Answer : Units of momentum are :
1. Kg m/s
2. N-s
Explanation:
The momentum of an object is given by the product of its mass and velocity with which it is moving. Mathematically, it is given by :
P = mv
Where
m is in kilogram
v is in m/s
Option (1) : N-m = It is not a unit of momentum. It includes the product of force and distance.
Option (2) : Kg s/m = It is again not a unit of momentum.
Option (3) : Kg m/s =
Since, p = mv
p = Kg × m/s
It can be the unit of momentum.
(4) Option (4) : N-s = The change in momentum is equal to the impulse applied on an object. It is given by the product of force and short duration of time. It can be the unit of momentum.
(5) Option (5) : Kg/m²/s² = It is not the unit of momentum.
Hence, the correct options are (c) and (d).
The quantities that are units of momentum among the options provided are C) kg m/s and D) N-s. The other options correspond to different physical quantities.
Explanation:The concept in question pertains to the momentum of an object, which, in physics, is a vector quantity defined as the product of an object's mass and its velocity. The standard international (SI) unit of momentum is kilogram meter per second (kg m/s).
Examining each giver option: A) Newton meter (N m) is a unit of work, not momentum. B) Kilogram second/meter (kg s/m) does not align with the definition of momentum. C) Kilogram meter/second (kg m/s), this is the correct SI unit for momentum. D) Newton-second (N-s) is also a correct unit for momentum as Newton is equivalent to kg m/s2. E) Kilogram meter2/second2 (kg m2/s2) is the unit for kinetic energy, not momentum.
So, C) kg m/s and D) N-s are the units of momentum among the given choices.
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Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as 1.55 × 105 m. Suppose further that a second group of engineers programmed the orbiter to go to 1.55 × 105 ft. What was the difference in kilometers between the two altitudes? How low did the probe go?
Answer:
108 km
Explanation:
The conversion factor between meters and feet is
1 m = 3.28 ft
So the second altitude, written in feet, can be rewritten in meters as
[tex]h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m[/tex]
or in kilometers,
[tex]h_2 = 47 km[/tex]
the first altitude in kilometers is
[tex]h_1 = 155 km[/tex]
so the difference between the two altitudes is
[tex]\Delta h = 155 km - 47 km = 108 km[/tex]
1. What do you need to change the momentum of a system?
2. What are the features of a typical modern running shoe? How does this change the ground reaction force during a heel strike when running with shoes compared to a heel strike when running without shoes?
3. When running with shoes how does the ground reaction force change from a heel strike run to a forefoot strike run?
A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field B⃗ =(1.63T)i^+(0.980T)j^?
Answer:
The magnitude and direction of the acceleration of the particle is [tex]a= 0.3296\ \hat{k}\ m/s^2[/tex]
Explanation:
Given that,
Mass [tex]m = 1.81\times10^{-3}\ kg[/tex]
Velocity [tex]v = (3.00\times10^{4}\ m/s)j[/tex]
Charge [tex]q = 1.22\times10^{-8}\ C[/tex]
Magnetic field [tex] B= (1.63\hat{i}+0.980\hat{j})\ T[/tex]
We need to calculate the acceleration of the particle
Formula of the acceleration is defined as
[tex]F = ma=q(v\times B)[/tex]
[tex]a =\dfrac{q(v\times B)}{m}[/tex]
We need to calculate the value of [tex]v\times B[/tex]
[tex]v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})[/tex]
[tex]v\times B=4.89\times10^{4}[/tex]
Now, put the all values into the acceleration 's formula
[tex]a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}[/tex]
[tex]a= -0.3296\ \hat{k}\ m/s^2[/tex]
Negative sign shows the opposite direction.
Hence, The magnitude and direction of the acceleration of the particle is [tex]a= 0.3296\ \hat{k}\ m/s^2[/tex]
The magnitude and direction of the particle’s acceleration produced by a uniform magnetic field [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^ is [tex]\bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]
Explanation:A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^?
A charged particle is a particle with an electric charge. Whereas electric charge is the matter physical property that causes to experience a force when placed in an electromagnetic field. Uniform magnetic field is the condition when magnetic field lines are parallel then magnetic force experienced by an object is same at all points in that field
From Newton's second law, the force is given by:
[tex]F=ma[/tex]
Magnetic force is
[tex]F= qv \times B[/tex]
[tex]ma = qv \times B[/tex]
[tex]a = \frac{qv \times B}{m}[/tex]
Subsituting with the givens above we get
[tex]a = \frac{(1.22 \times 10^{-8} C) (3 \times 10^{4} m/s) (1.63 T ) (\hat{j} \times \hat{i})}{1.81 \times 10^{-3} kg} = -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]
Therefore the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^ is [tex]\bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]
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When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 936 N and the drag force has a magnitude of 1032 N. The mass of the sky diver is 95.5 kg. Take upward to be the positive direction. What is his acceleration, including sign?
