3. A ray of light incident on one face of an equilateral glass prism is refracted in such a way that it emerges from the opposite surface at an angle of 900 to the normal. Calculate the i. angle of incidence. ii. minimum deviation of the ray of light passing through the prism [n_glass=1.52]
Answer:
i) angle of incidence;i = 29.43°
ii) δm = 38.92°
Explanation:
Prism is equilateral so angle of prism (A) = 60°
Refractive index of glass; n_glass = 1.52
A) Let's assume the incident angle = i and Critical angle = θc
We know that, sin θc = 1/n
Thus;
sin θc = 1/n_glass
θc = sin^(-1) (1/n_glass)
θc = sin^(-1) (1/1.52)
θc = 41.14°
Now, the angle of prism will be the sum of external angle that is critical angle and reflected angle.
Thus;
A = r + θc
r = A - θc
So;
r = 60° - 41. 14°
r = 18.86°
From, Snell's law. If we apply it to this question, we will have;
(sin i)/(sin r) = n_glass
Where;
i is angle of incidence and r is angle of reflection.
Let's make i the subject;
i = sin^(-1) (n_glass × sin r)
i = sin^(-1) (1.52 × sin 18.86)
i = sin^(-1) 0.4914
i = 29.43°
B) The formula to calculate minimum deviation would be from;
μ = [sin ((A + δm)/2)]/(sin A/2)
Where;
μ is Refractive index
δm is minimum angle of deviation
A is angle of prism
Now Refractive index is given by a formula; μ = (sin i)/(sin r)
So; μ = (sin 29.43)/(sin 18.86)
μ = 1.52
Thus;
1.52 = [sin ((60 + δm)/2)]/(sin 60/2)
1.52 * sin 30 = sin ((60 + δm)/2)
0.76 = sin ((60 + δm)/2)
sin^(-1) 0.76 = ((60 + δm)/2)
49.46 × 2 = (60 + δm)
98.92 - 60 = δm
δm = 38.92°
A merry-go-round is shaped like a uniform disk and has moment of inertia of 50,000 kg m 2 . It is rotating so that it has an angular momentum of 10,000 (kg m 2 radians/s) and its outer edge has a speed of 2 m/s. What is its radius, in m
Answer:
r = 20 m
Explanation:
The formula for the angular momentum of a rotating body is given as:
L = mvr
where,
L = Angular Momentum = 10000 kgm²/s
m = mass
v = speed = 2 m/s
r = radius of merry-go-round
Therefore,
10000 kg.m²/s = mr(2 m/s)
m r = (10000 kg.m²/s)/(2 m/s)
m r = 5000 kg.m ------------- equation 1
Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:
I = (1/2) m r²
where,
I = moment of inertia = 50000 kg.m²
Therefore,
50000 kg.m² = (1/2)(m r)(r)
using equation 1, we get:
50000 kg.m² = (1/2)(5000 kg.m)(r)
(50000 kg.m²)/(2500 kg.m) = r
r = 20 m
A glass sphere carrying a uniformly distributed charge of +Q is surrounded by an initially neutralspherical plastic shell. (Assume the charge +Q is uniformly distributed across thesurface of the glass sphere.)
Required:
a. Qualitatively, indicate the polarization of the plastic.
1. The plastic will polarize so as to have positive charge +Qon its inner surface and negativecharge −Q on its outer surface.
2. Dipoles in the plastic will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center.
4. Dipoles in the plastic will polarize and orient themselves radially, with their positiveendspointing toward the center.
b. Qualitatively, indicate the polarization of the inner glass sphere. Explain briefly.A net charge −Q from the dipoles will be uniformly distributed through the volume of the sphere.
1. There will be no polarization inside the glass sphere since the net electric field there iszero.
2. Dipoles in the glass will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the glass will polarize and orient themselves radially, with their positive endspointing toward the center.
c. Is the electric field at location Poutside the plastic shell larger, smaller, or the same as itwould be if the plastic weren't there? Explain briefly.
1. Larger, because a net positive charge is created from the polarization of the shell.
2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
3. Smaller, because the negative charges displaced during polarization are closer to Pthanthe positive charges.
4. Smaller, because the plastic shell shields location Pfrom the charge +Q, such that the netfield at Pis zero.
5. The same, because no net charge is created from the polarization of the field.
Answer:
(A) 3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center
(B) 2. There will be no polarization inside the glass sphere since the net electric field there is zero.
