Answer:
The y component of her velocity is zero
Explanation:
Given that North and South represent the positive and negative y component of the velocity vector.
And East and West represent the positive and negative x component of the velocity vector.
For a girl walking east at 2.7 m/s
The x component of velocity is;
vx = 2.7 m/s (east is positive x axis)
The y component of velocity is;
vy = 0 m/s (no movement in the northern and southern direction)
Therefore, The y component of her velocity is zero (0m/s)
A uniform solid disk and a uniform ring are place side by side at the top of a rough incline of height h.
a) If they are released from rest and roll without slipping, determine the velocity vring of the ring when it reaches the bottom.
b) Verify your answer by calculating their speeds when they reach the bottom in terms of h.
Explanation:
velocity of disc [tex]=\sqrt((gh)/0.75)[/tex]
lets call (h) 1 m to make it simple.
= 3.614 m/s
[tex]\sqrt((4/3) x 1 x 9.8) = 3.614[/tex] m/s pointing towards this:
[tex]4×V_d=\sqrt(4/3hg)[/tex]
[tex]V_h=\sqrt(hg)[/tex]
velocity of hoop=[tex]\sqrt(gh)[/tex]
lets call (h) 1m to make it simple again.
[tex]\sqrt(9.8 x 1) = 3.13[/tex] m/s
[tex]\sqrt(gh) = sqrt(hg)
so [tex]4×V_d= \sqrt(4/3hg)V_h=\sqrt(hg)[/tex]
The disc is the fastest.
While i'm on this subject i'll show you this:
Solid ball [tex]=0.7v^2= gh[/tex]
solid disc [tex]= 0.75v^2 = gh[/tex]
hoop [tex]=v^2=gh[/tex]
The above is simplified from linear KE + rotational KE, the radius or mass makes no difference to the above formula.
The solid ball will be the faster of the 3, like above i'll show you.
solid ball: velocity [tex]=\sqrt((gh)/0.7)[/tex]
let (h) be 1m again to compare.
[tex]\sqrt((9.8 x 1)/0.7) = 3.741[/tex] m/s
solid disk speed [tex]=\sqrt((gh)/0.75)[/tex]
uniform hoop speed [tex]=\sqrt(gh)[/tex]
solid sphere speed [tex]=\sqrt((gh)/0.7)[/tex]
What must the charge (sign and magnitude) of a 1.60 g particle be for it to remain balanced against gravity when placed in a downward-directed electric field of magnitude 680 N/C
Answer:
Explanation:
The charge must be negative so that force in a downward electric field will be upward so that its weight is balanced .
Let the charge be - q .
force on charge
= q x E where E is electric field
= q x 680
weight = 1.6 x 10⁻³ x 9.8
so
q x 680 = 1.6 x 10⁻³ x 9.8
q = 1.6 x 10⁻³ x 9.8 / 680
= 23 x 10⁻⁶ C
- 23 μ C .
Which of the following technologies is based on the work of Ibn al-Haytham?
A. Telescopes to observe the visible light of distant stars
B. Radiation treatments for breast cancer
C. Radar to detect the movement of storms
O D. An orbiting observatory to detect X-rays from space objects
Answer:
The answer is A
Explanation:
Its A because he created a telescope to be able to observe stars.
Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish really is? Does the fish see the person's face closer or farther than it really is? Explain your answer.
Answer:
In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer
In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away
Explanation:
This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.
In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer
1 sin θ₁ = 1.33 sin θ₂
θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)
In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away
Answer:
The fish appears closer than it really is because light from the fish is refracted away from the normal as it enters the air pocket in the goggles. This is because air has a smaller index of refraction than water. The person will trace rays back to an image point in front of the actual fish. The fish will see the person's face exactly where it actually is because the light from the face is not refracted as it travels through water only, and does not change from one medium to another.
Explanation:
An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put into a box having sides of length 0.5L without changing its temperature, the pressure it exerts on the walls of the larger box will be...
p.
2p.
4p.
8p.
12p.