Answer: [tex]1.0052m/s^{2}[/tex]
Explanation:
Assuming there is only force in the y-component, the total net force [tex]F_{y}[/tex] acting on the parachute and the sky diver is:
[tex]F_{y}=F_{D}-W[/tex] (1)
Where:
[tex]F_{D}=1032N[/tex] is the drag force acting upwards
[tex]W=936N[/tex] is the weight of the sky diver acting downwards, hence with negative sign
Then:
[tex]F_{y}=1032N-936N=96N[/tex] (2) This is the total net force excerted on the system parachute-sky diver, and the fact it is positive means is upwards
Now, according Newton's 2nd Law of Motion the force is directly proportional to the mass [tex]m[/tex] and to the acceleration [tex]a[/tex] of a body:
[tex]F_{y}=m.a[/tex] (3)
Where [tex]m=95.5kg[/tex] is the mass of the diver.
Substituting the known values and finding [tex]a[/tex]:
[tex]a=\frac{F_{y}}{m}[/tex] (4)
[tex]a=\frac{96N}{95.5kg}[/tex] (5)
Finally:
[tex]a=1.0052m/{s^{2}}\approx 1m/s^{2}[/tex] This is the acceleration of the sky diver. Note it has a positive sign, which means its direction is upwards.
A rope attached to a load of 175 kg bricks Ilifts the bricks with a steady acceleration of 0.12.m/s^2 straight up. What is the tension in the rope? (a)2028N (b)1645 N (c) 1894 N (d) 1976 N (e) 1736 N (f) 1792 N
Answer:
Tension, T = 1736 N
Explanation:
It is given that,
Mass of bricks, m = 175 kg
A rope is attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12 m/s² in vertically upwards direction. let T is the tension in the rope. Using second equation of motion as :
T - mg = ma
T = ma + mg
T = m(a + g)
T = 175 kg ( 0.12 m/s² + 9.8 m/s² )
T = 1736 N
Hence, the tension in the wire is 1736 N.
Answer:
The tension in the rope is 1736 N.
(e) is correct option.
Explanation:
Given that,
Mass of bricks = 175 kg
Acceleration = 0.12 m/s²
Let T is the tension in the rope.
A rope attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12.m/s^2 in vertically upward direction.
Using equation of balance
[tex]T-mg=ma[/tex]
[tex]T=mg+ma[/tex]
[tex]T=175(9.8+0.12)[/tex]
[tex]T= 1736\ N[/tex]
Hence, The tension in the rope is 1736 N.
The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7300 L of blood. Assume that the work done by the heart is equal to the work required to lift that amount of blood a height equal to that of the average citizen of Atlantic Falls, approximately 1.6 m. The density of blood is 1050 kg/m3. What is the heart's power output in watts?
Answer:
1.39 W
Explanation:
The volume of blood is
[tex]V=7300 L = 7.3 m^3[/tex]
the density is
[tex]\rho = 1050 kg/m^3[/tex]
So the total mass of blood lifted in one day is
[tex]m=\rho V=(1050 kg/m^3)(7.3 m^3)=7665 kg[/tex]
So the total work done is:
[tex]W=mgh=(7665 kg)(9.8 m/s^2)(1.6 m)=1.2\cdot 10^5 J[/tex]
The total time taken is one day, so
[tex]t=24 h = 86400 s[/tex]
So the power output is
[tex]P=\frac{W}{t}=\frac{1.2\cdot 10^5 J}{86400 s}=1.39 W[/tex]
A chain link fence should be cut quickly with a
Answer: it should be cut with a chainsaw
Explanation:
A bolt cutter is usually the preferred tool to use to cut through a chain link fence quickly, taking into account the thickness and hardness of the chain link fence. Safety precautions should be taken while using such tools.
Explanation:To cut through a chain link fence quickly without undue strain, the preferred tool is typically a bolt cutter. Bolt cutters possess the strength and design needed to snip through metal links easily. They come in various sizes, and the size needed would depend on the thickness and hardness of the chain link fence. Ideally, a medium-sized bolt cutter would be used for a standard fence. However, it's advisable to wear protective gear while using such tools, as the cut metal links might be razor-sharp and could cause injuries.
