Answer:
(a) F= 6.68*10¹¹⁴ N (-k)
(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Explanation
To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:
F=q* v X B Formula (1 )
q: charge (C)
v: velocity (m/s)
B: magnetic field (T)
vXB : cross product between the velocity vector and the magnetic field vector
Data
q= -1.24 * 10¹¹⁰ C
v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j
B =(1.40 T)i
B =(1.40 T)k
Problem development
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =
= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)
F= 6.68*10¹¹⁴ N (-k)
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =
=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s
F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
The force on a charged particle in a magnetic field is calculated using the Lorentz force equation. For the given particle with charge -1.24 x 10⁻⁹ C and velocity vector, the force is determined by the cross product of the velocity and the magnetic field, taking the charge into account.
Explanation:A particle with a charge of -1.24 x 10⁻⁹ C is moving with an instantaneous velocity of (4.19 X 10⁴ m/s)î + (-3.85 X 10⁴ m/s)ᴇ. To find the force exerted on this particle by a magnetic field we use the Lorentz force equation, which is F = q(v x B), where F is the force on the particle, q is the charge, v is the velocity of the particle, and B is the magnetic field.
For part (a) where the magnetic field B is (1.40 T)î, the velocity vector v is perpendicular to B since v has no i-component, thus the force can be found simply by calculating the magnitude as q*v*B since sin(θ) is 1 for θ = 90°. The direction of the force is given by the right hand rule, considering that both v and B are vectors and the charged particle is negatively charged.
For part (b) where the magnetic field B is (1.40 T)ᴅ, the velocity vector v has no k-component, thus v is perpendicular to B and the same principle applies.
The magnitude of the force in both scenarios is computed using the charge, the magnitude of velocity (which is the combination of both the i and j components), and the magnitude of the magnetic field. The direction will differ based on the cross product between v and B.
Learn more about Lorentz Force here:https://brainly.com/question/15552911
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A computer monitor accelerates electrons and directs them to the screen in order to create an image. If the accelerating plates are 1.45 cm apart, and have a potential difference of 2.50 x 10^4 V , what is the magnitude of the uniform electric field between them?
Answer:
Electric field at a distance of 1.45 cm will be [tex]172.41\times 10^4N/C[/tex]
Explanation:
We have given the distance d = 1.45 cm = 0.0145 m
And the potential difference [tex]V=2.5\times 10^4volt[/tex]
There is a relation between potential difference and electric field
Electric field at a distance d due to a potential difference is given by
[tex]E=\frac{V}{d}[/tex], here E is electric field, V is potential difference and d is distance
So [tex]E=\frac{V}{d}=\frac{2.5\times 10^4}{0.0145}=172.41\times 10^4N/C[/tex]
The magnitude of the uniform electric field between the plates is approximately [tex]\( 1.724 \times 10^4 \) V/m.[/tex]
The magnitude of the uniform electric field between the plates is [tex]\( E = \frac{\Delta V}{d} \)[/tex], where [tex]\( \Delta V \)[/tex] is the potential difference and ( d ) is the distance between the plates.
Now, we can calculate the electric field ( E ):
[tex]\[ E = \frac{\Delta V}{d} = \frac{2.50 \times 10^4 \text{ V}}{1.45 \times 10^{-2} \text{ m}} \][/tex]
[tex]\[ E = \frac{2.50 \times 10^4}{1.45} \times 10^2 \text{ V/m} \][/tex]
[tex]\[ E = 1.724 \times 10^4 \text{ V/m} \][/tex]
An electron is travelling in a straight line with a kinetic energy K = 1.60 x 10^-17J. What are (a) the magnitude and (b) the direction of the electric field that will stop the electron in a distance of 10.0 cm? For part (b), make a drawing showing the direction of motion of the electron, the direction of the electric field and the electric force on the electron
Answer:
(a) 1000 N/C
Explanation:
Kinetic energy of electron, K = 1.6 x 10^-17 J
distance, d = 10 cm = 0.1 m
Let the potential difference is V and the electric field is E.
(a) The relation between the kinetic energy and the potential difference is
K = e V
V = K / e
Where, e be the electronic charge = 1.6 x 10^-19 C
V = [tex]\frac{1.6\times 10^{-17}}{1.6\times 10^{-19}}[/tex]
V = 100 V
The relation between the electric field and the potential difference is given by
V = E x d
100 = E x 0.1
E = 1000 N/C
(b) The force acting on the electron, F = q E
where q be the charge on electron
So, F = -e x E
It means the direction of electric field and the force are both opposite to each other.
The direction of electric field and the force on electron is shown in the diagram.
A 100kg cannon at rest contains a 10kgcannonball. when fired,
the cannonball leaves the cannon with avelocity of 90m/s. what is
the magnitude of the recoil speed of thecannon.
Answer:
9 m/s
Explanation:
mass of cannon, M = 100 kg
mass of cannon ball, m = 10 kg
velocity of cannon ball, v = 90 m/s
Let the recoil velocity of cannon is V.
Us ethe conservation of linear momentum, as no external force is acting on the system, so the linear momentum of the system is conserved.