Answer:
2P
Explanation:
See attached file
4. Chloe has a vertical velocity of 3 m/s when she leaves the 1 m diving board. At this instant, her center of gravity is 2.5 m above the water. How high above the water will Chloe go
Answer:
2.95m
Explanation:
Using h= 2.5+ v²/2g
Where v= 3m/s
g= 9.8m/s²
h= 2.95m
A 2kg block is sitting on a hinged ramp such that you can increase the angle of the incline. The coefficient of static friction between the block and the ramp is 0.67 and the coefficient of kinetic friction is 0.25.
a. What angle do you have to tilt the ramp to get the block to slide?
b. What acceleration does the block experience at this angle when kinetic friction takes over?
Answer:
θ = 33.8
a = 3.42 m/s²
Explanation:
given data
mass m = 2 kg
coefficient of static friction μs = 0.67
coefficient of kinetic friction μk = 0.25
solution
when block start slide
N = mg cosθ .............1
fs = mg sinθ ...............2
now we divide equation 2 by equation 1 we get
[tex]\frsc{fs}{N} = \frac{sin \theta }{cos \theta }[/tex]
[tex]\frac{\mu s N }{N}[/tex] = tanθ
put here value we get
tan θ = 0.67
θ = 33.8
and
when block will slide then we apply newton 2nd law
mg sinθ - fk = ma ...............3
here fk = μk N = μk mg cosθ
so from equation 3 we get
mg sinθ - μk mg cosθ = ma
so a will be
a = (sinθ - μk cosθ)g
put here value and we get
a = (sin33.8 - 0.25 cos33.8) 9.8
a = 3.42 m/s²
in a certain region of space, the gravitational field is given by -k/r,where r=distance,k=const.if gravitational potential at r=r0 be v0,then what is the expression for the gravitational potential v?
options
1)k log(r/ro)
2)k log(ro/r)
3)vo+k log(r/ro)
4)vo+k log(ro/r)
plz help me out
I will mark u as brainliest if u answer correct
Answer:
The correct answer is option 3 .
Please check the answer once :)
f the mass of the block is 2 kg, the radius of the circle is 0.8 m, and the speed of the block is 3 m/s, what is the tension in the string at the top of the circle
Answer:
the size are components relative to the whole.
Explanation:
they are particularly good at showing percentage or proportional data
Point charges q1=50μCq1=50μC and q2=−25μCq2=−25μC are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge q3=20μCq3=20μC situated there?
Answer:
a) E = 2.7x10⁶ N/C
b) F = 54 N
Explanation:
a) The electric field can be calculated as follows:
[tex] E = \frac{Kq}{d^{2}} [/tex]
Where:
K: is the Coulomb's constant = 9x10⁹ N*m²/C²
q: is the charge
d: is the distance
Now, we need to find the electric field due to charge 1:
[tex] E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C [/tex]
The electric field due to charge 2 is:
[tex]E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C[/tex]
The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):
[tex]E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side[/tex]
Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.
b) The force on a charge q₃ situated there is given by:
[tex]E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}[/tex]
[tex] F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N [/tex]
Therefore, the force on a charge q₃ situated there is 54 N.
I hope it helps you!
(a) The electric field at a point midway between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained to be [tex]2.7\times 10^6 \,N/C[/tex].
(b) The electrostatic force on the third charge [tex]q_3[/tex] situated between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained as 54 N.
The answer can be explained as follows.
Electric FieldGiven that the two charges are;
[tex]q_1 = 50\times 10^{-6}\,C[/tex] and [tex]q_2 = -25\times 10^{-6}\,C[/tex](a) At the midpoint; [tex]r = 0.5\,m[/tex].
We know that the electric field due to charge [tex]q_1[/tex].
[tex]E_1 = k\,\frac{q_1}{r^2}[/tex]Where, [tex]k=9\times 10^9\,Nm^2/C[/tex]
[tex]E_1 = (9\times 10^9) \times\frac{(50 \times 10^{-6})}{(0.5)^2}=1.8\times 10^6N/C[/tex]The electric field due to charge [tex]q_2[/tex] is given by;
[tex]E_2 = (9\times 10^9) \times\frac{(-25 \times 10^{-6})}{(0.5)^2}=-9\times 10^5\,N/C[/tex]Therefore, the net electric field in the midpoint is given by;
[tex]E_{net} =E_2+E_1[/tex][tex]\implies E_{net}=1.8 \times 10^6 N/C + 9 \times 10^5\,N/C=2.7\times 10^6\,N/C[/tex]The direction is towards the right side.