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A test charge is A a very small negative charge with little miee B a point charge of q 100 C C a spbere of charge D. a very amall positive charge with little s
Answer:
D. a very small positive charge with little s
Explanation:
A test charge is a very small charge with positive value which do not disturb the electric field exist in the region
So test charge is to find out the strength of electric field that exist in the region.
If the magnitude of test charge is large then it will change the strength of the existing electric field and due to this the value of the force will be altered.
So here in this case the test charge must be small as well as it must be positive nature
If a converging lens forms a real, inverted image 24.0 cm to the right of the lens when the object is placed 48.0 cm to the left of a lens, determine the focal length of the lens
Answer:
Focal length, f = 16 cm
Explanation:
Image distance, v = 24 cm
Object distance, u = -48 cm
We need to find the focal length of the lens. It can be determined using the lens formula as :
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}[/tex]
[tex]\dfrac{1}{24\ cm}-\dfrac{1}{(-48\ cm)}=\dfrac{1}{f}[/tex]
f = 16 cm
So, the focal length of the converging lens is 16 cm. Hence, this is the required solution.
Answer:
f = 16 cm
Explanation:
If a converging lens forms a real, inverted image 24.0 cm to the right of the lens when the object is placed 48.0 cm to the left of a lens, the focal length of the lens is 16 cm.
A traveler pulls on a suitcase strap at an angle 36 above the horizontal with a force of friction of 8 N with the floor. If 752 J of work are done by the strap while moving the suitcase a horizontal distance of 15 m, what is the tension in the strap?
Answer:
71.8 N
Explanation:
T = Tension force in the strap
W = net work done = 752 J
f = force of friction = 8 N
d = displacement = 15 m
θ = angle between tension force and horizontal displacement = 36 deg
work done by frictional force is given as
W' = - f d
Work done by the tension force is given as
W'' = T d Cos36
Net work done is given as
W = W' + W''
W = T d Cos36 - f d
752 = T (15) Cos36 - (8) (15)
T = 71.8 N
A horizontal desk surface measures 1.7 m by 1.0 m. If the Earth's magnetic field has magnitude 0.42 mT and is directed 68° below the horizontal, what is the magnetic flux through the desk surface?
Answer:
The magnetic flux through the desk surface is [tex]6.6\times10^{-4}\ T-m^2[/tex].
Explanation:
Given that,
Magnetic field B = 0.42 T
Angle =68°
We need to calculate the magnetic flux
[tex]\phi=BA\costheta[/tex]
Where, B = magnetic field
A = area
Put the value into the formula
[tex]\phi=0.42\times10^{-3}\times1.7\times1.0\cos22^{\circ}[/tex]
[tex]\phi=0.42\times10^{-3}\times1.7\times1.0\times0.927[/tex]
[tex]\phi=6.6\times10^{-4}\ T-m^2[/tex]
Hence, The magnetic flux through the desk surface is [tex]6.6\times10^{-4}\ T-m^2[/tex].
An electron moves in a straight path with a velocity of +500,000 m/s. It undergoes constant acceleration of +500,000,000,000 meters per second per second. How far will the electron have traveled when it reaches a velocity of +675,000 m/s? A. +0.000000175 m B. +0.00000035 m C. +0.21 m D. +0.41 m
Answer:
Distance travelled, d = 0.21 m
Explanation:
It is given that,
Initial velocity of electron, u = 500,000 m/s
Acceleration of the electron, a = 500,000,000,000 m/s²
Final velocity of the electron, v = 675,000 m/s
We need to find the distance travelled by the electron. Let distance travelled is s. Using third equation of motion as :
[tex]v^2-u^2=2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{(675000\ m/s)^2-(500000\ m/s^2)^2}{2\times 500000000000\ m\s^2}[/tex]
s = 0.205 m
or
s = 0.21 m
So, the electron will travel a distance of 0.21 meters. Hence, this is the required solution.
A 10 Ω resistor is connected to a 120-V ac power supply. What is the peak current through the resistor?
Answer:
Peak current = 16.9 A
Explanation:
Given that
RMS voltage = 120 Volts
[tex]V_{rms} = 120 V[/tex]
AC is connected across resistance
[tex]R = 10 ohm[/tex]
now by ohm's law
[tex]V = i R[/tex]
[tex]120 = i (10)[/tex]
[tex]i_{rms} = \frac{120}{10} = 12 A[/tex]
now peak value of current will be given as
[tex]i_{peak} = \sqrt{2} i_{rms}[/tex]
[tex]i_{peak} = \sqrt2 (12) = 16.9 A[/tex]
What is the magnitude of the acceleration vector which causes a particle to move from velocity −5i−2j m/s to −6i+ 7j m/s in 8 seconds. Answer in m/s.