Momentum before the firing = momentum after the firing
M x 0 + m x 0 = M x V + m x v
0 = 100 x V + 10 x 90
V = - 9 m/s
Thus, the recoil velocity of cannon is 9 m/s.
Two 2.4 cm -diameter disks face each other, 1.0 mm apart. They are charged to ±11nC. What is the electric field strength between the disks?
Answer:
[tex]\rm 1.374\times 10^6\ N/C.[/tex]
Explanation:
Given:
Diameter of each disc, D = 2.4 cm.Distance between the discs, d = 1.0 mm.Charges on the discs, q = ±11 nC = [tex]\rm \pm 11\times 10^{-9}\ C.[/tex]The surface area of each of the disc is given by
[tex]\rm A=\pi \times Radius^2.\\\\Radius = \dfrac D2=\dfrac{2.4\ cm}{2} = 1.2\ cm = 1.2\times 10^{-2}\ m.\\A = \pi \times (1.2\times 10^{-2} )^2=4.524\times 10^{-4}\ m^2.[/tex]
For the case, when d<<D, the strength of the electric field at a point due to a charged sheet is given by
[tex]\rm E=\dfrac{|\sigma|}{2\epsilon_o}[/tex]
where,
[tex]\sigma[/tex] = surface charge density of the disc = [tex]\rm \dfrac qA[/tex].[tex]\epsilon_o[/tex] = electrical permittivity of free space = [tex\rm ]9\tiimes 10^9\ Nm^2/C^2[/tex].The electric field strength between the discs due to negative disc is given by
[tex]\rm E_1 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.[/tex]
Since, the electric field is directed from positive charge to negative charge, therefore, the direction of this electric field is towards the negative disc.
The electric field strength between the discs due to positive disc is given by
[tex]\rm E_2 = \dfrac{|\sigma|}{2\epsilon_o}=\dfrac{|q|}{2A\epsilon_o}.[/tex]
The direction of this electric field is away from the positive disc, i.e., towards negative disc.
The electric field between the discs due to both the disc is towards the negative disc, therefore the total electric field strength between the discs is given by
[tex]\rm E=E_1+E_2\\=\dfrac{|q|}{2A\epsilon_o}+\dfrac{|q|}{2A\epsilon_o}\\=\dfrac{|q|}{A\epsilon_o}\\=\dfrac{11\times 10^{-9}}{2\times (4.524\times 10^{-4})\times (8.85\times 10^{-12})}\\=1.374\times 10^6\ N/C.[/tex]
What is gravity at north pole, South pole and at different point on the equatorial regions. Give reasoning for your answers why do you think it is different or same. Can you imagine same concept for the electric charge, yes or No
Answer:
The gravity at Equator is 9.780 m/s2 and the gravity at poles is 9.832 m/s2. The gravity at poles are bigger than at equator, principally because the Earth is not totally round. The gravity is inversely proportional to the square of the radius, that is the reason for the difference of gravity (The radius at Poles are smaller than at Equator).
If Earths would have a net charge Q. The Electric field of Earth would be inversely proportional to the square of the radius of Earth (Electric field definition for a charge), the same case as for gravity. So there would be a difference between the electric field at poles and equator, too.
A certain reaction X + Y → Z is described as being first order in [X] and third order overall. Which of the following statements is or are true?: The rate law for the reaction is: Rate = k[X][Y]2. If the concentration of X is increased by a factor of 1.5, the rate will increase by a factor of 2.25. If the concentration of Y is increased by a factor of 1.5, the rate will increase by a factor of 2.25.
Answer:
The second statement is true: If the concentration of Y is increased by a factor of 1.5, the rate will increase by a factor of 2.25.
Explanation:
Hi there!
Let´s write the rate law for the original reaction and the reaction with X increased by 1.5:
rate 1 =k [X][Y]²
rate 2 = k[1.5 X][Y]²
Now we have to demonstrate if rate 2 = 2.25 rate 1
Let´s do the cocient between the two rates:
rate 2/ rate 1
if rate 2 = 2.25 rate 1
Then,
rate 2 / rate 1 = 2.25 rate 1 / rate 1 = 2.25
Let´s see if this is true using the expressions for the rate law:
rate 2 / rate 1
k[1.5 X][Y]² / k [X][Y]² = 1.5 k [X][Y]² / k[X][Y]² = 1.5
2.25 ≠ 1.5
Then the first statement is false.
Now let´s write the two expressions of the rate law, but this time Y will be increased by 1.5:
rate 1 = k[X][Y]²
rate 2 = k[X][1.5Y]²
Again let´s divide both expressions to see if the result is 2.25
rate 2 / rate 1
k[X][1.5Y]²/ k [X][Y]²
(distributing the exponent)
(1.5)²k [X][Y]² / k [X][Y]² = (1.5)² = 2.25
Then the second statement is true!
Answer:
a) True
b)False
c) True
Explanation:
The order of reactants decides the exponents of respective reactant concentrations.