Electrostatic Force(b) Now, there is another charge [tex]q_3=20\times 10^{-6}[/tex] in the midpoint.
So the force on the charge is ;
[tex]F=E_{net} \times q_3=(2.7 \times 10^6\,N/C) \times (20\times 10^{-6}\,C)=54\,N[/tex]Find out more about electrostatic force and fields here:
https://brainly.com/question/14621988
Value of g in CGS system
Answer:
in CGS system G is denoted as gram
Calculate the current through a 15.0-m long 20 gauge (having radius 0.405 mm) nichrome wire if it is connected to a 12.0-V battery. The resistivity of nichrome is 100 × 10-8 Ω ∙ m.
Given Information:
Radius of wire = r = 0.405 mm = 0.405×10⁻³ m
Length of wire = L = 15 m
Voltage = V = 12 V
Resistivity = ρ = 100×10⁻⁸ Ωm
Required Information:
Current = I = ?
Answer:
Current = I = 0.412 A
Explanation:
The current flowing through the wire can be found using Ohm's law that is
V = IR
I = V/R
Where V is the voltage across the wire and R is the resistance of the wire.
The resistance of the wire is given by
R = ρL/A
Where ρ is the resistivity of the wire, L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(0.405×10⁻³)²
A = 0.515×10⁻⁶ m²
So the resistivity of the wire is
R = ρL/A
R = (100×10⁻⁸×15)/0.515×10⁻⁶
R = 29.126 Ω
Finally, the current flowing through the wire is
I = V/R
I = 12/29.126
I = 0.412 A
Therefore, the current through a 15.0-m long 20 gauge nichrome wire is 0.412 A.
What fundamental frequency would you expect from blowing across the top of an empty soda bottle that is 24 cm deep, if you assumed it was a closed tube
Answer:
f = 357.29Hz
Explanation:
In order to calculate the fundamental frequency in the closed tube, you use the following formula:
[tex]f_n=\frac{nv}{4L}[/tex] (1)
n: order of the mode = 1
v: speed of sound = 343m/s
L: length of the tube = 24cm = 0.24m
You replace the values of the parameters in the equation (1):
[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]
The fundamental frequency of in the tube is 357.29Hz
Classify the bending of light as exhibited by the ray diagrams. According to your data, is light refracted away from or toward the normal as it passes at an angle into a medium with a higher index of refraction?
Answer:
the ray of light should approach normal
Explanation:
When light passes through two means of different refractive index, it fulfills the equation
n₁ sin θ₁ = n₂ sin θ₂
where index 1 and 2 refer to each medium
In this problem, they tell us that light passes to a medium with a higher index, which is why
n₁ <n₂
let's look for the angle in the second half
sinθ₂ = n₁ /n₂ sin θ₁
θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)
let's examine the angle argument the quantity n₁ /n₂ <1 therefore the argument decreases, therefore the sine and the angle decreases
Consequently the ray of light should approach normal
A proton moves at a speed 1.4 × 10^7 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.85 m. What is the field strength?
0.17T
Explanation:
When a charged particle moves into a magnetic field perpendicularly, it experiences a magnetic force [tex]F_{M}[/tex] which is perpendicular to the magnetic field and direction of the velocity. This motion is circular and hence there is a balance between the centripetal force [tex]F_{C}[/tex] and the magnetic force. i.e
[tex]F_{C}[/tex] = [tex]F_{M}[/tex] --------------(i)
But;
[tex]F_{C}[/tex] = [tex]\frac{mv^2}{r}[/tex] [m = mass of the particle, r = radius of the path, v = velocity of the charge]
[tex]F_{M}[/tex] = qvB [q = charge on the particle, B = magnetic field strength, v = velocity of the charge ]
Substitute these into equation (i) as follows;
[tex]\frac{mv^2}{r}[/tex] = qvB
Make B subject of the formula;
B = [tex]\frac{mV}{qr}[/tex] ---------------(ii)
Known constants
m = 1.67 x 10⁻²⁷kg
q = 1.6 x 10⁻¹⁹C
From the question;
v = 1.4 x 10⁷m/s
r = 0.85m
Substitute these values into equation(ii) as follows;
B = [tex]\frac{1.67 * 10 ^{-27} * 1.4 * 10^{7}}{1.6 * 10^{-19} * 0.85}[/tex]
B = 0.17T
Therefore, the magnetic field strength is 0.17T
7. Which statement is true about teens that are in Marcia’s final state of identity formation?
Answer:
D. All of the above
Explanation:
The last stage in the Marcia's identity formation theory is Identity achievement. In this last stage, teens have made a thorough search or exploration about their identity and have made a commitment to that identity. This identity represents their values, beliefs, and desired goals. At this point, they know want they want in life, and can now make informed decisions based on their belief and ideology.
