Answer:
a. 9.52 cm b. 4.34 × 10⁶ m/s
Explanation:
a. The horizontal distance traveled by the electron when it hits the plate.
The electric force F on the electron due to the electric field E of mass, m is
F = -eE = ma
a = -eE/m where a = acceleration of electron
The vertical distance moved by the electron is given by
Δy = ut +1/2at²
u = initial vertical velocity = 0. and take the top plate as y = 0 and bottom plate as y
So,
0 - y = 0 × t + 1/2at²
-y = 1/2at²
substituting a = -eE/m
-y = 1/2(-eE/m)t²
y = eEt²/2m
making t subject of the formula,
t = √(2ym/eE) where t is the time it takes to reach the bottom plate.
Since E = 4.0 × 10² N/C, y = distance between plates = 2.0 cm = 0.02 m, m = 9.109 × 10⁻³¹kg and e = 1.602 × 10⁻¹⁹ C
t = √[(2 × 0.02 m × 9.109 × 10⁻³¹kg)/(1.602 × 10⁻¹⁹ C × 4.0 × 10² N/C)]
t = √[(0.36436 × 10⁻³¹kgm)/(6.408 × 10⁻¹⁷ N)]
t = √[(0.0569 × 10⁻¹⁴kgm/N)t
t = 0.238 × 10⁻⁷ s
t = 23.8 × 10⁻⁹ s
t = 23.8 ns
The horizontal distance moved when it hits the plates x = vt where v = initial horizontal velocity = 4.0 × 10⁶ m/s
x = 4.0 × 10⁶ m/s × 23.8 × 10⁻⁹ s
= 0.0952 m
= 9.52 cm
b. The velocity of the electron as it strikes the plate.
To find the velocity of the electron as it strikes the plates, we calculate its final vertical velocity V as it strikes the plate. This is gotten from
v' = u + at since u = 0,
v' = at
= -eEt/m
= -(1.602 × 10⁻¹⁹ C × 4.0 × 10² N/C × 0.238 × 10⁻⁷ s)/9.109 × 10⁻³¹kg
= -1.525 × 10⁻²⁴ Ns/9.109 × 10⁻³¹kg
= -0.167 × 10⁷ m/s
= -1.67 × 10⁶ m/s
So, the resultant velocity as it strikes the plate v = √(v'² + v²)
= √((-1.67 × 10⁶ m/s)² + (4 × 10⁶ m/s)²)
= √(2.7889 + 16) × 10⁶ m/s
= √18.7889 × 10⁶ m/s
= 4.335 × 10⁶ m/s
≅ 4.34 × 10⁶ m/s
A 28-turn circular coil of radius 4.40 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies in time according to the expression B = 0.010 0t + 0.040 0t2, where B is in teslas and t is in seconds. Calculate the induced emf in the coil at t = 4.20 s.
Answer:
ε = -0.0589V = -58.9mV
Explanation:
In order to calculate the induced emf in the coil, you use the following formula:
[tex]\epsilon=-N\frac{d\Phi_B}{dt}=-N\frac{d(SBcos\alpha)}{dt}[/tex] (1)
ε: induced emf = ?
N: turns of the coil = 28
ФB: magnetic flux trough the coil
S: cross sectional area of the circular coil = π.r^2
r: radius of the cross sectional area of the coil = 4.40cm = 0.044m
B: magnetic field
α: angle between the direction of the magnetic field and the direction of the normal to the cross area of the coil = 0°
You take into account that the area is constant respect to the magnetic field that cross it. Only the magnetic field is changing with time. The magnetic field depends on time as follow:
[tex]B(t)=0.010t+0.040t^2[/tex] (2)
You replace the expression (2) into the equation (1), evaluate the derivative, and replace the values of the other parameters for t =4.20s:
[tex]\epsilon=-NS\frac{dB}{dt}=-NS\frac{d}{dt}[0.010t+0.040t^2]\\\\\epsilon(t)=-NS(0.010+0.080t)\\\\\epsilon(t)=-(28)(\pi(0.044m)^2)(0.010T/s+0.080T/s^2(4.20s))\\\\\epsilon(t)=-0.0589V=-58.9mV[/tex]
