Answer:
5.0 m/s south
Hope this Helps!
Answer:
5.0 m/s south
Explanation:
Determine the electrical force of attraction between two balloons
that are charged with the opposite type of charge but the same
quantity of charge. The charge on the balloons is 6.0 x 10-7 C and they
are separated by a distance of 0.50 m.
Answer:
F=1.3x10^-2N
Explanation:
Fe= k(6x10^-7C)^2/(0.5)^2
Electrical force of attraction between the balloons is F=1.3x10^-2N
The electric force of attraction between two balloons should be F=1.3x10^-2N.
Calculation of the electric force;Since The charge on the balloons is 6.0 x 10-7 C and they are separated by a distance of 0.50 m.
So, here the electric force is
Fe= k(6x10^-7C)^2/(0.5)^2
F=1.3x10^-2N
hence, The electric force of attraction between two balloons should be F=1.3x10^-2N.
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A ray is incident at at 50 degrees angle on a plane mirror. What will be the deviation after reflection from the mirror?
Answer:
Explanation:
If the ray were not deviated, it would travel straight through the mirror. Due to the mirror, the incident ray is reflected at 30°. The ray travels 30° + 30° = 60°. The angle of deviation is 180° - 60° = 120°.
How do I proton and and electron compared
How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns
Complete Question
How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns.
The output from the secondary coil is 12 V
Answer:
The value is [tex]N_s = 21 \ turns [/tex]
Explanation:
From the equation we are told that
The input voltage is [tex]V_{in} = 120 \ V[/tex]
The number of turns of the primary coil is [tex]N_p = 210 \ turn[/tex]
The output from the secondary is [tex]V_o = 12V[/tex]
From the transformer equation
[tex]\frac{N_p}{V_{in}} =\frac{N_s}{V_o}[/tex]
Here [tex]N_s[/tex] is the number of turns in the secondary coil
=> [tex]N_s = \frac{N_p}{V_{in}} * V_s[/tex]
=>[tex]N_s = \frac{210}{120} * 12[/tex]
=>[tex]N_s = 21 \ turns [/tex]
A person following a liberal ideology would likely approve of
A student creates an electromagnetic wave and then reverses the direction of the current. Which of the following will happen to the magnetic field?
Answer:
I believe the electromagnetic field should be reversed.
Explanation:
When a student creates an electromagnetic wave and then reverses the direction of the current, the direction of the magnetic field will be reversed.
What is an Electromagnetic wave?An electromagnetic wave may be defined as a type of wave that is significantly created as a result of vibrations between an electric field and a magnetic field. These waves are composed of oscillating magnetic and electric fields.
According to the context of this question, when an individual is constructing an electromagnetic wave and then reverses the direction of the current, it will eventually affect the direction of the magnetic field in the same direction with respect to the current. So, if the direction of the current is reversed, the direction of the magnetic field would also be reversed.
Therefore, when a student creates an electromagnetic wave and then reverses the direction of the current, the direction of the magnetic field will be reversed.
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Your question seems incomplete. The most probable complete question is as follows:
A student creates an electromagnetic wave and then reverses the direction of the current. Which of the following will happen to the magnetic field?
The direction of the magnetic field will be reversed. The magnetic field will expand.The magnetic field would be canceled out and disappear.The magnetic field will cause the voltage of the battery to be reduced.A "lovesick" individual wants to throw a bag of candy and love notes into the open window of their significant other’s bedroom 10.0 m above. Assuming it just reaches the window, they throw the gift at 60.0o to the ground: At what velocity should they throw the bag? How far from the house are they standing when they throw the bag? (Answer: A. 16.2m/s B. 11.5m)
Answer:
Explanation:
Let the velocity be v .
vertical component of the velocity = v sin 60 = √3 v /2
it reaches maximum height of 10 m .
v² = 2 gh
( √3 v/2 )² = 2 x 9.8 x 10
3 v² = 196 x 4
v² = 65.33 x 4
v = 16.2 m /s
Let time taken to reach height of 10 m
v = u - gt
v sin 60 = 9.8 t
16.2 x √3 /2 = 9.8 t
t = 1.43 s
horizontal distance covered = v cos 60 x t
16.2 x .5 x 1.43 = 11 .5 m
If vector A = 6i - 2j + 3k, determine
(a) A vector in the same direction as A with magnitude 2A
(b) A unit vector in the direction of A
(c) a vector opposite to A with magnitude of 4 m
Answer:
(a) [tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]
(b) [tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]
(c) [tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]
Explanation:
Vectors
Given a vector
[tex]\vec A=6\hat i-2\hat j+3\hat k[/tex]
We must determine the following:
a) A vector in the same direction as A with double magnitude 2A.