Explanation: charges are only distributed on the surface of the charged hollow conductor. The core must have zero charge.
(C) 2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
The temperature at the surface of the Sun is approximately 5,300 K, and the temperature at the surface of the Earth is approximately 293 K. What entropy change of the Universe occurs when 6.00 103 J of energy is transferred by radiation from the Sun to the Earth?
Answer:
The entropy change of the Universe that occurs is 19.346 J/K
Explanation:
Given;
temperature of the sun, [tex]T_s[/tex] = 5,300 K
temperature of the Earth, [tex]T_E[/tex] = 293 K
radiation energy transferred by the sun to the earth, E = 6000 J
The sun loses Q of heat and therefore decreases its entropy by the amount
[tex]\delta S_{sun} = \frac{-Q}{T_s}[/tex]
The earth gains Q of heat and therefore increases its entropy by the amount
[tex]\delta S_{Earth} = \frac{-Q}{T_E}[/tex]
The total entropy change is:
[tex]\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\ = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K[/tex]
Therefore, the entropy change of the Universe that occurs is 19.346 J/K
A string is stretched between fixed supports separated by 72.0 cm. It is observed to have resonant frequencies of 370 and 555 Hz, and no other resonant frequencies between these two.(a) What is the lowest resonant frequency for this string?(b) What is the wave speed for this string?
Answer:
(a) f = 185 Hz
(b) v = 266.4 m/s
Explanation:
(a) The lowest frequency can be calculated by using the following formula for the calculation of the modes (resonant frequencies) in a string:
[tex]f_n=\frac{nv}{2L}[/tex]
[tex]f_n=nf[/tex]
n: order of the mode
v: velocity of the waves in the string
L: length of the string = 72.0cm = 0.72m
fn: frequency of the n-th mode
With the information about two consecutive modes you can find the lowest resonant frequency. First you find the resonant mode n:
[tex]f_n=nf\\\\f_{n-1}=(n-1)f\\\\\frac{f_n}{f_{n-1}}=\frac{n}{n-1}[/tex]
you solve the previous equation for n:
[tex](n-1)f_n=nf_{n-1}\\\\555n-555=370n\\\\n=3[/tex]
With this information you can calculate the lowest resonant frequency:
[tex]f_n=nf\\\\f=\frac{f_n}{n}=\frac{555}{3}=185Hz[/tex]
b) You have information about two consecutive modes fn, fn-1. Then, you can calculate the velocity of the waves:
[tex]f_{n}-f_{n-1}=n\frac{v}{2L}-(n-1)\frac{v}{2L}\\\\f_n-f_{n-1}=\frac{v}{2L}\\\\v=2L(f_n-f_{n-1})[/tex]
fn = 555 Hz
fn-1: 370 Hz
[tex]v=2(0.72m)(555-370)Hz=266.4\frac{m}{s}[/tex]´
hence, the velocityof the waves in the string is 266.4 m/s
what statement is true according to newton’s first law of motion?
a. in the absence of unbalanced force an object at rest will stay at rest and an object in motion will come to a stop.
b. in the absence of an unbalanced force, an object will start moving and an object in motion will come to a stop.
c. in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
d. in the absence of an unbalanced force, an object will start moving and an object in motion will stay in motion.
Answer:
c. in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
Explanation:
First law: things keep doing what they are doing, unless force is applied.
What is the period of a wave if the frequency is? 5 Hz
Answer: If the woodpecker drums upon a tree 5 times in one second, then the frequency is 5 Hz; each drum must endure for one-fifth a second, so the period is 0.2 s.
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled has an acceleration of 2.10 m/s2. Part A By how much does the spring stretch if it pulls on the sled horizontally
Answer:
Stretch in the spring = 0.1643 (Approx)
Explanation:
Given:
Mass of the sled (m) = 9 kg
Acceleration of the sled (a) = 2.10 m/s ²
Spring constant (k) = 115 N/m
Computation:
Tension force in the spring (T) = ma
Tension force in the spring (T) = 9 × 2.10
Tension force in the spring (T) = 18.9 N
Tension force in the spring = Spring constant (k) × Stretch in the spring
18.9 N = 115 N × Stretch in the spring
Stretch in the spring = 18.9 / 115
Stretch in the spring = 0.1643 (Approx)
g A proton is held at rest in a uniform electric field. When it is released, the proton will lose... electrical potential energy. kinetic energy. both kinetic energy and electric potential energy. neither kinetic energy or electric potential energy.