Answer:
Acceleration, [tex]a=\dfrac{1}{8}(-i+9j)\ m/s^2[/tex]
Explanation:
Initial velocity of a particle in vector form, u = (-5i - 2j) m/s
Final velocity of particle in vector form, v = (-6i + 7j) m/s
Time taken, t = 8 seconds
We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{(-6i+7j)\ m/s-(-5i-2j)\ m/s}{8\ s}[/tex]
[tex]a=\dfrac{(-i+9j)}{8\ s}\ m/s^2[/tex]
or
[tex]a=\dfrac{1}{8}(-i+9j)\ m/s^2[/tex]
Hence, the value of acceleration vector is solved.
If the speed of light in a vaccum is c, the speed of light in a medium like glass with an index of refraction of 1.5 is : (a) 3c/2 (b) 3c (c) 2c/3 (d) 9c/4 (e) 4c/9 Please explain in detail why it is the answer you have chosen
Answer:
The speed of light in the medium is [tex]\dfrac{2c}{3}[/tex]
(c) correct option.
Explanation:
Given that,
Speed of light in vacuum = c
Refraction index = 1.5
We need to calculate the value of speed of light in the medium
The refractive index is equal to the speed of light in vacuum divide by the speed of light in medium.
Using formula of refractive index
[tex]\mu = \dfrac{c}{v}[/tex]
[tex]v=\dfrac{c}{\mu}[/tex]
Where, c = speed of light in vacuum
v = speed of light in medium
Put the value into the formula
[tex]v=\dfrac{c}{1.5}[/tex]
[tex]v=\dfrac{2c}{3}[/tex]
Hence, The speed of light in the medium is [tex]\dfrac{2c}{3}[/tex]
A sample of a material has 200 radioactive particles in it today. Your grandfather measured 400 radioactive particles in it 60 years ago. How many radioactive particles will the sample have 60 years from today?
Answer:
Amount of radioactive particles left after 60 years = 100 particles.
Explanation:
Amount of radioactive particles before 60 years = 400
Amount of radioactive particles present today = 200
That is radio active particles reduced to half. That is 60 years is half life of this radio active material.
After 60 years this 200 radio active particles will reduce to half.
Amount of radioactive particles left after 60 years = 0.5 x 200 = 100 particles.
Final answer:
The sample will have 100 radioactive particles remaining 60 years from today, based on the half-life of the material being 60 years.
Explanation:
The question concerns the concept of radioactive decay and specifically the half-life of a radioactive sample. In this case, the sample's quantity of radioactive particles was observed to decrease from 400 to 200 over a span of 60 years. Thus, the half-life of the material is 60 years, which is the time it takes for half of the radioactive atoms (parent nuclei) to decay into their decay products (daughter elements).
Given that the sample has 200 particles today, we can predict that in another 60 years, the number of radioactive particles will again be halved. Therefore, after 60 years from today, we expect there to be 100 radioactive particles remaining in the sample.
A 12.5 kg box sliding on a frictionless flat surface tuns into a fixed spring which compresses at a distance x= 14.1 cm. the spring constant is 94.5 kN/m Find the initial speed of the box
Answer:
12.3 m/s
Explanation:
m = mass of the box sliding on frictionless flat surface = 12.5 kg
x = compression of the spring = 14.1 cm = 0.141 m
k = spring constant = 94.5 kN/m = 94500 N/m
v = initial speed of the box
Using conservation of energy
Kinetic energy of the box = Spring potential energy
(0.5) m v² = (0.5) k x²
m v² = k x²
Inserting the values
(12.5) v² = (94500) (0.141)²
v = 12.3 m/s
If a ball is thrown vertically upward with a velocity of 144 ft/s, then its height after t seconds is s = 144t − 16t2. (a) What is the maximum height reached by the ball? ft (b) What is the velocity of the ball when it is 320 ft above the ground on its way up? (Consider up to be the positive direction.) ft/s What is the velocity of the ball when it is 320 ft above the ground on its way down? ft/s
(a) 168.2 ft/s
The vertical position of the ball is given by
[tex]s = 144t - 16t^2[/tex]
where t is the time.