If the vector goes in the same direction but has a different magnitude, we only need to multiply each component by a common factor, in this case, by 2. Thus, the required vector is:
[tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]
b) A unit vector in the same direction of A.
The unit vector needs to compute the magnitude of the vector:
[tex]\mid A\mid=\sqrt{6^2+2^2+3^2}[/tex]
[tex]\mid A\mid=\sqrt{36+4+9}=\sqrt{49}=7[/tex]
[tex]\mid A\mid=7[/tex]
The unit vector is:
[tex]\displaystyle \vec{U_A}=\frac{\vec A}{\mid \vec A\mid}[/tex]
[tex]\displaystyle \vec{U_A}=\frac{12\hat i-4\hat j+6\hat k}{7}[/tex]
[tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]
c) A vector opposite to A with magnitude 4 m. We assume the original vector is also expressed in m.
The opposite vector to A is obtained simply by multiplying the unit vector by -1. To make its magnitude equal to 4, also multiply by 4. In all, we multiply the unit vector by -4:
[tex]-4\vec{U_A}=-4(12/7\hat i-4/7\hat j+6/7\hat k)[/tex]
[tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 7.63 m/s in 3.94 s. What is the magnitude of the linear impulse experienced by a 73.7 kg passenger in the car during this time? Submit Answer Tries 0/20 What is the average force experienced by the passenger?
Answer:
1. p = 562.3 kg*m/s
2. F = 142.7 N
Explanation:
1. The linear impulse (p) is given by:
[tex] p = mv [/tex]
Where:
m: is the passenger's mass = 73.7 kg
v: is the speed = 7.63 m/s
[tex] p = mv = 73.7 kg*7.63 m/s = 562.3 kg*m/s [/tex]
Hence, the magnitude of the linear impulse experienced by a passenger is 562.3 kg*m/s.
2. The average force can be calculated using the following equation:
[tex] F = \frac{m(v_{f} - v_{0})}{t} = \frac{73.7 kg(7.63 m/s - 0)}{3.94 s} = 142.7 N [/tex]
Therefore, the average force experienced by the passenger is 142.7 N.
I hope it helps you!
(iii) Why do right angle mirrors produce three images of the object?
Explanation:
The two mirrors inclined to each other formed the first two images with are of the same size as the object while the third mirror is produced from the intersection of rays that emanated during the production of the first two images to produce a third image which is smaller than the object and there making the total number of images to be 3.
Hence this mirrors produces 3 images due to the third image formed from the intersection of the rays that produces the first two images.
The formula that relates the image produced by inclined mirror and the angle of inclination is expressed as:
number of images n = 360/θ - 1
θ is the angle of inclination of the two mirrors
n is the number of images
If the mirrors are inclined at right angles, then θ = 90°
Substitute into the formula;
n = 360/90 -1
n = 36/9 -1
n = 4-1
n = 3
The equation that governs the period of a pendulum’s swinging. T=2π√L/g
Where T is the period, L is the length of the pendulum and g is a constant, equal to 9.8 m/s2. The symbol g is a measure of the strength of Earth’s gravity, and has a different value on other planets and moons.
On our Moon, the strength of earth’s gravity is only 1/6th of the normal value. If a pendulum on Earth has a period of 4.9 seconds, what is the period of that same pendulum on the moon?
Answer:
The period of that same pendulum on the moon is 12.0 seconds.
Explanation:
To determine the period of that same pendulum on the moon,
First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be [tex]g_{M}[/tex].
From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²
∴ [tex]g_{M}[/tex] = [tex]\frac{1}{6} \times 9.8 m/s^{2}[/tex]
[tex]g_{M}[/tex] = 1.63 m/s²
From the question, T=2π√L/g
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
We can write that,
[tex]T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }[/tex] .......... (1)
Where [tex]T_{E}[/tex] is the period of the pendulum on Earth and [tex]g_{E}[/tex] is the measure of the strength of Earth's gravity
and
[tex]T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }[/tex] .......... (2)
Where [tex]T_{M}[/tex] is the period of the pendulum on Moon and [tex]g_{M}[/tex] is the measure of the strength of Earth's gravity on the Moon.