The induced emf in the coil is -58.9mV
1. Two blocks (5.0 kg and 6.0 kg) are connected through a frictionless system. The angle of
the ramp is 45°
A. Find the tension in the string.
B. Determine the normal force on the 5.0 kg block.
Answer: A) 45.5N b) 34.6N
Explanation:
Given the following :
Mass of block A(m1) = 5kg
Mass of block B (m2) = 6kg
Angle of incline (Θ) = 45°
Force pulling the down the incline = m1×a×sinΘ
Tension (T) = (m2×a) - (m2)×g
Where g = acceleration due to gravity
a = acceleration
Net force = (5×9.8×sin45°) - (6×9.8)
Net force = 34.648232 - 58.8
The net force acting on the body = 24.4N
Therefore,
Acceleration, a = Net force/ total mass
a = 24.4 / (6+5)
a = 2.22ms^-2
T = (m2 ×g) - (m2×a)
T = (6 × 9.8) - (6 × 2.22)
T = 58.8 - 13.32
T = 45.48
T = 45.5N
B) Normal reaction:
Horizontal component:
m1gCosΘ = 5 × 9.8 × cos45°
= 5 × 9.8 × 0.7071067
= 34.648232
= 34.6N
The tension on the string is 45.62N and the normal force is 34.65N
Data;
m1 = 5.0kgm2 = 6.0kgangle = 45°Tension in The StringTaking the two mass into consideration
for the 6kg mass
[tex]6g - 6a[/tex]
for the 5kg mass
[tex]5gsin45 + 5a\\[/tex]
Equating the tension in both equations
[tex]6g - 6a = 5gsin45 + 5a[/tex]
Let's solve for a
[tex]6a + 5a = 6g - 5gsin45\\11a = 6g - 5gsin45\\a = \frac{6g - 5gsin45}{11}\\ a = \frac{6* 9.8 - 5*9.8*sin45}{11}\\ a = 2.1956 m/s^2\\T = 6g - 6a\\T = 6*9.8 - 6*2.1956\\T = 45.62N[/tex]
b)
The normal force on the 5kg block.
The normal force on the 5kg block can be calculated as
[tex]N = 5 * 9.8 cos 45 = 34.65N[/tex]
The tension on the string is 45.62N and the normal force is 34.65N
Learn more on tension and normal force here;
https://brainly.com/question/24994188
As the distance between a satellite in a circular orbit and the central object increases, the period of the satellite .
Answer:
As the distance between a satellite in a circular orbit and the central object increases, the period of the satellite increases .
Explanation:
Answer:
Increases
Explanation:
A box of mass 4.5 kg is pushed across a rough surface (μK = 0.18) for a distance of 2.0 m by a constant force of 10 N. If the object reaches a speed of 2.0 m/s by the end of the push, what was its speed at the beginning of the push?
Answer:
Explanation:
Work done by force applied = force x displacement
= 10 x 2 = 20 J
Negative work done by frictional force
= μ mg x d where μ is coefficient of kinetic friction , m is mass and d is displacement
= - .18 x 4.5 x 9.8 x 2
= - 15.87 J
Net positive work done on the mass = 4.13 J
If v was the initial velocity
increase in kinetic energy = positive work done on mass
= 1 / 2 m v² - 1/2 m u² = 4.13 where v is final and u is initial velocity
1 /2 x 4.5 x 2 ² - 1/2 x 4.5 u² = 4.13
9 - 2.25 u² = 4.13
2.25 u² = 4.87
u² = 2.16
u = 1.47 m /s .
5. A bus starting from rest accelerates in a straight line at a constant rate of 3m/s2 for 8s. Calculate the distance travelled by the bus during this time interval.
Answer:
d = 96 meters
Explanation:
a = acceleration
t = time
d = distance
[tex]d = \frac{1}{2} \times a \times {t}^{2} [/tex]
[tex]d = \frac{1}{2} \times 3 \times {8}^{2} [/tex]
[tex]d = 96[/tex]
A wind blows at 100 km / h from the east. An aircraft points due north and sets off. In still air its speed is 150 km / h. What will the aircraft's actual velocity be?
Answer:
180.28 km/h
Explanation:
Let north and south represent the positive and negative y axis and east and west represent the positive and negative x axis.
Given;
Velocity of wind vx= 100 km/s
Velocity of aircraft in still air vy = 150 km/s
The resultant velocity Vr is;
Vr = √(vx^2 + vy^2)
Vr = √(100^2 + 150^2)
Vr = 180.2775637731 km/h
Vr = 180.28 km/h
Is Einstein's equation E=mc^2 is applicable on our earth if it is not why?is it applicable in space can a body moves through speed of light
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◦•●◉✿What does E mc2 actually mean?
E = mc2. It's the world's most famous equation, but what does it really mean? "Energy equals mass times the speed of light squared." On the most basic level, the equation says that energy and mass (matter) are interchangeable; they are different forms of the same thing.✿◉●•◦
Un automovil circula a 126km/h por una autopista. Su conductor observa que a 150 m delante de el, se encuentra un árbol caído que ocupa toda la calzada. Inmediatamente presiona los frenos con una aceleración de -3,5 m/s². Determinar si logra evitar el choque.