Since X is first order, exponent is = 1
Overall third order, Exponent X + Exponent Y = 3
Hence, exponent Y = 2.
Hence,
Rate = k[X][Y]^2
b)
If X conc is increased by 1.5 Rate should increase by 1.5 because k proportional to [X]. Hence, statement is False
c)
If Y conc is increased by 1.5 Rate should increase by 2.25 because k proportional to [Y]^2 = (1.5)^2 = 2.25. Hence, statement is True
When sound travels through the ocean, where the bulk modulus is 2.34 x 10^9 N/m^2, the wavelength associated with 1000.0 Hz waves is 1.51 m. 1) Calculate the density of seawater. (Express your answer to three significant figures.)
Answer:
1.03×10³ kg/m³
Explanation:
β = Bulk modulus = 2.34×10⁹ N/m²
f = Frequency = 1000 Hz
λ = Wavelength = 1.51 m
ρ = Density
Speed of the wave
v = fλ
⇒v = 1000×1.51
⇒v = 1510 m/s
Speed of sound
[tex]v=\sqrt{\frac{\beta}{\rho}}\\\Rightarrow \rho=\frac{\beta}{v^2}\\\Rightarrow \rho=\frac{2.34\times 10^9}{1510^2}\\\Rightarrow v=1026.27\ kg/m^3[/tex]
Density of seawater is 1.03×10³ kg/m³ or 1.03 g/cm³ (rounding)
Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the pressure P of the gas is inversely proportional to the volume V of the gas. (a) Suppose that the pressure of a sample of air that occupies 0.106 m3 at 25°C is 50 kPa. Write V as a function of P. (b) Calculate dVydP when P − 50 kPa. What is the meaning of the derivative? What are its units?
(a) V = [tex]\frac{8.3}{P}[/tex]
(b) (i) the value of [tex]\frac{dV}{dP}[/tex] when P = 50kPa is - 0.00332 [tex]\frac{m^{3} }{kPa}[/tex]
(ii) the meaning of the derivative [tex]\frac{dV}{dP}[/tex] is the rate of change of volume with pressure.
(iii) and the units are [tex]\frac{m^{3} }{kPa}[/tex]
Explanation:Boyle's law states that at constant temperature;
P ∝ 1 / V
=> P = k / V
=> PV = k -------------------------(i)
Where;
P = pressure
V = volume
k = constant of proportionality
According to the question;
When;
V = 0.106m³, P = 50kPa
Substitute these values into equation (i) as follows;
50 x 0.106 = k
Solve for k;
k = 5.3 kPa m³
(a) To write V as a function of P, substitute the value of k into equation (i) as follows;
PV = k
PV = 8.3
Make V subject of the formula in the above equation as follows;
V = 8.3/P
=> V = [tex]\frac{8.3}{P}[/tex] -------------------(ii)
(b) Find the derivative of equation (ii) with respect to V to get dV/dP as follows;
V = [tex]\frac{8.3}{P}[/tex]
V = 8.3P⁻¹
[tex]\frac{dV}{dP}[/tex] = -8.3P⁻²
[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{P^{2} }[/tex]
Now substitute P = 50kPa into the equation as follows;
[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{50^{2} }[/tex] [[tex]\frac{kPam^{3} }{(kPa)^{2} }[/tex]] ----- Write and evaluate the units alongside
[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{2500 }[/tex] [[tex]\frac{kPam^{3} }{k^{2} Pa^{2} }[/tex]]
[tex]\frac{dV}{dP}[/tex] = - 0.00332 [[tex]\frac{m^{3} }{kPa}[/tex]]
Therefore,
(i) the value of [tex]\frac{dV}{dP}[/tex] when P = 50kPa is - 0.00332 [tex]\frac{m^{3} }{kPa}[/tex]
(ii) the meaning of the derivative [tex]\frac{dV}{dP}[/tex] is the rate of change of volume with pressure.
(iii) and the units are [tex]\frac{m^{3} }{kPa}[/tex]
Boyle’s Law describes the inverse relationship between the pressure and volume of a gas at constant temperature. To express volume as a function of pressure, use the formula V = k/P. The derivative of volume with respect to pressure indicates the rate at which volume changes as pressure changes, measured in cubic meters per kilopascal (m3/kPa).
Explanation:Boyle’s Law states that for a given amount of gas at a constant temperature, the volume V is inversely proportional to the pressure P. This relationship can be mathematically represented as PV = k, where k is a constant.
(a) Given a sample of air at 25°C with a volume of 0.106 m3 and a pressure of 50 kPa, we can write the volume as a function of pressure by rearranging the formula to V = k/P. Using the initial conditions, we find that k = P × V = 50 kPa × 0.106 m3 = 5.3 kPa · m3, so V(P) = 5.3 kPa · m3 / P.
(b) To calculate dV/dP when P = 50 kPa, we differentiate the function V(P) with respect to P, resulting in dV/dP = -5.3 kPa · m3 / P2. At P = 50 kPa, this becomes dV/dP = -5.3 / (502) = -0.00212 m3/kPa. The derivative represents the rate of change of volume with respect to pressure, which in this case means that for each increase of 1 kPa in pressure, the volume decreases by 0.00212 m3. The units of the derivative are m3/kPa.