Since we are to determine the period of the same pendulum on the moon, then, [tex]2\pi[/tex] and [tex]L[/tex] are constants.
Dividing equation (1) by (2), we get
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
From the question,
[tex]T_{E} = 4.9secs[/tex]
[tex]g_{E}[/tex] = 9.8 m/s²
[tex]g_{M}[/tex] = 1.63 m/s²
[tex]T_{M}[/tex] = ??
From,
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
[tex]\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }[/tex]
[tex]\frac{4.9}{T_{M} } = 0.40783[/tex]
[tex]T_{M} =\frac{4.9}{0.40783 }[/tex]
[tex]T_{M} = 12.01 secs[/tex]
∴ [tex]T_{M} = 12.0secs[/tex]
Hence, the period of that same pendulum on the moon is 12.0 seconds.
Answer:
The period of that same pendulum on the moon is 12.0 s
Explanation:
Given;
period of a pendulum’s swinging, T=2π√L/g
the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)
period of pendulum on Earth, T₁ = 4.9 s
period of pendulum on moon, T₂ = ?
The length of the pendulum is constant, make it the subject of the formula;
[tex]T = 2\pi \sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\\\\(\frac{T}{2\pi} )^2 =\frac{L}{g}\\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\ L = \frac{gT^2}{4\pi^2}\\\\L_1 = L_2\\\\\frac{g_1T_1^2}{4\pi^2}= \frac{g_2T_2^2}{4\pi^2}\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = \frac{g_1T_1^2}{g_2} \\\\T_2 = \sqrt{\frac{g_1T_1^2}{g_2}}\\\\ T_2 = \sqrt{\frac{g_1T_1^2}{g_1/6}}\\\\ T_2 = \sqrt{\frac{6*g_1T_1^2}{g_1}}\\\\T_2 = \sqrt{6T_1^2}\\\\ T_2 = T_1\sqrt{6} \\\\T_2 = (4.9)\sqrt{6}\\\\ T_2 = 12.0 \ s[/tex]
Therefore, the period of that same pendulum on the moon is 12.0 s
Bird A, with a mass of 2.2 kg, is stationary while Bird B, with a mass of 1.7 kg, is moving due north from Bird A at 3 m/s. What is the velocity of the center of mass for this system of two birds
Answer:
1.3 m/s
Explanation:
It is given that,
Mass of bird A, [tex]m_A=2.2\ kg[/tex]
Mass of bird B, [tex]m_B=1.7\ kg[/tex]
Initial speed of bird A is 0 as it was at rest
Initial speed of bird B is 3 m/s
We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,
[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]
So, the center of mass for this system is 1.3 m/s.
I WILL MARK YOU AS BRAINLIEST IF RIGHT
What is the magnitude of the net force acting on this object?
Answer:
The net force on an object is the total force applied on the object after adding up all the forces
In the given diagram,
we can see that the 2 forces of 4N and 4N will cancel each other out since they are equal and in the opposite direction
Now, we are left with a force of 2N and 10N,
the net force will be the difference of these forces:
Net force = 10N - 2N
Net force = 8N downwards
Another way to do it:
The two 4N forces will be cancelled out,
and we are left with a 2N and a 10N force
(notice how we cancelled equal and opposite forces for the 4N)
We can divide the 10N force into (2N + 8N)
Since the 2N forces are equal and opposite, they will be cancelled out
and we will be left with a net force of 8N downwards
Calculate the effective charges on the H and F atoms of the HF molecule in units of the electronic charge, e.
Answer:
Explanation:
Hydrogen fluoride (HF) is an ionic/electrovalent compound that dissociates into ions when dissolved in water. It's dissociation is as seen below
HF ⇄ H⁺ + F⁻
There is a transfer of electron from the hydrogen atom which produces the hydrogen ion (H⁺), while the fluorine atom receives the donated ion to become negatively charged (F⁻). The amount of charge in one electron is generally given as 1.602 × 10⁻¹⁹ coloumbs.
The required value of effective charge on HF molecule, due to H and F is 1.602 × 10⁻¹⁹ Coulombs.