Answer:
El conductor no puede evitar el choque.
Explanation:
Primero, convierta la velocidad del conductor a m / s:
1 km/h = 0.277778 m/s
126 km/h = 126 * 0.277778 = 35 m/s
La velocidad del automóvil es de 35 m / s.
El conductor presiona los frenos con una aceleración de -3.5 m / s² para evitar un choque a 150 m por delante.
Veamos qué distancia se moverá el automóvil después de que comience a desacelerar.
Utilizaremos una de las ecuaciones de movimiento lineal de Newton:
[tex]v^2 = u^2 + 2as[/tex]
donde v = velocidad final = 0 m / s (el automóvil debe detenerse)
u = velocidad inicial = 35 m / s
a = aceleración = -3.5 m / s².
s = distancia recorrida
Por lo tanto:
[tex]0 = 35^2 + (2 * -3.5 * s)\\\\=> 1225 = 7.0s\\\\s = 1225 / 7 = 175 m[/tex]
Esto significa que el automóvil se detendrá a 175 m.
Por lo tanto, a esa velocidad y aceleración, el conductor chocará contra el árbol caído porque el automóvil no podrá detenerse antes de alcanzar la posición del árbol.
Ratio of acceleration due to grabity and universal gravitational constant
Answer:
acceleration due to gravity = 9.8m/s^2
universal gravitational constant= 6.67×10 ^_11 nm^2 kg_2
now, ratio=9.8/6.67×10^_11.
An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x 102 N/C and separation between the charged plates is 2.0 cm. Determine the horizontal distance travelled by the electron when it hits the plate.
Answer:
xmax = 9.5cm
Explanation:
In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.
In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.
You use the Newton second law and the electric force on the electron:
[tex]F_e=qE=ma[/tex] (1)
q: charge of the electron = 1.6*10^-19 C
m: mass of the electron = 9.1*10-31 kg
E: magnitude of the electric field = 4.0*10^2N/C
You solve the equation (1) for a:
[tex]a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}[/tex]
Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:
[tex]x_{max}=v_o\sqrt{\frac{2d}{a}}[/tex] (2)
Here, the height d is the distance between the plates d = 2.0cm = 0.02m
vo: initial velocity of the electron = 4.0*10^6m/s
You replace the values of the parameters in the equation (2):
[tex]x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm[/tex]
The horizontal distance traveled by the electron is 9.5cm
a shot putter shoots a 7.3kg shot from rest to 14m/s in 1.5s. what was the average power?
Answer: 477W
Explanation:
Given the following :
Mass (m) = 7.3kg
Initial Velocity (u) = 0
Final velocity (v) = 14m/s
time (t) = 1.5s
Power = workdone (W) / time (t)
The workdone can be calculated as the change in kinetic energy (KE) :
Recall ;
KE = 0.5mv^2
Therefore, change in KE is given by:
0.5mv^2 - 0.5mu^2
Change in KE = 0.5(7.3)(14^2) - 0.5(7.3)(0^2)
Change in KE = 715.4J
Therefore ;
Average power = Workdone / time
Workdone = change in KE = 715.4N
Average power = 715.4 / 1.5
Average power = 476.93333 W
= 477W
A force of 6 N is used to open a door, wherein the distance of the force to the axis of rotation is 80 cm. If the angular acceleration it acquires is 0.5 rad/s 2, determine the moment of inertia of the door.
Explanation:
∑τ = Iα
(6 N) (0.80 m) = I (0.5 rad/s²)
I = 9.6 kg m²
Where is the energy in a glucose molecule stored?
Answer:
Energy is stored in the bonds between atoms
Answer:
Energy is stored in the bonds between atoms
Explanation:
Ape-x
Samuel applies a horizontal force of 35.0 N to a sleigh over a distance of 1.50 m along a level surface. Calculate the work done on the sleigh by samuel. show your work.
Answer:
52.5 J
Explanation:
Work done (W) is the product of the force (F) applied on a body and the distance (s) moved in the direction of the force.
i.e W = F × s
It is a scalar quantity and measured in Joules (J).
Given that: F = 35.0 N and s = 1.50 m, then;
W = F × s
W = 35.0 × 1.5
= 52.5 J
Therefore, the work done on the sleigh by Samuel is 52.5 J.
How do you calculate the speed of a 3.1eV photon and a 3.1eV electron?
Explanation:
We need to find the speed of a 3.1eV proton and a 3.1eV electron.
