity of a 55.0-kg person hitting the ground, assuming no drag contribution in such a short distance?Generally, the equation for is mathematically given as
mass of squirrel,
[tex]m=550 \mathrm{~g}\\\\Surface area, $A=945 \mathrm{~cm}^{2}=88 \times 10^{-3}$\\\\Height, $h-4 \mathrm{~m}$\\[/tex]
Terminal velocity is given by:
[tex]$v_{i}=\sqrt{\frac{2 m g}{\rho A C}}$[/tex]
where \rho is the density of fluid that is falling and it is given by
[tex]$\rho=\frac{m}{V}$[/tex]
since, volume =area * height
[tex]^{\rho=} \frac{0.55 \mathrm{Kg}}{0.0945 \mathrm{~m}^{2} \times 4.0 \mathrm{~m}}\\\\$\rho=0.1455 \mathrm{Kg} / \mathrm{m}^{3}$[/tex]
A is the surface area of squirrels.
C is the drag coefficient.
The surface area facing the fluid is given by:
[tex]A_{f}=\frac{0.0945 \mathrm{~m}^{2}}{2} \\\\\\ A_{f}=0.04725 \mathrm{~m}^{2}[/tex]
so, terminal velocity is :
[tex]$v_{t}=\sqrt{\frac{2 \times 0.55 \mathrm{Kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2}}{0.1455 \mathrm{Kg} / \mathrm{m}^{3} \times 0.04725 \mathrm{~m}^{2} \times 1}}$[/tex]
Vt=39.5983
In conclusion, the terminal velocity of the squirrel is 39.5983m/s
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What is the symbol for a variable resistor?
A. Horizontal wire
B. 2 parallel wires
C. Line with a zig zag pattern with a diagonal line drawn across it
D. Circle with an X through it
The symbol for a variable resistor is a line with a zig zag pattern with a diagonal line drawn across it.
What is a variable resistor ?
A resistor whose electric resistance value may be altered is known to as a variable resistor. A variable resistor, which works usually by sliding a contact (wiper) over a variable resistor, is basically an electro-mechanical transducer.
The flow of electrical current is restricted by a resistor. The resistivity of a fixed resistor is steady. By adjusting a slider's position, the resistance of this resistor can be altered. Some volume controls and dimmer switches employ variable resistors.
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Even if all stars were the same distance from Earth, their absolute magnitude and
apparent magnitude would be very different.
True
False
Answer: True
hope this helps!
a)Calculate the effective value of g, the acceleration of gravity, at 7900 m above the Earth's surface.
b)Calculate the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.
The solution for the acceleration of gravity is given as
[tex]g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex][tex]g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]This is further explained below.
What is the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.?Generally,
Mass of earth [tex]$M=5.97 \times 10^{24} \mathrm{~kg}$[/tex]
Radius of earth [tex]$R=6371 \mathrm{~km}$[/tex]
Gravitational const. [tex]$G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$[/tex]
height [tex]$h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$[/tex]
[tex]R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}[/tex]
In conclusion, acceleration due to gravity at this point will be
[tex]g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex]
for [tex]$h_{2}=7900 \mathrm{~km}$[/tex]
[tex]R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]
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The noise from a power mower was measured at 104 dB. The noise level at a rock concert was measured at 121 dB. Find the ratio of the intensity of the rock music to that of the power mower.
Based on the calculations, the ratio of the intensity of Ir to Ip is equal to 50.12.
How to find the ratio of the intensity?Let the intensity of the power mower be Ip.Let the intensity of the rock music be Ir.Mathematically, sound intensity level can be calculated by using this formula:
[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]
Where:
I is the intensity of the sound.