The given problem is based on the concept of effective charges. The net positive charge carried out by the electrons of atomic species, after forming a polyelectronic atom is known as Effective charge.
As per the given problem, the Hydrogen fluoride (HF) is an ionic/electrovalent compound that dissociates into ions when dissolved in water. It's dissociation is given as,
HF ⇄ H⁺ + F⁻
There is a transfer of electron from the hydrogen atom which produces the hydrogen ion (H⁺), while the fluorine atom receives the donated ion to become negatively charged (F⁻). The amount of charge in one electron is generally given as 1.602 × 10⁻¹⁹ Coulombs.
Thus, we can conclude that the required value of effective charge on HF molecule, due to H and F is 1.602 × 10⁻¹⁹ Coulombs.
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A fountain shoots a jetof water straight up. The nozzle is 1 cm in diameter and the speed of the water exiting the nozzle is 30 m/s. What is the force exerted by the water jet
Answer:
Explanation:
mass of water coming out per second = A x v where A is area of cross section of the nozzle and v is velocity of water
A = 3.14 x .005²
= 785 x 10⁻⁷ m²
mass of water coming out per second = 785 x 10⁻⁷ x 30 = 23.55 x 10⁻⁴ kg
momentum of this mass = 23.55 x 10⁻⁴ x 30 = 706.5 x 10⁻⁴ kg m /s .
Rate of change of momentum = 706.5 x 10⁻⁴
Let force be F
F - mg = 706.5 x 10⁻⁴
F = mg + 706.5 x 10⁻⁴
F = 23.55 x 10⁻⁴ x 9.8 + 706.5 x 10⁻⁴
= 937.3 x 10⁻⁴ N .
A chef places an open sack of flour on a kitchen scale. The scale reading of
40 N indicates that the scale is exerting an upward force of 40 N on the sack. The magnitude of this force equals the magnitude of the force of Earth’s gravitational attraction on the sack. The chef then exerts an upward force of
10 N on the bag and the scale reading falls to 30 N.Draw a free-body diagram of the latter situation.
Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:
40N
Explanation:
trust
Please help
A student plans an investigation to determine the refractive index of glass. The student uses this equipments.
- a ray box
- a rectangular glass block
- a protractor
- a pencil
Describe how the student collect her data.
Peter is running laps around a circular track with a diameter of 100 meters. If it takes Peter 12 minutes to run 4 laps, how quickly is he running (in meters per second)?
Answer:
v = 1.74 m/s
Explanation:
Given that,
Diameter of a circular track, d = 100 m
Distance covered for the 4 laps,
[tex]D=4\pi d\\\\D=4\pi \times 100\\\\D=1256.63\ m[/tex]
Time, t = 12 minutes = 720 s
We need to find the velocity of the peter. It can be calculated as follows :
[tex]v=\dfrac{D}{t}\\\\v=\dfrac{1256.63\ m}{720\ s}\\\\v=1.74\ m/s[/tex]
So, the speed is running with a velocity of 1.74 m/s.
Peter is running at 1.7453 m/sec.
Given to us,
Diameter of the circular track, D = 100 meters,
Number of laps Peter run, L = 4 laps,
Time taken by Peter, t = 12 minutes,
1 lap = circumference of the circle,
4 laps = 4 x circumference of the circle,
As we know, the circumference of a circle is given by πD.
So, 4 laps = 4 x circumference of the circle,
[tex]\begin{aligned}4 laps &= 4\times \pi \times D\\&= 4 \times \pi \times 100\\& = 1,256.6370\ meters\\\end{aligned}[/tex]
Also, we know that 1 minute has 60 sec.
so, 4 minutes = (4 x 60) seconds
Further, speed is given [tex]\bold{(\dfrac{Distance}{Time} )}[/tex]
Thus,
[tex]\begin{aligned}speed &= \dfrac{Distance\ coverd\ by\ Peter}{Time\ taken\ by\ Peter}\\&=\dfrac{1,256.6370}{12\times 60}\\&=1.7453\ m/sec \end{aligned}[/tex]
Hence, Peter is running at 1.7453 m/sec.
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Converting compound units
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?
Answer:
Dividing the silicon density by 1000 and then multiply it by 1000000.