What is the repulsive force between two pith balls that are 11.0 cm apart and have equal charges of −38.0 nC?
Answer:
[tex]F=1.07\times 10^{-3}\ N[/tex]
Explanation:
Given that,
Charge on two balls, [tex]q_1=q_2=-38\ nC=-38\times 10^{-9}\ C[/tex]
Distance between charges, r = 11 cm = 0.11 m
We need to find the repulsive force between balls. Mathematically, it is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{(38\times 10^{-9})^2}{(0.11)^2}[/tex]
F = 0.001074 N
or
[tex]F=1.07\times 10^{-3}\ N[/tex]
So, the magnitude of repulsive force between the balls is [tex]1.07\times 10^{-3}\ N[/tex].
What is the electric force (with direction) on an electron in a uniform electric field of strength 3400 N/C that points due east? Take the positive direction to be east.
Answer:
So, the force its [tex] \ 5.4468 \ 10 ^{-16} [/tex] N to the west.Explanation:
The force [tex]\vec{F}[/tex] on a charge q made by an electric field [tex]\vec{E}[/tex] its
[tex]\vec{F} = q \vec{E}[/tex]
The electric charge of the electron its
[tex]q \ = \ - \ 1.602 \ 10 ^{-19} \ C[/tex].
Taking the unit vector [tex]\hat{i}[/tex] pointing towards the east, the electric field will be:
[tex]\vec{E}= 3400 \ \frac{N}{C} \ \hat{i}[/tex].
So, the force will be:
[tex]\vec{F} = \ - \ 1.602 \ 10 ^{-19} \ C \ * \ 3400 \ \frac{N}{C} \ \hat{i} [/tex]
[tex]\vec{F} = \ - \ 5446.8 \ 10 ^{-19} \ N \ \hat{i} [/tex]
[tex]\vec{F} = \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]
[tex]\vec{F} = \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]
[tex]\vec{F} = \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]
So, the force its [tex] \ 5.4468 \ 10 ^{-16} [/tex] N to the west.
A man stands on top of a 25 m tall cliff. The man throws a rock upwards at a 25 degree angle above the horizontal with an initial velocity of 7.2 m/s. What will be the vertical position of the rock at t = 2 seconds? What will be the horizontal position of the rock at that time?
Answer:
a) vertical position at 2s: 50.68 m
b) horizontal position at 2s: 13.05 m
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which the travel of the rock has two components: x-component and y-component. Being the equations to find the position as follows:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Where:
[tex]V_{o}=7.2 m/s[/tex] is the rock's initial speed
[tex]\theta=25\°[/tex] is the angle at which the rock was thrown
[tex]t=2 s[/tex] is the time
y-component:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (2)
Where:
[tex]y_{o}=25 m[/tex] is the initial height of the rock
[tex]y[/tex] is the height of the rock at 2 s
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
Knowing this, let's begin with the answers:
a) Vertical position at 2 s:
[tex]y=25 m+(7.2 m/s)sin(25\°) (2 s)-\frac{(9.8m/s^{2})(2)^{2}}{2}[/tex] (3)
[tex]y=25 m+6.085 m-19. 6 m[/tex] (4)
[tex]y=50.68 m[/tex] (5) This is the vertical position at 2 s
b) Horizontal position at 2 s:
[tex]x=(7.2 m/s) cos(25\°) (2 s)[/tex] (5)
[tex]x=13.05 m[/tex] (6) This is the horizontal position at 2 s
Explanation:
It is given that,
Position of the man, h = 25 m
Initial velocity of the rock, u = 7.2 m/s
The rock is thrown upward at an angle of 25 degrees.
(a) Let y is the vertical position of the rock at t = 2 seconds. Using the equation of kinematics to find y as :
[tex]y=-\dfrac{1}{2}gt^2+u\ sin\theta\ t+h[/tex]
[tex]y=-\dfrac{1}{2}\times 9.8(2)^2+7.2\ sin(25)\times 2+25[/tex]
y = 11.48 meters
(b) Let x is the horizontal position of the rock at that time. It can be calculated as :
[tex]x=u\ cos\theta \times t[/tex]
[tex]x=7.2\ cos(25)\times 2[/tex]
x = 13.05 meters
Hence, this is the required solution.
A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.
A) If the paintball stops completely as a result of striking its target, what is the magnitude of the change in the paintball’s momentum?
B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, what is the magnitude of the change in the paintball’s momentum?
C) The strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. Use this idea along with your answers to (a) and (b) to explain why a paintball bouncing off your skin hurts more than a paintball exploding upon your skin.