Making I the subject of formula, we have:
I = Io × [tex]10^{\frac{\beta }{10} }[/tex]
For Ip, we have:
Ip = Io × [tex]10^{\frac{104 }{10} }[/tex]
Ip = Io × [tex]10^{10.4}[/tex]
For Ir, we have:
Ir = Io × [tex]10^{\frac{121 }{10} }[/tex]
Ir = Io × [tex]10^{12.1}[/tex]
Now, we can find the ratio of the intensity:
Ir/Ip = Io × [tex]10^{12.1}[/tex]/Io × [tex]10^{10.4}[/tex]
Ir/Ip = [tex]10^{12.1}[/tex]/[tex]10^{10.4}[/tex]
Ir/Ip = [tex]10^{12.1 -10.4}[/tex]
Ir/Ip = [tex]10^{1.7}[/tex]
Ir/Ip = 50.12.
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An air-filled 39.1 μH solenoid has a length 4.0 cm and a cross-sectional area 0.60 cm2. How many turns are in this solenoid?
a.21000
b.144
c.12
d.120
e.1200
No. of turns in the solenoid is an option (b) 144.
The self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current.
Self-inductance of solenoid = 39.1 μH
= 39.1 × [tex]10^-^6[/tex] H
Length of the solenoid = 4.0 cm
Cross-sectional area = 0.60 cm²
Expression for the self-inductance of a coil ;
L = µ₀N²A / [tex]l[/tex]
where,
L = Self- Inductance
N = No. of turns.
A = Cross-sectional area
[tex]l=[/tex] Length of the solenoid
L =( 4π × [tex]10^-^7[/tex] × N² × 0.60 ) / 4.0
39.1× 4.0 / 4π × [tex]10^-^7[/tex] × 0.60 = N²
N² = 2.07 × [tex]10^6[/tex]
N = 144
Therefore, the no. of turns of the solenoid is 144.
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An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
a) Find the magnitude of the magnetic field this electron produces at the point A .
b) Find the magnitude of the magnetic field this electron produces at the point B .
c) Find the magnitude of the magnetic field this electron produces at the point C .
d) Find the magnitude of the magnetic field this electron produces at the point D
Hi there!
We can use Biot-Savart's Law for a moving particle:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }[/tex]
B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)
q = charge of particle (1.6 × 10⁻¹⁹ C)
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
r = distance from particle (2.10 μm)
There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }[/tex]
Where 'θ' is the angle between the velocity and radius vectors.
a)
To find the angle between the velocity and radius vector, we find the complementary angle:
θ = 90° - 60° = 30°
Plugging 'θ' into the equation along with our other values:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }[/tex]
[tex]B = \boxed{7.07 *10^{-10} T}[/tex]
b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.
c)
In this instance, the radius vector and the velocity vector are perpendicular so
'θ' = 90°.
[tex]B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}[/tex]
d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.
Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.
[tex]\boxed{B = 0 T}[/tex]
A tennis player tosses a tennis ball straight up and then catches it after 1.64 s at the same height as the point of release.
(a) What is the acceleration of the ball while it is in flight?
magnitude ______ m/s2
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(b) What is the velocity of the ball when it reaches its maximum height?
magnitude _______________ m/s
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(c) Find the initial velocity of the ball.
______m/s upward
(d) Find the maximum height it reaches.
___________m
A. The acceleration of the ball while it is in flight?
magnitude is 0 m/s² (magnitude is zero)
B. The velocity of the ball when it reaches its maximum height is 0 m/s (magnitude is zero)
C. The initial velocity of the ball 8.036 m/s upward
D. The maximum height reached by the ball is 3.29 m
A. How to determine the acceleration in the flightConsidering that the ball came to rest after 1.64s, it means the entire acceleration of the flight is zero as the ball was not moving in any form again.
B. How to determine the velocity at maximum heightAt maximum height, the velocity of the ball is zero as it no longer has magnitude to keep going upwards. Hence the ball begins to ball down.
C. How to determine the initial velocityAcceleration due to gravity (g) = 9.8 m/s²Final velocity (v) = 0 m/sTime of flight (T) = 1.64 sTime to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 sInitial velocity (u) =?v = u - gt (since the ball is going against gravity)
0 = u - (9.8 × 0.82)
0 = u - 8.036
Collect like terms
u = 0 + 8.036
u = 8.036 m/s upward
D. How to determine the maximum height reached by the ballTime to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 sAcceleration due to gravity (g) = 9.8 m/s²Maximum height (h)h = ½gt²
h = ½ × 9.8 × 0.82²
h = 3.29 m
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What is double-slit experiment?