James Marcia is a psychologist known mainly for his research and theories in human identity. Identity according to him is the sum total of a person's beliefs, values, and ideologies that shape what a person actually becomes and is known for. Occupation and Ideologies primarily determine identity. The four stages of Identity status include, Identity diffusion, foreclosure, moratorium, and achievement.
"On earth, you have a pendulum of length L that oscillates with period T. Your friend lives on a planet where the acceleration of gravity is four times as big as it is on the earth. What should be the length of a pendulum on your friend s planet so that it also oscillates with the same period T
Answer:
4L
Explanation:
Data provided in the question according to the question is as follows
Length = L
Gravity = G
For friend
Length = ?
Growth = 4G
Moreover,
[tex]T_1 = T_2[/tex]
Based on the above information ,
Now we have to apply the simple pendulum formula which is shown below:
[tex]T = 2\pi \frac{L}{G}[/tex]
Now equates these equations in both sides
[tex]2\pi \frac{L}{G} = 2\pi \frac{L}{4G}[/tex]
So after solving this, the length of the pendulum is 4L
Answer:
the length of a pendulum on your friend s planet should be 4 times than that on earth
Explanation:
We know that time period of simple pendulum is given by
[tex]T= 2\pi\sqrt{\frac{L}{g} }[/tex]
L= length of pendulum
g= acceleration due to gravity
therefore, Let T_1 and T_2 be the time period of the earth and other planet respectively.
[tex]\frac{T_1}{T_2} =\sqrt(\frac{L_1}{L_2}\times\frac{g_2}{g_1})[/tex]
ATQ
T_1=T_2=T, g_2=4g_1
Putting the values in above equation and solving we get
[tex]\frac{L_1}{L_2} =\frac{1}{4}[/tex]
A dielectric material such as paper is placed between the plates of a capacitor holding a fixed charge. What happens to the electric field between the plates
Answer:
Majorly the electric field is reduced among other effect listed in the explanation
Explanation:
In capacitors the presence of di-electric materials
1. decreases the electric fields
2. increases the capacitance of the capacitors.
3. decreases the voltage hence limiting the flow of electric current.
The di-electric material serves as an insulator between the metal plates of the capacitors
Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?
Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?
a-velocity
b-mass
c-momentum
d-direction
Answer:
b. Mass
Explanation:
This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.
As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.
Therefore, the correct answer is B. Mass.
This question involves the concept of the law of conservation of momentum.
Jerome should always keep the "mass" of each object constant after the objects collide and bounce apart.
The law of conservation of momentum states that the momentum of a system of objects must remain constant before and after the collision has taken place.
Mathematically,
[tex]m_1u_1+m_2u_2=m_1v_1+m_v_2[/tex]
where,
m₁ = mass of the first object
m₂ = mass of the second object
u₁ = velocity of the first object before the collision
u₂ = velocity of the second object before the collision
v₁ = velocity of the first object after the collision
v₂ = velocity of the second object after the collision
Hence, it is clear from the formula that the only thing unchanged before and after the collision is the mass of each object.
Learn more about the law of conservation of momentum here:
brainly.com/question/1113396?referrer=searchResults
The attached picture illustrates the law of conservation of momentum.
An ac series circuit contains a resistor of 20 ohms, a capacitor of 0.75 microfarads of 120 x 10-3 H. If an effective (rms) voltage of 120 V is applied, what is the effective (rms) current when the circuit is in resonance
Answer:
The effective (rms) current when the circuit is in resonance is 6 A
Explanation:
Given;
resistance of the resistor, R = 20 ohms
capacitance of the capacitor, C = 0.75 microfarads
inductance of the inductor, L = 0.12 H
effective rms voltage, [tex]V_{rms}[/tex] = 120
At resonance, the impedance Z = R, Since the capacitive reactance (Xc) is equal to inductive reactance (XL).