Answer:
(A) - 0.273 kg m /s
(B) - 0.546 kg m /s
Explanation:
mass of paintball, m = 0.0032 kg
initial velocity, u = 85.3 m/s
(A) Momentum is defined as the product of mass of body and the velocity of body.
initial momentum, pi = mass x initial velocity
pi = 0.0032 x 85.3 = 0.273 kg m /s
Finally the ball stops, so final velocity, v = 0 m/s
final momentum, pf = mass x final velocity = 0.0032 x 0 = 0 m/s
change in momentum = pf - pi = 0 - 0.273 = - 0.273 kg m /s
(B) initial velocity, u = + 85.3 m/s
final velocity, v = - 85.3 m/s
initial momentum, pi = mass x initial velocity
pi = 0.0032 x 85.3 = 0.273 kg m /s
final momentum, pf = mass x final velocity = - 0.0032 x 85.3 = - 0.273 kgm/s
change in momentum = pf - pi = - 0.273 - 0.273 = - 0.546 kg m /s
(C) According to the Newton's second law, the rate of change of momentum is directly proportional to the force exerted.
As the change in momentum in case (B) is more than the change in momentum in case (A), so the force exerted on the skin is more in case (B).
You need to figure out how high it is from the 3rd floor of Keck science building to the ground. You know that the distance from the ground to the 2nd floor is 5m. Your friend drops a ball from rest from the 3 rd floor and you start to time the ball when it passes by you at the 2nd floor. You stop the timer when the ball hits the ground. The time recorded is 0.58 seconds.
Answer:1.624 m
Explanation:
Given
height of 2nd floor =5 m
time recorded is 0.58 sec
Let h be the height of third floor above 2 nd floor
[tex]h=ut+\frac{at^2}{2}[/tex]
here u=0
t=time taken to cover height h
[tex]h=\frac{gt^2}{2}[/tex] ----------1
Now time taken to complete whole building length
[tex]h+5=\frac{g(t+0.58)^2}{2}[/tex] ----------2
Subtract 1 form 2
[tex]5=\frac{g(2t+0.58)(0.58)}{2}[/tex]
Taking gravity [tex]g=1 m/s^2[/tex]
1=(2t+0.58)(0.58)
t=0.57 s
Substitute the value of t in 1
h=1.624 m
A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be lifted if one pound of fat contains approximately 3500 Calories. If the weight can be lifted once every two seconds, how long would this activity take? 1 Cal=4184 J.
Answer:
37357 sec
or 622 min
or 10.4 hrs
Explanation:
GIVEN DATA:
Lifting weight 80 kg
1 cal = 4184 J
from information given in question we have
one lb fat consist of 3500 calories = 3500 x 4184 J
= 14.644 x 10^6 J
Energy burns in 1 lift = m g h
= 80 x 9.8 x 1 = 784 J
lifts required [tex]= \frac{(14.644 x 10^6)}{784}[/tex]
= 18679
from the question,
1 lift in 2 sec.
so, total time = 18679 x 2 = 37357 sec
or 622 min
or 10.4 hrs
A cart is given an initial velocity of 5.0 m/s and
experiencesa constant acceleration of 2.0 m/s ^2. What is the
magnitude of thecart's displacement during the first 6.0 s of its
motion?
Answer:
cart displacement is 66 m
Explanation:
given data
velocity = 5 m/s
acceleration = 2 m/s²
time = 6 s
to find out
What is the
magnitude of cart displacement
solution
we will apply here equation of motion to find displacement that is
s = ut + 0.5×at² .............1
here s id displacement and u is velocity and a is acceleration and time is t here
put all value in equation 1
s = ut + 0.5×at²
s = 5(6) + 0.5×(2)×6²
s = 66
so cart displacement is 66 m
A charge if 2 nC is placed 10cm to the right of a conducting sphere with a diameter of 2 cm. A charge of 5 nC is placed 10 nC to the left of the same sphere. determine the charge at the center of the sphere.
Answer:
The charge at the center of the conducting sphere is zero.
Explanation:
A principle of conductor materials is that the electric field inside a conductor in electrostatic state is always zero. The gauss law says that the flux of a electric field in a closed surface is proportional to the charge enclosed by the surface. Then if the Electric field inside of a conductor is zero, imperatively the charge anywhere inside the conductor is zero too, so the charge at the center of the sphere is zero.
Two charges q1 and q2 have a total charge of 10 mu or micro CC. When they are separated by 4.1 m, the force exerted by one charge on the other has a magnitude of 8 mN. Find q1 and q2 if both are positive so that they repel each other, and q1 is the smaller of the two. (For the universal constant k use the value 8.99 times 109 N m2/C2.)
Answer:
q₂ = 6.8 x 10⁻⁶C
q₁ = 3.2 x 10⁻⁶ C
Explanation:
First charge is q₂ and second charge will be ( 10 x10⁻⁶ - q₂ )
Force between them F = 8 X 10⁻³ N distance between them d = 4.1 m
F = k ( q₁ xq₂ ) / d²
8 x 10⁻³ = [tex]\frac{8.99\times10^9\times q_2\times (10^{-5}-q_2)}{(4.1)^2}[/tex]
14.95 x 10⁻¹² = 10⁻⁵q₂ - q₂²
q₂²-10⁻⁵q₂ + 14.95 x 10⁻¹² =0
This is a quadratic equation .The solution of it gives the value of q₂
q₂ = 6.8 x 10⁻⁶C
q₁ = 3.2 x 10⁻⁶ C
q₁ = 3.2 x 10⁻⁶ C
q₂ = 6.8 x 10⁻⁶C
What is Charge in Physics?Charge is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.