The double-slit experiment shows that both matter and light can exhibit properties of conventionally defined waves and particles.
The double-slit experiment is a part of a class of "double path" experiments in which a wave is split into two separate waves that later combine to form a single wave (the wave is typically composed of many photons and is better known as a wave front, which should not be confused with the wave properties of the individual photon).
Isaac Newton's corpuscular theory of light, which had previously prevailed as the accepted explanation of light transmission in the 17th and 18th centuries, was defeated by double-slit experiment , which was conducted in the early 1800s.
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Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
The minimum necessary height of the IV bag above the position of the needle, if the density of the solution is 1.308kg/L, and the pressure in the vein is 35.7 mmHg is, 2.78m.
To find the answer, we need to know more about the pressure exerted by a liquid column.
How to find the height of IV bag above the position of needle?Consider a liquid of density ρ contained in a vessel of height h, the pressure exerted by the liquid column at the bottom of the vessel is given by ,[tex]P=[/tex] ρgh
In our question, it is given that,[tex]density=1.308 kg/L\\\\P=35.7 mmHg.\\[/tex]
Thus, the height of the bag h will be,[tex]h=\frac{P}{density*g} =\frac{35.7}{1.308*9.8}\\\\ h=2.78 m[/tex]
Thus, we can conclude that, the height of the Iv bag is 2.78 m above the position of needle.
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If the solution has a density of 1.308 kg/L and the vein pressure is 35.7 mmHg, the minimum height of the IV bag above the needle position is 2.78 m.
We need to learn more about the pressure that a liquid column exerts in order to determine the solution.
How can I determine the height of IV bag above where the needle is?The pressure exerted by the liquid column at the bottom of the vessel is given by for a liquid with density enclosed in a vessel of height h.[tex]P=[/tex] ρgh
In our inquiry, it is assumed that,[tex]Density=1.308kg/L\\P=35.7mmHg[/tex]
As a result, the bag's height will be,[tex]h=\frac{P}{density*g}\\\\ h=2.78m[/tex]
As a result, we may say that the I.V. bag is 2.78 m above the ground.
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In Milikan’s experiment, a drop of radius of 1.64 μm and density 0.851 g/cm3 is suspended in the lower chamber when a downward-pointing electric field of 1.92 * 105 N/C is applied.
What is the weight of the drop?
Find the charge on the drop, in terms of e.
How many excess or deficit electrons does it have?
(a) The weight of the drop is 1.54 x 10⁻²⁵ N,
(b) The the charge on the drop, in terms of e is 5 x 10⁻¹²e and
(c) The excess electrons is 5 x 10⁻¹² electron.
Weight of the dropThe weight of the drop is calculated as follows;
Volume of the drop; V = ⁴/₃πr³
V = ⁴/₃π(1.64 x 10⁻⁶)³ = 1.845 x 10⁻¹⁷ m³
mass = density x volume
mass = 0.851 g/cm³ x 1.845 x 10⁻²³ cm³ = 1.572 x 10⁻²³ g = 1.57 x 10⁻²⁶ kg.
Weight = 1.57 x 10⁻²⁶ kg x 9.8 m/s² = 1.54 x 10⁻²⁵ N.
Charge on the dropq = F/E
q = (1.54 x 10⁻²⁵ N)/(1.92 x 10⁵)
q = 8.01 x 10⁻³¹ C
1.6 x 10⁻¹⁹ C = 1e
8.01 x 10⁻³¹ C = ?
= 5 x 10⁻¹²e
Excess electron on the drop1.6 x 10⁻¹⁹ C ------- 1 electron
8.01 x 10⁻³¹ C ------- ?
= 5 x 10⁻¹² electron
Thus, the weight of the drop is 1.54 x 10⁻²⁵ N, the the charge on the drop, in terms of e is 5 x 10⁻¹²e and the excess electrons is 5 x 10⁻¹² electron.
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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet? (Planet-X doesn't have any atmosphere.)
2. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
3. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
(a) The speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.
(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.
(c) The radius of the synchronous orbit of a satellite is 69,801 km .
Speed of the satellitev = √GM/r
where;
M is mass of the planetr is radius of the planetv = √[(6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 7,338.93 m/s
Escape velocity of the satellitev = √2GM/r
v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 10,378.82 m/s
Speed of the satellite at the given periodv = 2πr/T
r = vT/2π
r = (7,338.93 x 16.6 x 3600 s) / (2π)
r = 69,801 km
Thus, the speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.
The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.
The radius of the synchronous orbit of a satellite is 69,801 km .
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The three displacement vectors have magnitude of A=5.00m,B=5.00m and C=4.00m.Find the resultant (magnitude and directional angle) of the three vectors?
3.00 m is the magnitude of the resultant vector
137.1° is the directional angle of the resultant vector.
1) To find the magnitude of the resultant
Discover each vector's individual components first.
We must consider the signs of the components because the angles in the figure are measured in various ways.
In this case, both the y component of vector C and the x component of vector A are negative.
The vectors' components are as follows, using a little trigonometry:
Magnitude of A = 5 m
[tex]A_{x}[/tex] = - (5.00m) cos20° = -5 ×0.408 = -4.698 m
[tex]A_{y}[/tex] = + (5.00m)sin20° = +5 × 0.342 = +1.710 m
Magnitude of B = 5m
[tex]B_{x}[/tex] = +(5.00m)cos60° = 5 × 0.5 = +2.5m
[tex]B_{y}[/tex] = +(5.00m)sin60° = 5 × √3/2 = +4.33 m
Cx = 0
Cy = -4.00m
The sum of all three vectors, which we refer to as R, produces components.
Rₓ = Aₓ + Bₓ + Cₓ
= -4.698 + 2.5 + 0
= -2.198 m
[tex]R_{y}[/tex] = [tex]A_{y} + B_{y} + C_{y}[/tex]
= +1.71 +4.33 - 4.00
= 2.040 m
R = [tex]\sqrt{R_{x^{2} }+R_{y^{2} }[/tex]
= [tex]\sqrt{(-2.198)^{2 } + (2.040)^{2} }[/tex]
= 3.00 m
2) To find the directional angle of resultant
tan θ = 2.040/-2.198 = -0.928
θ = -42.9°
Such a vector would be in the so-called "fourth quadrant," as it is well known. However, we discovered that R has a negative x component and a positive y component, indicating that such a vector must reside in the "second quadrant.
The calculator accidentally returned an angle that is 180 degrees off, thus we must add 180 degrees to the naïve angle in order to get the right angle.
Therefore, R's actual direction is determined by-
θ = -42.9° + 180° = 137.1°
Hence, the magnitude of resultant vector is 3.00 m and directional angle is 137.1°
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The following table lists the speed of sound in various materials. Use this table to answer the question.
Substance Speed (m/s)
Glass 5,200
Aluminum 5,100
Iron 4,500
Copper 3,500
Salt water 1,530
Fresh water 1,500
Mercury 1,400
Hydrogen at 0°C 1,284
Ethyl Alcohol 1,125
Helium at 0°C 965
Air at 100°C 387
Air at 0°C 331
Oxygen at 0°C 316
Sound will travel fastest in air at _____.
-5°C
0°C
10°C
15°C
Sound will travel fastest in air at 15°C.
Speed of sound in air
The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;
Air at 100°C 387 m/s
Air at 0°C 331 m/s
From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.
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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×103 km, mass of Earth = 5.98×1024 kg, G = 6.67×10-11 Nm2/kg2.)
The speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.
Speed of the satellitev = √(GM/r)
where;
G is universal gravitation constantM is mass of Earthr is radius of the satellitev = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/3.57 x 6.37x 10³)
v = 1.32 x 10⁵ m/s
Thus, the speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.
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Forces always act in ________. A. Solitude B. Pairs C. Unpredictable ways D. Interesting ways
I need help with my homework
C. The center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.
What is center of mass?The center of mass of an object is the unique point where the weighted relative position of the distributed mass sums to zero.