Explanation:
A kilogram equals 1000 grams and a cubic meter equals 1000000 cubic centimeters. Hence, we must divide the silicon density by 1000 and then multiply itby 1000000 to convert the value into kilograms per cubic centimeter. That is:
[tex]x = 2.33\,\frac{g}{cm^{3}}\times \frac{1\,kg}{1000\,g}\times \frac{1000000\,cm^{3}}{1\,m^{3}}[/tex]
[tex]x = 2330\,\frac{kg}{m^{3}}[/tex]
In a nutshell, we must multiply the density of silicon by 1000 to obtains its value in kilograms per cubic meter.
An object, initially at rest, is subject to an acceleration of 45 m/s^2. How long will it take that object to travel 1000m? Round to one decimal place.
Answer:
6.7 seconds
Explanation:
d=(1/2)at^2
equation
1000=(1/2)45t^2.
substitute
2000=45t^2.
multiply by 2 for both sides
44.44=t^2.
divide both sides by 45
6.7=t
take the square root of both sides
The Intensity level of a loud saw is 100 db at a distance of 5m. At what distance would the level be 80 db
Answer:
50 m
Explanation:
The relationship between the intensity of sound in dB and distance is given by the formula:
[tex]B_2=B_1+20log(\frac{R_1}{R_2} )\\\\Where \ B_2\ is \ the\ sound\ intensity\ at\ distance\ R_2\ and\\B_1\ is \ the\ sound\ intensity\ at\ distance\ R_1\ \\\\Given\ that: B_1=100\ dB, R_1=5\ m, B_2=80\ dB\\\\B_2=B_1+20log(\frac{R_1}{R_2} )\\\\80=100+20log(\frac{5}{R_2} )\\\\-20=20log(\frac{5}{R_2} )\\\\log(\frac{5}{R_2} )=-1\\\\\frac{5}{R_2}=10^{-1}\\\\\frac{5}{R_2}=0.1\\\\R_2=5/0.1=50\ m[/tex]
Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus.
What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?
Answer:
1. [tex] E = 1.14 \cdot 10^{13} N/C [/tex]
2. [tex]E_{k} = 9.1 \cdot 10^{-17} J[/tex]
Explanation:
1. The strength of the nucleus' electric field (E):
[tex]E = \frac{kq}{r^{2}}[/tex]
Where:
k: is the Coulomb constant = 9x10⁹ Nm²/C²
q: is the proton charge = 1.6x10⁻¹⁹ C
r: is the radius = 10⁻¹⁰ m
[tex]E = \frac{kq}{r^{2}} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*79*1.6 \cdot 10^{-19} C}{(10^{-10} m)^{2}} = 1.14 \cdot 10^{13} N/C[/tex]
2. The kinetic energy (Ek) of an electron is the following:
[tex] E_{k} = \frac{1}{2}mv^{2} [/tex]
Where:
m is the electron's mass = 9.1x10⁻³¹ kg
v: is the speed of the electron
We can find the speed of the electron by equaling the centripetal force (Fc) and the electrostatic force (Fe):
[tex] F_{c} = F_{e} [/tex]
[tex] \frac{mv^{2}}{r} = \frac{kq^{2}}{r^{2}} = qE [/tex]
[tex] v^{2} = \frac{qEr}{m} = \frac{1.6 \cdot 10^{-19} C*1.14 \cdot 10^{13} N/C*10^{-10} m}{9.1 \cdot 10^{-31} kg} = 2.00 \cdot 10^{14} m^{2}/s^{2} [/tex]
Now, we can find the kinetic energy:
[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*2.00 \cdot 10^{14} m^{2}/s^{2} = 9.1 \cdot 10^{-17} J [/tex]
I hope it helps you!
Please provide explanation!!!
Thank you.