Answer:
It will lose electrical potential energy.
Explanation:
A photon held at rest in a uniform electrical field will lose electrical potential energy when it is released this is because the electrical potential energy is the energy posses by the photon at rest or by virtue of the position is converted to kinetic energy which is energy posses by a body in motion.
Since the photon is released and set in motion , it now has kinetic energy and has lost the potential energy because it is set in motion.
An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object
Answer:
v = 25.45 m/s
Explanation:
In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:
[tex]h_{max}=\frac{v_o^2}{g}[/tex] (1)
vo: initial speed of the object = 18 m/s
g: gravitational acceleration = 9.8 m/s²
Furthermore you use the following formula for the final speed of the object:
[tex]v^2=v_o^2-2gh[/tex] (2)
h: height
You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:
[tex]v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2[/tex]
Then, you solve the previous result for vo:
[tex]v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}[/tex]
The initial speed of the object was 25.45 m/s
When the play button is pressed, a CD accelerates uniformly from rest to 430 rev/min in 4.0 revolutions. If the CD has a radius of 7.0 cm and a mass of 17 g , what is the torque exerted on it?
Answer:
The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].
Explanation:
As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot \Delta n[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]\ddot n[/tex] - Angular acceleration, measured in revolution per square minute.
[tex]\Delta n[/tex] - Change in angular position, measured in revolutions.
The angular acceleration is cleared and calculated:
[tex]\ddot n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \Delta n}[/tex]
Given that [tex]\dot n_{o} = 0\,\frac{rev}{min}[/tex], [tex]\dot n = 430\,\frac{rev}{min}[/tex] and [tex]\Delta n = 4\, rev[/tex], the angular acceleration is:
[tex]\ddot n = \frac{\left(430\,\frac{rev}{min} \right)^{2}-\left(0\,\frac{rev}{min} \right)^{2}}{2\cdot (4\,rev)}[/tex]
[tex]\ddot n = 23112.5\,\frac{rev}{min^{2}}[/tex]
The angular accelaration measured in radians per square second is:
[tex]\alpha = \left(23112.5\,\frac{rev}{min^{2}} \right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}} \right)[/tex]
[tex]\alpha \approx 40.339\,\frac{rad}{s^{2}}[/tex]
Net torque experimented by the CD during its accleration is equal to the product of its moment of inertia with respect to its axis of rotation and angular acceleration:
[tex]\tau = I \cdot \alpha[/tex]
Where:
[tex]I[/tex] - Moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
In addition, a CD has a form of a uniform disk, whose moment of inertia is:
[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]
Where:
[tex]m[/tex] - Mass of the CD, measured in kilograms.
[tex]r[/tex] - Radius of the CD, measured in meters.
If [tex]m = 0.017\,kg[/tex] and [tex]r = 0.07\,m[/tex], then:
[tex]I = \frac{1}{2}\cdot (0.017\,kg)\cdot (0.07\,m)^{2}[/tex]
[tex]I = 4.165\times 10^{-5}\,kg\cdot m^{2}[/tex]
Now, the net torque exerted on CD is:
[tex]\tau = (4.165\times 10^{-5}\,kg\cdot m^{2})\cdot \left(40.339\,\frac{rad}{s^{2}} \right)[/tex]
[tex]\tau = 1.680\times 10^{-3}\,N\cdot m[/tex]
The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].
Zinc is added to a breaker containing hydrochloric acid and the beaker gets warm what type os reaction is this
Answer:
Exothermic
Explanation:
Depending on the unit you are in, the answer may vary.
This is an exothermic reaction because it produces heat (the beaker gets warm).
Why do some nucleus emit electrons?
Answer:
In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
Explanation:
Hope this helps!
The Great Lakes are all part of what? The Mississippi River The St. Lawrence Seaway A large body of salt lakes The Missouri River
Answer:
St Lawrence Sea way
Explanation:
The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.
An arrow is shot from a height of 1.55 m toward a cliff of height H. It is shot with a velocity of 26 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.99 s later.