For proton, using conservation of energy such that,
[tex]\dfrac{1}{2}m_pv_p^2=eV\\\\v_p^2=\dfrac{2eV}{m_p}\\\\v_p=\sqrt{\dfrac{2eV}{m_p}}[/tex]
[tex]m_p[/tex] is mass of proton
[tex]v_p=\sqrt{\dfrac{2\times 3.1\times 1.6\times 10^{-19}\ J}{1.67\times 10^{-27}\ kg}}\\\\v_p=2.43\times 10^5\ m/s[/tex]
For electron,
[tex]\dfrac{1}{2}m_ev_e^2=eV\\\\v_e^2=\dfrac{2eV}{m_e}\\\\v_e=\sqrt{\dfrac{2eV}{m_e}}[/tex]
[tex]m_e[/tex] is mass of proton
[tex]v_e=\sqrt{\dfrac{2\times 3.1\times 1.6\times 10^{-19}\ J}{9.1\times 10^{-31}\ kg}}\\\\v_e=1.44\times 10^6\ m/s[/tex]
Hence, this is the required solution.
Derive the following equations for uniformly accelerated motion by graphical method. a) Velocity -time relation b) Position - time relation 3) Position – velocity relation.
Answer:
a) velocity - time realation
that graph's gradient gives the uniform acceleration
Why are fuel cells used on space shuttles instead of batteries?
Answer:
Fuel cells are used in the space shuttle as one component of the electrical power system. ... The fuel cell power plants generate heat and water as by-products of electrical power generation. The excess heat is directed to fuel cell heat exchangers, where the excess heat is rejected to Freon coolant loops
A drop
of
oil volume 10 m
Spreads out on water to make a
Circular fils of diameter 10
What is that thickness?
Answer:
[tex]27.5\ m[/tex]
Explanation:
As we know that volume of cylinder is
[tex]v=\pi r^{2} *h[/tex]
Where v=volume , h= height or thickness and r= radius
Here,
[tex]v= 10 m ,\ diameter= 10, \ r=\frac{diameter}{2} \ r=\frac{10}{2}\\ r=5[/tex]
Putting these values in the previous equation , we get
[tex]10\ = \frac{22}{7} *5 *5*h\\ 14\ =\ 110*h\\h=\frac{110}{14} \\h=\frac{55}{2} \\\\h=27.5\ m[/tex]
Therefore thickness is 27.5 m
What is electronic configuration?
Answer:
Electronic Configuration is the distribution of electrons in sub shells (s,p,d,f).
For Example,
The electronic configuration of Carbon (6 electrons) is [tex]1s^22s^22p^2[/tex]
0.0000302 in scientific notation
Answer:
3.02x10^-5
Explanation:
A scientific notation consists of
c x 10^n
the c must be a number between 1-9, while n must be an integer.
it indicates the c being multiplied by the nth power of 10.
From 0.0000302, we need to move the decimal to after 3 so that 3.02 can be a number between 1-9. When moving the decimal point to the right side, each digit moved counts as -1 power of 10.
So, to give 3.02, we need to move the decimal by 5 digits. hence, we can conclude n = -5.
0.0000302 = 3.02x10^-5
Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on a level surface. It travels in the air for 13.6 seconds before it strikes the ground 760 m from the shooter. At what angle above the horizontal was the bullet fired? Round to the nearest whole number and include units in your answer Use g= -9.8 m/s2 for the acceleration of gravity.
Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
Δx=760 mV₀=87 m/st=13.6 sa=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!With that we can plug the values in to get:
[tex]760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}[/tex]
[tex]760=(1183.2)(cos\theta)[/tex]
[tex]cos\theta=\frac{760}{1183.2}[/tex]
[tex]\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}[/tex]
A 22.8 kg rocking chair begins to slide across the carpet when the push reaches 57.0 N. What is the coefficient of static friction?
Answer:
0.255
Explanation:
The following data were obtained from the question:
Force (F) = 57 N
Mass (m) = 22.8 Kg
Coefficient of static friction (µ) =...?
Next, we shall determine the normal reaction (R). This is illustrated below:
Mass (m) = 22.8 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal reaction (R) =?
R = mg
R = 22.8 x 9.8
R = 223.44 N
Finally, we can obtain the coefficient of static friction (µ) as follow:
Force (F) = 57 N
Normal reaction (R) = 223.44 N
Coefficient of static friction (µ) =...?
F = µR
57 = µ x 223.44
Divide both side by 223.44
µ = 57/223.44
µ = 0.255
Therefore, the coefficient of static friction (µ) is 0.255.