By differentiating this expression, we find the velocity:
[tex]v = 144-32 t[/tex]
The maximum height is reached when the velocity is zero, so:
[tex]0 = 144 - 32 t[/tex]
From which we find
[tex]t = \frac{144}{32}=4.5 s[/tex]
And substituting this value into the equation for s, we find the maximum height:
[tex]s = 144(4.5 s)-16(4.5 s)^2=324 ft[/tex]
(b) 16 ft/s
We want to find the velocity of the ball when the position of the ball is
s = +320 ft
Substituting into the equation for the position,
[tex]320 = 144t-16t^2[/tex]
[tex]16t^2 -144t +320 = 0[/tex]
Solving for t, we find two solutions:
t = 4 s
t = 5 s
The first one corresponds to the instant in which the ball is still on its way up: Substituting into the equation for the velocity, we find the velocity of the ball at that time
[tex]v = 144 - 32 t=144- 32(4 s)=16 ft/s[/tex]
(c) -16 ft/s
Now we want to find the velocity of the ball when the position of the ball is
s = +320 ft
but on its way down. In the previous part, we found
t = 4 s
t = 5 s
So the second time corresponds to the instant in which the ball is at s = 320 ft but on the way down.
Substituting t = 5 s into the equation for the velocity, we find:
[tex]v = 144 - 32 t=144- 32(5 s)=-16 ft/s[/tex]
And the negative sign means the direction is downward.
The answers for the ball thrown vertically upward with a velocity of 144 ft/s and with a height after t seconds of s = 144t - 16t² are:
a) The maximum height reached by the ball is 324 ft.
b) The velocity of the ball when it is 320 ft above the ground on its way up is 16 ft/s.
c) The velocity of the ball when it is 320 ft above the ground on its way down is -16 ft/s.
a) The maximum height reached by the ball can be calculated with the given equation:
[tex] s = 144t - 16t^{2} [/tex] (1)
Where:
s: is the height
t: is the time
We can find the time with the following equation:
[tex] v_{f} = v_{i} - gt [/tex] (2)
Where:
[tex] v_{f} [/tex]: is the final velocity = 0 (at the maximum height)
[tex] v_{i} [/tex]: is the initial velocity = 144 ft/s
g: is the acceleration due to gravity = 32 ft/s²
Solving equation (2) for t and entering into equation (1), we can find the maximum height:
[tex]s = 144t - 16t^{2} = 144(\frac{v_{i}}{g}) - 16(\frac{v_{i}}{g})^{2} = 144(\frac{144 ft/s}{32 ft/s^{2}}) - 16(\frac{144 ft/s}{32 ft/s{2}})^{2} = 324 ft[/tex]
Hence, the maximum height is 324 ft.
b) To find the velocity of the ball when it is 320 ft above, we can use the following equation:
[tex] v_{f}^{2} = v_{i}^{2} - 2gs [/tex]
[tex]v_{f}^{2} = (144 ft/s)^{2} - 2*32 ft/s^{2}*320 ft[/tex]
The above equation has two solutions:
[tex]v_{f_{1}} = 16 ft/s[/tex]
[tex]v_{f_{2}} = -16 ft/s[/tex]
Since the question is for the velocity of the ball on its way up and considering the way up as the positive direction, the answer is the positive value [tex]v_{f_{1}} = 16 ft/s[/tex].
c) The velocity of the ball when it is 320 ft above the ground on its way down is -16 ft/s (we take the negative value calculated above, [tex] v_{f_{2}}[/tex]). We consider the way down as the negative direction.
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g a stone with mass m=1.60 kg IS thrown vertically upward into the air with an initial kinetic energy of 470 J. the drag force acting on the stone throughout its flight is constant, independent of the velocity of the stone, and has a magnitude of 0.900 N. what is the maximum height reached by the stone?
Answer:
Height reached will be 28.35 m
Explanation:
Here we can use the work energy theorem to find the maximum height
As we know by work energy theorem
Work done by gravity + work done by friction = change in kinetic energy
[tex]-mgh - F_f h = 0 - \frac{1}{2}mv_i^2[/tex]
now we will have
[tex]-1.60(9.8)(h) - 0.900(h) = - 470[/tex]
[tex]-16.58 h = -470[/tex]
[tex]h = 28.35 m[/tex]
so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction
You are driving directly behind a pickup truck, going at the same speed as the truck. A crate falls from the bed of the truck to the road. (a) Will your car hit the crate before the crate hits the road if you neither brake nor swerve? (b) During the fall, is the horizontal speed of the crate more than, less than, or the same as that of the truck?
Final answer:
The crate will hit the road before your car hits the crate. The horizontal speed of the crate is the same as that of the truck.
Explanation:
(a) If you are driving directly behind a pickup truck at the same speed and neither brake nor swerve, the crate will hit the road before your car hits the crate. This is because the crate and your car are both traveling at the same horizontal speed, and the crate will have a shorter distance to fall than your car would have to travel to reach the crate.