(a) Draw a sketch of the given example. Include the x-y coordinate system.
(b) What is the height of the cliff?
(c) What is the maximum height reached by the arrow along its trajectory?
(d) What is the arrow's impact speed just before hitting the cliff?
Answer:
Explanation:
vertical component of the velocity of arrow
= 26 sin 60 = 22.516 m
height reached by it after 3.99 s
h = ut - 1/2 g t²
= 22.516 x 3.99 - .5 x 9.8 x 3.99²
= 89.83 - 78
11.83 m
Total height of cliff = 1.55 + 11.83
= 13.38 m
c ) maximum height covered s
v² = u² - 2gs
0 = u² - 2gs
s = u² / 2g
= 22.516² / 2 x 9.8
= 25.86
maximum height reached
= 25.86 + 1.55
= 27.41 m
d )
vertical speed after 3.99 s
v = u - gt
= 22.516 - 9.8 x 3.99
= -16.586
Horizontal component will remain unchanged
Horizontal component = 26 cos 60
= 13 m /s
Resultant of two velocities
= √ 13²+ 16.568²
= 21 m /s
A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6.34). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s.
(a) What is the tension in the cord in the original situation when the block has speed v = 0.70 m/s? (b) What is the tension in the cord in the final situation when the block has speed v = 2.80 m/s? (c) How much work was done by the person who pulled on the cord?
Answer:
a) 0.147 N
b) 9.408 N
c) 9.261 N
Explanation:
The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force [tex]\frac{mv^{2} }{r}[/tex]. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.
1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a turn b. a child swinging around a pole c. a person sitting on a bench facing the center of a carousel d. a rock swinging on a string e. the Earth orbiting the Sun.
Complete Question
For each of the following scenarios, describe the force providing the centripetal force for the motion:
a. a car making a turn
b. a child swinging around a pole
c. a person sitting on a bench facing the center of a carousel
d. a rock swinging on a string
e. the Earth orbiting the Sun.
Answer:
Considering a
The force providing the centripetal force is the frictional force on the tires \
i.e [tex]\mu mg = \frac{mv^2}{r}[/tex]
where [tex]\mu[/tex] is the coefficient of static friction
Considering b
The force providing the centripetal force is the force experienced by the boys hand on the pole
Considering c
The force providing the centripetal force is the normal from the bench due to the boys weight
Considering d
The force providing the centripetal force is the tension on the string
Considering e
The force providing the centripetal force is the force of gravity between the earth and the sun
Explanation:
At a time when mining asteroids has become feasible, astronauts have connected a line between their 3220-kg space tug and a 6240-kg asteroid. They pull on the asteroid with a force of 362 N. Initially the tug and the asteroid are at rest, 311 m apart. How much time does it take for the ship and the asteroid to meet
-- F = m a ... ==> a = F/m
-- The tension in the rope is 362 N. That same force acts on the asteroid and on the tug, pulling them together.
-- The asteroid's acceleration is 362N / 6240 kg = 0.058 m/s², headed for a point on the rope somewhere between the asteroid and the tug.
-- The tug's acceleration is 362 N / 3220 kg = 0.112 m/s², also headed for a point on the rope somewhere between the tug and the asteroid.
-- So now we have a gap between them, initially 311 m long, closing with a speed that starts at zero and accelerates at 0.170 m/s² .
-- D = (1/2) a T²
311 m = (1/2) (0.170 m/s²) (T²)
T² = 311 m / 0.085 m/s²
T = √(311/0.085) seconds
T = 60.41 seconds
The answer I get is so durn near 60 seconds (1 minute) that it suggests two things to me: ==> That's where the weird numbers of 362N and 311m came from, and ==> there's a good chance that my answer is correct.
Note: It's important to me that you know that 5 points for this one is really cheap and chintzy, and the reason I decided to try it was only to see whether I could.
What is the answer for this question
EASY! WILL GIVE BRAINLIEST!
Find the conductivity of a conduit with a cross-sectional area of 0.50 cm2 and a length of 15 cm if its conductance G is 0.050 ohm-1.