What is center of gravity?Center of gravity is the point from which the weight of a body or system may be considered to act.
Thus, the center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.
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A solid cylinder of uniform density of 0.85 g/cm3 floats in a glass of water tinted light blue by food coloring.
Its circular surfaces are horizontal. What effect will the following changes, each made to the initial system, have on X, the height of the upper surface above the water? The liquids added do not mix with the water, and the cylinder never hits the bottom.
1. The cylinder is replaced with one that has the same density and diameter, but with half the height.
2. Some of the water is removed from the glass.
3. A liquid with a density of 1.06 g/cm3 is poured into the glass.
4. The cylinder is replaced with one that has the same height and diameter, but with density of 0.83 g/cm3.
5. A liquid with a density of 0.76 g/cm3 is poured into the glass.
6. The cylinder is replaced with one that has the same density and height, but 1.5× the diameter.
Options are: Increase, Decrease, No change
The buoyant force acting on the cylinder is, [tex]Fb = \rho Ahg[/tex]. Here A is the cross-sectional area of the cylinder, h is the height of the cylinder, ρ is the density of the cylinder, and g is the acceleration due to gravity.
This buoyant force is also equal to the volume of the fluid displaced. [tex]Fb = \sigma h(A-x)g[/tex]. Here σ is the density of the fluid.
Equate the above two equations and solve for x.
[tex]\rho Ahg = \sigma A(h-x)g\\\rho h = \sigma h - \sigma x\\x = \frac{(\sigma - \rho)h}{\sigma}[/tex]
So, the distance x depends on the density of the fluid, density of the cylinder and the height of the cylinder.
1. The density of the cylinder is same and distance x is independent of the diameter of the cylinder. Therefor, there will be no change in the distance x. Hence, the correct answer is No change.
2. Now the height is changing keeping the density same. As the distance x is directly proportional to the height, the distance x will increase.
3. The density of the added liquid is greater than of the water and it does not mix with the water. So, the liquid will settle down and there will be no change in the distance x.
4. The density of the added liquid is less than that of the water and it does not mix with the water. So, the liquid will not settle down and the distance x will change. The change in distance x can be determined as follow:
[tex]\rho Agh = \sigma' Axg + \sigma A(h-x)g\\\rho h=\sigma' x + \sigma h - \sigma x\\x=(\frac{\sigma - \rho}{\sigma - \sigma'})h[/tex]
Here, σ' is the density of the added liquid.
From the above relation it is clear, that on adding the liquid of the density less than that of water, the denominator term become small ando so the value of x will increase.
5. On removing some of the water inside the glass, the height of the water column will decrease, but the value of x does not depend on the height of the water column. So, there will be no change in the distance x.
6. The density of the new cylinder is smaller than that of the earlier one. So, the numerator term will increase. Therefore, the value of x will increase.
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if you dip your hand in cold water after having dipped in warm water, will you feel the water colder than it actually is?
Answer:
Yes.
Explanation:
When you move your hand from the cold water to the “warmer” (room temp) water, one hand feels warm.
As you move your hand from the warm water to the “colder” (room temp) water, that hand feels colder.
Although both hands experience the last bowl of water at the same temperature, your brain senses two
separate sensations. So the water feels “warm” or “cold” relative to the water your hand was in previously.
The greater the difference in temperature, the easier it is to sense a difference.
What is grandfather Paradox?
A grandfather paradox is a situation where individual travels to the past and then introduces a change which affects or contradicts the present.
What is a grandfather paradox?A paradox is a situation or statement which involves two contradictions.
A grandfather paradox is a situation which is defined by the ability of an individual to travel to a time in the past usually before the birth of their grandfather and then introduces a change which affects or contradicts the present. For example, killing the grandfather to prevent their birth.
In conclusion, a grandfather paradox is is an event which contradicts the present as a result of a change done to the past.
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Motorcycle safety helmet extend the time of collision hence decreasing,
a:chance of collision
b:force acting
c: velocity
c:Impulse
Answer:
D. Impulse
Explanation: Hope this helps