Answer:
(a) 102 cm/s
(b) 0.490 cm²
Explanation:
(a) Use Bernoulli equation.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0
½ ρ v₁² + ρgh₁ = ½ ρ v₂²
½ v₁² + gh₁ = ½ v₂²
½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²
v = 102 cm/s
(b) The flow rate is constant.
v₁ A₁ = v₂ A₂
(25.0 cm/s) (2.00 cm²) = (102 cm/s) A
A = 0.490 cm²
3 For this force system the equivalent system at P is ___________ A FRP 40 lb along x dir and MRP 60 ft lbB FRP 0 lb and MRP 30 ft lbC FRP 30 lb along y dir and MRP 30 ft lbS FRP 40 lb along x dir and MRP 30 ft lb
This question is incomplete, the complete question is;
For this force system the equivalent system at P is ___________
A) FRP = 40 lb (along +x-dir.) and MRP = +60 ft.lb
B) FRP = 0 lb and MRP = +30 ft.lb
C) FRP 30 lb (along +y-dir.) and MRP = -30 ft.lb
D) FRP 40 lb (along +x-dir.) and MRP = +30 ft.lb
Answer:
D) FRP 40 lb (along +x-dir.) and MRP = +30 ft.lb
Explanation:
From the figure in the image i uploaded along this answer;
FRP = ( 40 lb i + 30 lb j ) + [30 lb (-j)]
Where i and j are the unit vectors along X & Y axis respectively.
So, FRP = 40 lb i
that is, FRP = 40 lb along +X direction
MRP = [ 30 lb x ( 1 ' + 1' ) ] +( -30 lb x 1 ' )
= (30 lb x 2 ' )- 30 lb ft
= 60 lb ft - 30 lb ft
= 30 lb ft
Therefore option(D) is correct
An electron moving in the direction of the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the
Answer:
-z
Explanation:
The force on a moving charge due to a magnetic field follows the right hand rule, so a positive charge, experiencing a magnetic deflection in the -y direction, while it moves in the direction of the x-axis, will do it due to a magnetic field pointing in the +z direction.
As the electron has a negative charge, the magnetic field will point in the opposite direction, i.e., in the -z direction.
A car starts from rest and accelerates for 7.2 s with an acceleration of 1.4 m/s2. How far does it travel? Answer in units of m.
Answer:
xn = 36.28 [m]
Explanation:
To solve this problem we must use the following equation of kinematics, which is ideal for a body that moves with constant acceleration.
[tex]x=x_{0}+(v_{o} *t)+(\frac{1}{2} )*a*t^{2}[/tex]
where:
x - xo = displacement of the car [m]
Vo = initial velocity = 0
t = time = 7.2 [s]
a = acceleration = 1.4 [m/s^2]
The initial velocity is zero, as the car begins its movement from rest.
xn = (x - xo), Now replacing
xn = (0*7.2) + 0.5*1.4*(7.2^2)
xn = 36.28 [m]
A recipe gives the instructions below
After browning the meat pour off fat from the pan to further reduce fat use a strainer.
what type lf separation methods are described in the recipe
A decantation and screening
B distillation and screening
C decantation and centrifugation
D distillation and filtration
Answer:
A. decantation and screening
Explanation:
Decantation is the one of the process of separating the mixture. In this process the precipitated liquid is separated from the solid. According to the given instruction for the recipe, the fat which is in liquid state is separated from meat. In the process of screening, more liquid is separated by placing the mixture on the screen. Here, the gravity plays an important role for the process of separation.
Answer:
a
Explanation:
A car at the top of a ramp starts from rest and rolls to the bottom of the ramp, achieving a certain final speed. If you instead wanted the car to achieve twice as much speed at the bottom of the ramp, how high should the ramp be compared to the first case
Answer:
It must be 4 times high.
Explanation:
Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.This means, that at any time, the following must be true:ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)⇒ [tex]m*g*h = \frac{1}{2} * m*v^{2}[/tex]
Let's call v₁, to the final speed of the car, and h₁ to the height of the ramp.So, at the bottom of the ramp, all the gravitational potential energy
must be equal to the kinetic energy of the car (Defining the bottom of
the ramp as our zero reference for the gravitational potential energy):
[tex]m*g*h_{1} = \frac{1}{2} * m*v_{1} ^{2}[/tex] (1)
Now, let's do v₂ = 2* v₁Replacing in (1) we get:[tex]m*g*h_{2} = \frac{1}{2} * m*(2*v_{1}) ^{2}[/tex] (2)
Dividing (2) by (1), and rearranging terms, we get:h₂ = 4* h₁waht is science
wjwissbsskdldmndndnd
Answer:
the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.
Explanation:
how are s waves and p waves simuliar?
A.they shake the ground
B.they travel through liquids
C. they arrive at the same time
D.they shake the ground from side to side
Answer:
A
Explanation:
hope this helps