σ = _____ ohm-1cm-l
3
75
1.5
0.0017
the answer is 1.5 hope this helps
Answer:
1.5
Explanation:
0.5=σ/(15/0.5)
σ=3/2 or 1.5
A 2 kg object is subjected to three forces that give it an acceleration −→a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj. If two of the three forces, are −→F1 = (30.0N)ˆi + (16.0N)ˆj and −→F2 = −(12.0N)ˆi + (8.00N)ˆj, find the third force.
Answer:
[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]
Explanation:
You have three forces F1, F2 an F3 that produce the following acceleration:
a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj
you know that force F1 and F2 are:
F1 = (30.0N)ˆi + (16.0N)ˆj
F2 = −(12.0N)ˆi + (8.00N)ˆj
and the force F3 is unknown:
F3 = F3x ˆi + F3y ˆj
The second Newton law is given by the following equation:
[tex]\vec{F}=m\vec{a}[/tex]
F: the total force = F1 +F2 + F3
m: mass of the object = 2 kg
By the properties of vectors you have:
[tex]\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_{3x})N\hat{i}+(16.0+8.00+F_{3y})N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_{3x})N\hat{i}+(24.0+F_{3y})\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}[/tex]
Both x and y component must be equal in the previous equality, then you have:
[tex]18.0N+F_{3x}=-16.00N\\\\F_{3x}=-34.00N\\\\24.0N+F_{3y}=12.0N\\\\F_{3y}=-12.00N[/tex]
Hence, the vector F3 is:
[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]
Q) Suppose, you are in a sporting event. You notice that everyone stands up when it’s his turn,
creating a wave that moves through the crowd and they sit back down again after a while. This wave
move around the stadium without moving the people around it. Considering this situation, justify
your answer about nature of wave.
Answer:
The nature of the wave formed is a transverse progressive wave.
Explanation:
A wave is a disturbance that travels through a material medium without permanent displacement of the particles of the medium. The two major types are: transverse and longitudinal.
A transverse wave is one in which the direction of vibration of the particles of the medium is perpendicular to the direction of propagation of the wave. Examples are: water wave, light wave etc. While a longitudinal wave is one in which the direction of vibrations of the particles of the medium is parallel with the direction of propagation of the wave, creating a region of rarefaction and compression. Examples are; sound wave, wave in a rope, wave in a slinky etc.
The cited wave formed in the given question is a transverse wave because each person stands and sits after some time to create a moving (progressive) wave without them moving from their positions.
Sr-90 has a half-life of T1/2 = 2.85 a (years). How much Sr-90 will remain in a 5.00 g sample after 5.00 a? Show all of your work. (2 marks)
Answer:
1.48 g
Explanation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
A = (5.00 g) (½)^(5.00 a / 2.85 a)
A = 1.48 g
The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2 where xf is the final position, xi is the initial position, vi is the initial velocity, a is the acceleration, and t is the time. Let's say a car starts with an initial speed of 15 m/s, and moves between the 1000 m and 5000 m marks on a roadway in a time of 60 s. What is its acceleration?
Answer:
a = 1.72 m/s²
Explanation:
The given kinematic equation is the 2nd equation of motion. The equation is as follows:
xf = xi + (Vi)(t) + (1/2)(a)t²
where,
xf = the final position = 5000 m
xi = the initial position = 1000 m
Vi = the initial velocity = 15 m/s
t = the time taken = 60 s
a = acceleration = ?
Therefore,
5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²
5000 m = 1000 m + 900 m + a(1800 s²)
5000 m = 1900 m + a(1800 s²)
5000 m - 1900 m = a(1800 s²)
a(1800 s²) = 3100 m
a = 3100 m/1800 s²
a = 1.72 m/s²
Question 7 of 10
The coefficient of kinetic friction between a couch and the floor is 0.4. If the
couch has a mass of 35 kg and you push it with a force of 200 N. what is the
net force on the couch as it slides?
O A. 337 N
B. 143 N
O C. 343 N
O D. 63 N
Answer:
D
Explanation:
Now the net force is the applied force minus the frictional force; this is expressed mathematically as:
Fnet= Fappplied - Ffrictional
Now the frictional force is given as ;
Coefficient of friction × normal reaction
Normal reaction is the weight of the human acting in opposite direction.