It is also known as electric charge, electrical charge, or electrostatic charge and symbolized q.
Given,
First charge is q₂ and second charge will be ( 10 x10⁻⁶ - q₂ )
F = 8 X 10⁻³ N
d = 4.1 m
By using the Formula,
[tex]F =\frac{k ( q_{1} * q_{2} ) }{d^{2} }[/tex]
8 x 10⁻³ = [tex]\frac{8.99*10^{9}*q_{2} *10^{-5} -q_{2} }{(4.1)^{2} }[/tex]
14.95 x 10⁻¹²
= 10⁻⁵q₂ - q₂²
= q₂²-10⁻⁵q₂ + 14.95 x 10⁻¹²=0
Therefore,
The above equation is a quadratic equation , it gives the value of q₂
q₂ = 6.8 x 10⁻⁶C
q₁ = 3.2 x 10⁻⁶ C
Learn more about Charges here:https://brainly.com/question/14467306
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What should a graph of the velocity vs time look like if it shows an object moving toward the motion detector at a constant speed? a. An upwards curved parabola.
b. A downwards curved parabola.
c. A straight line with zero slope.
d. A straight line sloping upward.
Answer:
option C
Explanation:
the correct answer is option C.
Graph between velocity time which shows the constant velocity is straight line with slope zero.
constant velocity means velocity is not changing with respect to time hence this condition will be only when graph is straight line with slope zero.
area under velocity time graph shows the displacement of the object.
And where as slope of velocity time graph show acceleration of the object.
Final answer:
The correct option is (d) A straight line sloping upward. A graph of velocity vs time that shows an object moving toward a motion detector at a constant speed is represented as a straight line sloping upward, indicating a constant positive velocity.
Explanation:
When considering an object moving toward a motion detector at a constant speed, the velocity vs. time graph should depict a constant velocity. This is visualized as a straight line on such a graph, since the speed of the object does not change over time.
The correct choice to represent an object moving at a constant speed towards a motion detector is (d) A straight line sloping upward.
The key concept here is that when velocity is constant, the slope of the velocity vs. time graph remains constant. If the object is moving toward the motion detector, the velocity is positive, therefore, the graph slopes upwards, indicating a positive velocity that does not change.
This scenario clearly shows the relationship between velocity and time when an object moves at a constant speed.
How many electrons would have to be removed from a coin to leave it with a charge of +2.2 x 10^- 8 C?
Answer:
[tex]N=1.375*10^{11}[/tex] electrons
Explanation:
The total charge Q+ at the coin is equal, but with opposite sign, to the charge negative Q- removed of it. Q- is the sum of the charge of the N electrons removed from the coin:
[tex]Q_{-}=N*q_{e}[/tex]
[tex]Q_{-}=-Q_{+}[/tex]
q_{e}=-1.6*10^{-19}C charge of a electron
We solve to find N:
[tex]N=-Q_{+}/q_{e}=-2.2*10^{-8}/(-1.6*10^{-19})=1.375*10^{11}[/tex]
In reaching her destination, a backpacker walks with an average velocity of 1.32 m/s, due west. This average velocity results, because she hikes for 5.21 km with an average velocity of 3.49 m/s due west, turns around, and hikes with an average velocity of 0.687 m/s due east. How far east did she walk (in kilometers)?
Answer:
distance in east is 1273.78 m
Explanation:
given data
average velocity = 1.32 m/s west =
hike = 5.21 km = 5.21 × 10³ m
average velocity = 3.49 m/s west
average velocity = 0.687 m/s east
to find out
distance in east
solution
we consider here distance in east is = x
so distance from starting point = 5.21 × 10³ - x ...................1
and we can say time required to reach end
time required = distance / speed
time required = [tex]\frac{5.21 *10^3 - x}{1.32}[/tex] ................2
and
time required for 6.44 km west
time required = [tex]\frac{5.21 *10^3 - x}{3.49}[/tex] ................3
and time required for distance x
time required = [tex]\frac{x}{0.687}[/tex] ................4
so from equation 2 , 3 and 4
[tex]\frac{5.21 *10^3 - x}{1.32}[/tex] = [tex]\frac{5.21 *10^3 - x}{3.49}[/tex] + [tex]\frac{x}{0.687}[/tex]
x = 1273.78 m
so distance in east is 1273.78 m
A road goes down a slope. For every 12 km measured along the ground, the road drops 500 m in elevation. a) What is the angle of slope on the road?
b) What is the map distance along the road for every 12 km actually traveled? (measured in km)
c) What is the map distance along the road for every 1 mile actually traveled? (measured in miles)
Answer:
a) 2.24°
b) 11.99 km
c) 0.9992 miles
Explanation:
We can think of the road as a triangle with a 12 km hypotenuse and a 0.5 km side. Then:
l = 12
h = 0.5
and
sin(a) = h / l
a = arcsin(h / l)
a = arcsin(0.5 / 12)
a = 2.24°
That is the slope.