Answer:
.255
Explanation:
I'm also on acellus and it's the right answer
A gas is held at atmospheric pressure, approximately 100kPa, and room temperature, 298K,
in a container closed with a piston. The piston is initially in a fixed position so the container has a volume of 2.0dm3.
The gas is heated to a temperature of 100∘C
What is the pressure in the container?
Answer:
125 KPa
Explanation:
Data obtained from the question include:
Initial pressure (P1) = 100 KPa
Initial temperature (T1) = 298 K
Final temperature (T2) = 100°C = 100°C + 273 = 373 K
Final pressure (P2) =..?
Since the volume of the container is fixed, the final pressure in the container can be obtained as follow:
P1/T1 = P2/T2
100/298 = P2/373
Cross multiply
298 x P2 = 100 x 373
Divide both side by 298
P2 = (100 x 373) / 298
P2 = 125.2 ≈ 125 KPa
Therefore, the final pressure in the container is approximately 125 KPa.
A car speeds over a hill past point A, as shown in the figure. What is the maximum speed the car can have at point A such that its tires will not leave the track? Round to one decimal place and include units. Image:
Answer:
see explanations below
Explanation:
At the point when the car leaves the track, the reaction on the road is zero, meaning that the centrifugal force equals the gravitation force, namely
mv^2/r = mg
Solve for v in SI units
v^2 = gr = 9.81 m/s^2 * 14.2 m = 139.302 m^2/s^2
v = sqrt(139.302) = 11.8 m/s
Answer: at 11.8 m/s (26.4 mph) car will leave the track.
A body is acted upon by a force of 4N.
As a result, the velocity of the body has
changed from 0.3 m/s to 0.1 m/s passing
through a certain distance. If the mass
of the body is 1 kg, find the distance travelled.
[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Kinematics.
so here we get as,
V^2 = U^2 + 2as
so here, a = -0.2 m/s^2
(0.1)^2 = (0.3)^2 + (-0.2)(s)
=> 0.01 = 0.09 - 0.2s
=> 0.2s = 0.08
=> s = 0.08/0.2
=> s = 0.4 m
a disc starts from rest with an angular acceleration completes 10 revolutions in 2 seconds. the time taken by it to complete 10 more revolution from that instant is
Starting from rest, the disc completes [tex]\theta[/tex] revolutions after [tex]t[/tex] seconds according to
[tex]\theta=\dfrac\alpha2t^2[/tex]
with angular acceleration [tex]\alpha[/tex]. It completes 10 rev in 2 s, which means
[tex]10\,\mathrm{rev}=\dfrac\alpha2(2\,\mathrm s)^2\implies\alpha=5\dfrac{\rm rev}{\mathrm s^2}[/tex]
Find the time it takes to complete 20 rev with this acceleration:
[tex]20\,\mathrm{rev}=\dfrac12\left(5\dfrac{\rm rev}{\mathrm s^2}\right)t^2\implies t=\sqrt8\,\mathrm s\approx2.83\,\mathrm s[/tex]
so it takes approximately 0.83 s to complete 10 more rev.
which glass is used in astrology?
Answer:
The answer is either looking glass or stained glass.
Explanation:
The popular glass used in astrology is stained glass which is popular in symbolization of huge astrology figures. As for looking glass, it's more popular in astrological souvenirs or objects.
Answer:
i think answer is pisces horoscope
Why is it advisable not to look directly at the Sun? Please Give scientific reason for your answer
Answer:
your eyes would get seriously damaged
Explanation:
when you look directly at the sun, the sun shines it's rays directly to your eyes , which damage them. that's why you should always wear sunglasses in that type of weather, and never point directly at the sun. if you find this answer helpful, mark it as brainliest.
how does an electric iron work when the power is on
Answer:
The basic principle on which the electric iron works is that when a current is passed through a piece of wire, the wire heats up. This heat is distributed to the sole (base) plate of the electric iron through conduction.
Which describes refraction? *
10 points
O
Refracted rays travel through a boundary into a new medium.
O
The angle of incidence is the same for angle of refraction
Refracted rays change direction and go back to the original medium
Answer:
Refracted rays travel through a boundary into a new medium.
Explanation:
Refracted rays travel through a boundary into a new medium. is only true for refraction.
The angle of incidence is the same for angle of refraction, is not true for refraction. Refraction follows Snell's law, states that ratio of the sine of the angle of refraction and the sine of the angle of incidence is always constant and equivalent to the ratio of phase velocities of the two mediums it is passing through.
Refracted rays change direction and go back to the original medium is false for refraction however, it is true for reflection.