(b) During the fall, the horizontal speed of the crate is the same as that of the truck. This is because both the truck and the crate are moving at the same speed horizontally, and gravity acts only vertically on the falling crate.
Rubbing a balloon on your hair results in the balloon having a -1.93C charge. If you move the balloon 25 cm away from your hair, what is the electric field acting between the balloon and your hair? O 6.17 x 10 N/C O 2.78 x 10 N/C O 4.01 x 10s N/C O 4.67 x 10 N/C
Explanation:
It is given that,
Charge acquired on rubbing a balloon on your hair, q = -1.93 C
If you move the balloon 25 cm away from your hair, r = 25 cm = 0.25 m
Electric field acting between the balloon and your hair is given by :
[tex]E=\dfrac{kq}{r^2}[/tex]
k = electrostatic constant
[tex]E=\dfrac{9\times 10^9\ Nm^2/C^2\times 1.93\ C}{(0.25\ m)^2}[/tex]
[tex]E=2.779\times 10^{11}\ N/C[/tex]
or
[tex]E=2.78\times 10^{11}\ N/C[/tex]
Hence, this is the required solution.
A disk with a radius of R is oriented with its normal unit vector at an angle\Theta with respect to a uniform electric field. Which of the following represent the electric flux through the disk? A: E(πR^2)cosϕ B: E(πR^2)sinΘ C: E(πR^2)cosΘ D: E(2πR)sinΘ E: E(2πR)cosΘ F: E(πR^2)sinϕ
Answer:
option (A)
Explanation:
electric flux is defined as the number of electric field lines which crosses through any area.
It is given by
Ф = E . A (It is the dot product of electric field vector and area vector)
According to the question, the angle between electric filed vector and area vector is θ.
So, electric flux
Ф = E x π R^2 Cosθ
The electric flux through a disk in a uniform electric field is represented by E(πR^2)cosΘ, so the correct answer is C: E(πR²)cosΘ.
The question is asking about the electric flux through a disk when the disk's normal is oriented at an angle Θ with respect to a uniform electric field. Electric flux is given by the product of the electric field strength, the area through which the field is passing, and the cosine of the angle between the field and the normal to the surface. The formula for the electric flux through a surface is Φ = E * A * cos(Θ), where E is the electric field strength, A is the area of the surface, and Θ is the angle between the electric field and the normal to the surface. For a disk with radius R, the area is πR². Thus, the correct answer for the electric flux through the disk is C: E(πR²)cosΘ.
A car that is traveling in a straight line at 40 km/h can brake to a stop within 20 m. If the same car is traveling at 120 km/h what would be its stopping distance in this case? Assume the braking force is the same in both cases and ignore air resistance.
Answer:
180 m
Explanation:
Case 1.
U = 40 km/h = 11.1 m/s, V = 0, s = 20 m
Let a be the acceleration.
Use third equation of motion
V^2 = u^2 + 2 as
0 = 11.1 × 11.1 - 2 × a × 20
a = 3.08 m/s^2
Case 2.
U = 220 km/h = 33.3 m/s, V = 0
a = 3.08 m/s^2
Let the stopping distance be x.
Again use third equation of motion
0 = 33.3 × 33.3 - 2 × 3.08 × x
X = 180 m
A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at constant speed? Important: Assume that the pushing force is to the right
Answer:
(a) 446.88 N
(b) 241.08 N
Explanation:
m = 60 kg, μs = 0.760, μk = 0.410
(a) To just start the crate moving: the coefficient of friction is static.
F = μs m g
F = 0.760 x 60 x 9.8
F = 446.88 N
(b) To slide the crate with constant speed: the coefficient of friction is kinetic.
F = μk m g
F = 0.410 x 60 x 9.8
F = 241.08 N
Static friction is the friction that occurs in the body when the body is just about to move. The static friction force will be 446.88 N.While the kinetic friction force will be 241.08 N.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
m is the mass of crate = 60 kg
μs is the coefficient of static friction = 0.760
μk is the coefficient of kinetic friction= 0.410
(a) Static friction is the friction that occurs in the body when the body is just about to move.
[tex]\rm F_s = \mu_s m g\\\\\rm F_s = 0.760\times60\times9.8\\\\\rm F_s = 446.88 N[/tex]
Hence static friction force will be 446.8 N.
(b) When the body is moving in a straight and inclined plane the value of friction force acting on the body is known as kinetic friction.