The map distance travelled would be the other side of the triange.
cos(a) = d / l
d = l* cos(a)
d = 12 * cos(2.24) = 11.99 km
And for miles
d = 1 mile * cos(2.24) = 0.9992 miles
If an object moves twice as fast how do you think its energy changes? halves
doesn't change
doubles
quadruples
Answer:
If an object moves twice as fast its kinetic energy quadruples.
Explanation:
The kinetic energy (K₁) of a body of mass (m) that moves with speed (v) is:
K₁= 1/2 * m* v²
If we double the speed of the body, its kinetic energy (K₂) will be:
K₂= 1/2 * m*( 2v)²
K₂= 1/2 * m* 4 *v²
K₂= 4(1/2 * m *v²)
K₂= 4*K₁
Nerve impulses in a human body travel at a speed of about 100 m/s. Suppose a man accidentally stubs his toe. About how much time does it take the nerve impulse to travel from the foot to the brain (in s)? Assume the man is 1.80 m tall and the nerve impulse travels at uniform speed.
Answer:
t = 0.018 s
Explanation:
given,
nerve impulse in human body travel at a speed of = 100 m/s
height of the man = 1.80 m
time taken by the impulse to travel from foot to brain = ?
distance = speed × time
1.80 = 100 × t
t =[tex]\dfrac{1.80}{100}[/tex]
t = 0.018 s
hence, the time taken by the nerve to reach brain from toe is 0.018 s
A car travels in a straight line from a position 50 m to your right to a position 210 m to your right in 5 sec. a. What is the average velocity of the car?b. Construct a position vs. time graph and calculate the slope of the line of best fit.Please give me the exact coordinates to plot. Thank you!
Answer:
a) The average velocity is 32 m/s
b) See the attached figure. Slope of the line = 32.
Graphic: a line that passes through the points (0; 50) and (5; 210)
Explanation:
a) The average velocity is calculated as the displacement over time:
v = ΔX / Δt
where
ΔX : displacement ( final position - initial position)
Δt : time (final time - initial time)
Considering the origin of the reference system as the position where the observer is:
ΔX = 210 m - 50 m = 160 m
Δt = 5 s
v = 160 m / 5 s = 32 m/s
b) In the graphic position vs time, plot a line that passes through the points (0, 50) (because at time 0, the car is 50 m away from you, the center of the reference system) and (5, 210). The x-axis, time, is in seconds and the y-axis, position, in meters. The slope of the line is calculated as:
slope = (X₂ - X₁) /(t₂ - t₁) = (210 - 50) / (5 - 0) = 32. Then, the velocity is equal to the slope of the line. See the attached figure.
Problem 1 The Van de Graaff electrostatic generator develops a charge of approximately −1 × 10−5C and a pith ball has charge of approximately 1 × 10−9C. Which of the following describes the relationship between the size of the force the Van de Graaff exerts on the pith ball and the size of the force the pith ball exerts on the Van de Graaff? Select One of the Following: (a) The Van de Graaff exerts a much larger force on the pith ball. (b) The pith ball exerts a much larger force on the Van de Graaff. (c) The forces have the same magnitude.
Answer:
(c) The forces have the same magnitude.
Explanation:
The electric Force between the two charges are proportional to the product of both charges:
[tex]F_{electric}=k*q_{1}q_{2}/r^{2}[/tex]
Both charges feels the same force but with opposite direction.
(c) The forces have the same magnitude.
You have been working on a new, strong, lightweight ceramic which will be used to replace steel bearing balls. One cubic meter of the steel alloy currently in use has a mass of 8.08 ⨯ 10^3 kg, whereas a cubic meter of your new material has a mass of 3.14 ⨯ 10^3 kg. If the balls currently in use have a radius of 1.9 cm and for this application to keep balls of the same mass instead of the same size, what would the radius replacement ball of the new alloy be?
Answer:
r=2.6 cm
Explanation:
Hi!
Lets call steel material 1, and the new alloy material 2. You know their densities:
[tex]\rho_1=8.08*10^3\frac{kq}{m^3}\\\rho_2=3.14*10^3\frac{kq}{m^3}[/tex]
The volume of a sphere with radius r is given by:
[tex]V=\frac{4}{3}\pi r^3[/tex]
Then the masses of the bearings are:
[tex]m_1=\rho_1V_1=\frac{4}{3}\pi r_1^3 \\m_1=\rho_2V_2=\frac{4}{3}\pi r_2^3[/tex]
For the masses to be the same:
[tex]\rho_1 V_1 = \rho_2 V_2\\\frac{V_2}{V_1} =\frac{\rho_1}{\rho_2}\\(\frac{r_2}{r_1})^3 = \frac{8.08}{3.14}=2.57\\r_2=\sqrt[3]{2.57}\;r1 =1.37r_1=1.37*1.9cm=2.6cm[/tex]
A boy uses a slingshot to launch a pebble straight up into the air. The pebble reaches a height of 37.0 m above the launch point 2.3 seconds later. Assume air resistance is negligible (a) What was the pebble's initial speed (just after leaving the slingshot)? m/s (b) How much time did it take for the pebble to first reach a height of 18.5 m above its launch point? s
The pebble's initial speed is approximately 22 m/s. It takes approximately 1.45 seconds for the pebble to first reach a height of 18.5 m.