Normal reaction of the human is ;
35 × 9.8 = 343N { note that weight = m× g and g= 9.8m/S2, a known standard }
Hence the Frictional force =343×0.4 =137.20N
Hence Fnet = 200-137.20 = 62.8N
Fnet = 63N to the nearest whole
The net force on the couch as it slides is 63N.
What is frictional force?
When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.
The friction force is given by
f = coefficient of friction x Normal force
Given, the coefficient of kinetic friction between a couch and the floor is 0.4. If the couch has a mass of 35 kg and you push it with a force of 200 N.
Normal reaction is the weight of the human acting in opposite direction.
Normal reaction N =35 × 9.8 = 343N
Frictional force f =0.4 x 343
f =137.20N
The net force will be
Fnet= Fappplied - Ffrictional
Fnet = 200-137.20 = 62.8N
Fnet = 63N
Thus, the net force on the couch as it slides is 63N.
Learn more about friction force.
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A population _____ follows a period of
Answer:
a population increase
Explanation:
During the 20th century, the world population increased from 1.65 billion to 6 billion. In 1970, the world's population was half that of today. In less than 15 years, 47% of the population will live in areas already under heavy water stress. In Africa, between 75 and 250 million people will face growing shortages in 2020 due to climate change. The scarcity of some arid and semi-arid regions will have a decisive impact on migration.
If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens would receive our signals at speed of:_______.
a. 0.99c
b. 1.10c
c. 1.00c
d. 0.90c
e. 0.10c
Answer:
3×10^7 m/s or 0.10c (e)
Explanation: If the actual value of the speed of light were to be put into consideration.
Given that the speed of light is c = 3.0×10^8m/s
The alien spaceship is approaching at the rate of 10% of the speed of light.
10% of 3.0×10^8m/s
10/100 × 3.0×10^8m/s
0.1 ×3.0×10^8m/s
3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c
Answer: 1.00c
Explanation: I got it correct on the homework
A freight car moves along a frictionless level railroad track at constant speed. The freight car is open on top. A large load of coal is suddenly dumped into the car. What happens to the speed of the freight car
Answer:
The speed of the freight car decreases.
Explanation:
According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received
In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.
Therefore the speed of the freight car decreased by applying the law of conservation of momentum i
Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Simultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 3.90 m/s , and puck B moves with a speed of 4.30 m/s . What is the distance covered by puck A by the time the two pucks collide
Answer:
The distance covered by puck A before collision is [tex]z = 8.56 \ m[/tex]
Explanation:
From the question we are told that
The label on the two hockey pucks is A and B
The distance between the two hockey pucks is D 18.0 m
The speed of puck A is [tex]v_A = 3.90 \ m/s[/tex]
The speed of puck B is [tex]v_B = 4.30 \ m/s[/tex]
The distance covered by puck A is mathematically represented as
[tex]z = v_A * t[/tex]
=> [tex]t = \frac{z}{v_A}[/tex]
The distance covered by puck B is mathematically represented as
[tex]18 - z = v_B * t[/tex]
=> [tex]t = \frac{18 - z}{v_B}[/tex]
Since the time take before collision is the same
[tex]\frac{18 - z}{V_B} = \frac{z}{v_A}[/tex]
substituting values
[tex]\frac{18 -z }{4.3} = \frac{z}{3.90}[/tex]
=> [tex]70.2 - 3.90 z = 4.3 z[/tex]
=> [tex]z = 8.56 \ m[/tex]
During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a velocity of 2.60 m/s, grabs and holds onto him so that they move off together with a velocity of 1.30 m/s. If the mass of the tackler is 122 kg, determine the mass of the receiver. Assume momentum is conserved.
Answer:
122kgExplanation:
Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.
Momentum = Mass * Velocity
Before collision;
Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)
Momentum of the tackler
m2u2 = 2.60*122 = 317.2 kgm/s
where m2 and u2 are the mass and velocity of the tacker respectively.
Sum of momentum before collision = 0+317.2 = 317.2 kgm/s
After collision
Momentum of the bodies = (m1+m2)v
v = their common velocity
m1 = mass of the receiver
Momentum of the bodies = (122+m1)(1.30)
Momentum of the bodies = 158.6+1.30m1
According to the law above;
317.2 = 158.6+1.30m1
317.2-158.6 = 1.30m1
158.6 = 1.30m1
m1 = 158.6/1.30
m1 = 122kg
The mas of the receiver is 122kg