[tex]\rm F_k = \mu_km g\\\\\rm F_k = 0.410\times60\times9.8\\\\\rm F_k= 241.08\; N[/tex]
Hence kinetic friction force will be 241.08 N.
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A hot air balloon is ascending at a rate of 12 m/s. when it is 80m above the ground, a package is dropped over the side of he passenger basket. What is the speed of the package just before it hits the ground?
Answer:
41.4 m/s
Explanation:
Consider downward direction of motion as negative
v₀ = initial velocity of the package as it is dropped over the side = 12 m/s
v = final velocity of the package just before it hits the ground
y = vertical displacement of the package = - 80 m
a = acceleration = - 9.8 m/s²
Using the kinematics equation
v² = v₀² + 2 a y
v² = 12² + 2 (- 9.8) (- 80)
v² = 144 + 1568
v = - 41.4 m/s
The negative sign indicates the downward direction of motion.
Hence the speed of package comes out to be 41.4 m/s
A rod 10.0 cm long is uniformly charged and has a total charge of -21.0 µC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 34.0 cm from its center.
The magnitude of the electric field is -1.39 x 10^6 N/C and it is directed inward.
Explanation:To determine the magnitude and direction of the electric field along the axis of the rod at a point 34.0 cm from its center, we can use the formula for the electric field due to a uniformly charged rod. The formula is given by:
E = (k * Q * L) / (x^2 * sqrt(L^2 + x^2))
where E is the electric field, k is the Coulomb's constant, Q is the total charge on the rod, L is the length of the rod, and x is the distance from the center of the rod to the point where we want to find the electric field.
Substituting the given values:
E = (9.0 x 10^9 Nm^2/C^2 * (-21.0 x 10^-6 C) * 0.10 m) / (0.34 m)^2 * sqrt((0.10 m)^2 + (0.34 m)^2) = -1.39 x 10^6 N/C
The negative sign indicates that the electric field is directed inward.
A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the ground? B) What is the maximum height of the ball?
Answer:
A) The ball hits the ground 74.45 m far from the hitting position.
B) Maximum height of the ball = 18.57 m
Explanation:
There are two types of motion in this horizontal and vertical motion.
We have velocity = 27 m/s at 45° above the horizontal
Horizontal velocity = 27cos45 = 19.09 m/s
Vertical velocity = 27sin45 = 19.09 m/s
Time to reach maximum height,
v = u + at
0 = 19.09 - 9.81 t
t = 1.95 s
So total time of flight = 2 x 1.95 = 3.90 s
A) So the ball travels at 19.09 m/s for 3.90 seconds.
Horizontal distance traveled = 19.09 x 3.90 = 74.45 m
So the ball hits the ground 74.45 m far from the hitting position.
B) We have vertical displacement
S = ut + 0.5 at²
H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m
Maximum height of the ball = 18.57 m
Diagnostic ultrasound of frequency 3.82 MHz is used to examine tumors in soft tissue. (a) What is the wavelength in air of such a sound wave? (b) If the speed of sound in tissue is 1650 m/s, what is the wavelength of this wave in tissue? (Take the speed of sound in air to be 343 m/s.)
Explanation:
It is given that,
Frequency of diagnostic ultrasound, f = 3.82 MHz = 3820 Hz
The speed of the sound in air, v = 343 m/s
(a) We need to find the wavelength in air of such a sound wave. Let it is given by λ₁
i.e. [tex]\lambda=\dfrac{v}{\nu}[/tex]
[tex]\lambda_1=\dfrac{343\ m/s}{3820\ Hz}[/tex]
[tex]\lambda_1=0.089\ m[/tex]
(b) If the speed of sound in tissue is 1650 m/s .
[tex]\lambda_2=\dfrac{v}{\nu}[/tex]
[tex]\lambda_2=\dfrac{1650\ m/s}{3820\ Hz}[/tex]
[tex]\lambda_2=0.43\ m[/tex]
Hence, this is the required solution.