Explanation:To determine the pebble's initial speed, we can use the equation for projectile motion:
h = v0yt - (1/2)gt2
Where h is the height, v0y is the initial vertical speed, t is the time, and g is the acceleration due to gravity.
Since the pebble is launched straight up, the final height is equal to the initial height. Plugging in the given values, we have:
37 m = v0y(2.3 s) - (1/2)(9.8 m/s2)(2.3 s)2
Simplifying this equation gives us the value of v0y, the initial vertical speed. To find the pebble's initial speed, we can use the Pythagorean theorem:
v0 = √(v0x2 + v0y2)
Where v0 is the initial speed and v0x is the initial horizontal speed. Since the pebble is launched straight up, v0x = 0. Plugging in the calculated value of v0y, we can solve for v0.
(a) The pebble's initial speed is approximately 22 m/s.
(b) To find the time it takes for the pebble to first reach a height of 18.5 m, we can use the equation for height:
18.5 m = v0yt - (1/2)gt2
Solving for t gives the time it takes for the pebble to reach the desired height.
(b) It takes approximately 1.45 seconds for the pebble to first reach a height of 18.5 m.
The proton with initial v1 = 3.00 x 10^5 m/s enters a region 1.00 cm long where it is accelerated. Its final velocity is v = 6.40 x 10^6 m/s . What was its constant acceleration. Answer in 10^15 m/s^2
Answer:
The constant acceleration, [tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]
Solution:
As per the question:
[tex]v_{1} = 3 x 10^5 m/s[/tex]
[tex]v_{2} = 6.40 x 10^6 m/s[/tex]
length of the re region, l = 1 cm = 0.01 m
Now,
Using the third equation of motion:
[tex]v_{2}^2 = v_{1}^2 + 2a_{c}l[/tex]
[tex]a_{c} = \frac{v_{2}^2 - v_{1}^2}{2l}[/tex]
[tex]a_{c} = \frac{(6.40 x 10^6)^2 - (3 x 10^5)^2}{2\times 0.01}[/tex]
[tex]a_{c} = 2.044\times 10^15 m/s^2[/tex]
A positive point charge on a 2-D plane starts at the position (2 cm, 7 cm). It moves along an equipotential line to the position (10 cm, 7 cm). The plane is oriented parallel to a
uniform electric field of 120 N/C. How much work does the electric field do in this scenario? (Answer in Joules, J. You do not need to type the units, only the number.)
Answer:
W = 0
Explanation:
given data:
starting position is ( 2cm , 7 cm)
ending position is (10 cm, 7 cm)
we know thta work done is given as
[tex] W = q\Delta v[/tex]
here , q is particle charge
[tex]\Delta v[/tex] is change in potential
as we know that intial and final position are having uniform eleectric potential, therefore it have same potential i.e. V_1 = V_2, thus
one more reason of being work done equal to zero is,
Since the distance covered by the item and the direction of motion is always perpendicular to each other during the round journey, no work is therefore carried out.
W = 0
You drop a rock into a deep well and hear the sounds of it hitting the bottom 5.50 s later. If the speed of sound is 340 m/s, determine the depth of the well.
Answer:
The depth of well equals 120.47 meters.
Explanation:
The time it takes the sound to reach our ears is the sum of:
1) Time taken by the rock to reach the well floor.
2) Time taken by the sound to reach our ears.
The Time taken by the rock to reach the well floor can be calculated using second equation of kinematics as:
[tex]h=ut+\frac{1}{2}gt^2[/tex]
where,
'u' is initial velocity
't' is the time to cover a distance 'h' which in our case shall be the depth of well.
'g' is acceleration due to gravity
Since the rock is dropped from rest hence we infer that the initial velocity of the rock =0 m/s.
Thus the time to reach the well base equals
[tex]h=\frac{1}{2}gt_1^2\\\\\therefore t_{1}=\sqrt\frac{2h}{g}[/tex]
Since the propagation of the sound back to our ears takes place at constant speed hence the time taken in the part 2 is calculated as
[tex]t_{2}=\frac{h}{Speed}=\frac{h}{340}[/tex]
Now since it is given that
[tex]t_{1}+t_{2}=5.50[/tex]
Upon solving we get
[tex]\sqrt{\frac{2h}{g}}+\frac{h}{340}=5.5\\\\\sqrt{\frac{2h}{g}}=5.5-\frac{h}{340}\\\\(\sqrt{\frac{2h}{g}})^{2}=(5.5-\frac{h}{340})^{2}\\\\\frac{2h}{g}=5.5^2+\frac{h^2}{(340^2)}-2\times 5.5\times \frac{h}{340}[/tex]
Solving the quadratic equation for 'h' we get
h = 120.47 meters.