A stone is dropped from the upper observation deck of a tower, 250 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the distance (in meters) of the stone above ground level at time t. h(t) = (b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.) s (c) With what velocity does it strike the ground? (Round your answer to one decimal place.) m/s (d) If the stone is thrown downward with a speed of 2 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.) s
(a) [tex]y(t)=250 - 4.9 t^2[/tex]
For an object in free-fall, the vertical position at time t is given by:
[tex]y(t) = h + ut - \frac{1}{2}gt^2[/tex]
where
h is the initial vertical position
u is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
t is the time
In this problem,
h = 250 m
u = 0 (the stone starts from rest)
So, the vertical position of the stone is given by
[tex]y(t) = 250 - \frac{1}{2}(9.8) t^2 = 250 - 4.9 t^2[/tex]
(b) 7.14 s
The time it takes for the stone to reach the ground is the time t at which the vertical position of the stone becomes zero:
y(t) = 0
Which means
[tex]y(t) = h - \frac{1}{2}gt^2=0[/tex]
So for the stone in the problem, we have
[tex]250 - 4.9 t^2 = 0[/tex]
Solving for t, we find:
[tex]t=\sqrt{\frac{250}{4.9}}=7.14 s[/tex]
(c) -70.0 m/s (downward)
The velocity of an object in free fall is given by the equation
[tex]v(t) = u - gt[/tex]
where
u is the initial velocity
g = 9.8 m/s^2 is the acceleration of gravity
t is the time
Here we have
u = 0
So if we substitute t = 7.14 s, we find the velocity of the stone at the time it reaches the ground:
[tex]v=0-(9.8 m/s^2)(7.14 s)=-70.0 m/s[/tex]
The negative sign means the direction of the velocity is downward.
(d) 6.94 s
In this situation, the stone is thrown downward with an initial speed of 2 m/s, so its initial velocity is
u = -2 m/s
So the equation of the vertical position of the stone in this case is
[tex]y(t) = h + ut - \frac{1}{2}gt^2=250 - 2t - 4.9 t^2[/tex]
By solving the equation, we find the time t at which the stone reaches the ground.
We find two solutions:
t = -7.35 s
t = 6.94 s
The first solution is negative, so it has no physical meaning, therefore we discard it. So, the time it takes for the stone to reach the ground is:
t = 6.94 s
An object in free fall, like a stone dropped from a tower, has its motion governed by the acceleration due to gravity. Using the physics of motion, we can find the height of the stone at any given time, the time it takes to reach the ground, the velocity it strikes the ground, and the time taken if it is initially thrown downward.
Explanation:To solve these types of questions, we need to use the physics of motion. In the case of an object in free fall like a stone dropped from a tower, the only force acting on it is gravity, which pulls it downwards.
(a) The formula h(t) = 250 - 1/2gt^2 represents the height of the stone above ground level at any time t. Here, g is the acceleration due to gravity (9.8 m/s^2).
(b) The stone will reach the ground when h(t) = 0. Solving the equation 250 - 1/2*9.8*t^2 = 0 gives t ≈ 7.18 seconds (rounded to two decimal places).
(c) The velocity v with which the stone strikes the ground can be found using v = gt. Substituting g = 9.8 m/s^2 and t = 7.18 s gives v ≈ 70.4 m/s (rounded to one decimal place).
(d) If the stone is thrown downward with initial velocity of 2 m/s, the equation for time becomes 250 -2t- 1/2*9.8*t^2 = 0. Solving this gives t ≈ 7.04 seconds (rounded to two decimal places).
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On an airplane's takeoff, the combined action of the air around the engines and wings of an airplane exerts a 8420-N force on the plane, directed upward at an angle of 69.0° above the horizontal. The plane rises with constant velocity in the vertical direction while continuing to accelerate in the horizontal direction. (a) What is the weight of the plane? N (b) What is its horizontal acceleration?
(a) 7861 N
Along the vertical direction, the plane is moving at constant velocity: this means that the net vertical acceleration is zero, so the vertical component of the 8420 N upward force is balanced by the weight (pointing downward).
The vertical component of the upward force is given by:
[tex]F_y = F sin \theta[/tex]
where
F = 8420 N is the magnitude of the force
[tex]\theta=69.0^{\circ}[/tex] is the angle above the horizontal
Substituting,
[tex]F_y = (8420 N)(sin 69.0^{\circ}) =7861 N[/tex]
This means that the weight of the plane is also 7861 N.
(b) 3.87 m/s^2
From the weight of the plane, we can calculate its mass:
[tex]m=\frac{W}{g}=\frac{7861 N}{9.8 m/s^2}=802 kg[/tex]
Where g = 9.8 m/s^2 is the acceleration due to gravity.
Along the horizontal direction, the 8420 N is not balanced by any other backward force: so, there is a net acceleration along this direction.
The horizontal component of the force is given by
[tex]F_x = F cos \theta = (8420 N)(cos 69.0^{\circ})=3107 N[/tex]
According to Newton's second law, the net force along the horizontal direction is equal to the product between the plane's mass and the horizontal acceleration:
[tex]F_x = m a_x[/tex]
so if we solve for a_x, we find:
[tex]a_x = \frac{F_x}{m}=\frac{3107 N}{802 kg}=3.87 m/s^2[/tex]