The effective (rms) current, = [tex]V_{rms}[/tex] / R
= 120 / 20
= 6 A
Therefore, the effective (rms) current when the circuit is in resonance is 6 A
A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:
Answer:
The ratio is [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
Explanation:
Generally the Moment of inertia of a spherical object (shell) is mathematically represented as
[tex]I = \frac{2}{3} * m r^2[/tex]
Where m is the mass of the spherical object
and r is the radius
Now the the rotational kinetic energy can be mathematically represented as
[tex]RE = \frac{1}{2}* I * w^2[/tex]
Where [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{v}{r}[/tex]
=> [tex]w^2 = [\frac{v}{r}] ^2[/tex]
So
[tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]
[tex]RE = \frac{1}{3} * mv^2[/tex]
Generally the transnational kinetic energy of this motion is mathematically represented as
[tex]TE = \frac{1}{2} mv^2[/tex]
So
[tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]
[tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
A total electric charge of 2.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm . The potential is zero at a point at infinity.
a) Find the value of the potential at 45.0 cm from the center of the sphere.
b) Find the value of the potential at 26.0 cm from the center of the sphere.
c) Find the value of the potential at 16.0 cm from the center of the sphere.
Answer:
a) 40 V
b) 69.23 V
c) 69.23 V
Explanation:
See attachment for solution
key points that can be found in the realist philosophical position
Answer:
Key points that can be found in the realist philosophical position are as follows:
The view that we observe or identify is real, truly out there.The objects which are identified are independent of someone's perceptions, linguistic practices, conceptual scheme, and beliefs.Quantum mechanics is an example of philosophical realism that claims world is mind-independent.Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t = 0 and subsequently experiences a constant angular acceleration α = 1.3 rad/s2. Determine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time t = 1.7 s.
Given that,
Angular velocity = 0.17 rad/s
Angular acceleration = 1.3 rad/s²
Time = 1.7 s
We need to calculate the angular velocity
Using angular equation of motion
[tex]\omega=\omega_{0}+\alpha t[/tex]
Put the value in the equation
[tex]\omega=0.17+1.3\times1.7[/tex]
[tex]\omega=2.38(k)\ m/s[/tex]
We need to calculate the angular displacement
Using angular equation of motion
[tex]\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}[/tex]
Put the value in the equation
[tex]\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}[/tex]
[tex]\theta=2.1675\times\dfrac{180}{\pi}[/tex]
[tex]\theta= 124.18^{\circ}[/tex]
We need to calculate the velocity at point A
Using equation of motion
[tex]v_{A}=v_{0}+\omega\times r[/tex]
Put the value into the formula
[tex]v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))[/tex]
[tex]v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i[/tex]
[tex]v_{A}=(-0.267j-0.393i)\ m/s[/tex]
We need to calculate the acceleration at point A
Using equation of motion
[tex]a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)[/tex]
Put the value in the equation
[tex]a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]
[tex]a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]
[tex]a_{A}=-0.146j-0.215i−0.636i+0.937j[/tex]
[tex]a_{A}=0.791j-0.851i[/tex]
[tex]a_{A}=-0.851i+0.791j\ m/s^2[/tex]
Hence, (a). The velocity at point A is [tex](-0.267j-0.393i)\ m/s[/tex]
(b). The acceleration at point A is [tex](-0.851i+0.791j)\ m/s^2[/tex]
I WILL MARK YOU AS BRAINLIEST!!! An object is launched straight up into the air with an initial velocity of 40 meters per second, from a height 30 m above the ground. Assuming that gravity pulls it down, changing its position by about 4.9 /2, after how many seconds will the object hit the ground? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answer:
8.9 seconds
Explanation:
The height of the object at time t is:
y = h + vt − 4.9t²
where h is the initial height, and v is the initial velocity.
Given h = 30 and v = 40:
y = 30 + 40t − 4.9t²
When y = 0:
0 = 30 + 40t − 4.9t²
4.9t² − 40t − 30 = 0
Solving with quadratic formula:
t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)
t = [ 40 ± √(1600 + 588) ] / 9.8
t = 8.9
It takes 8.9 seconds for the object to land.
6a. A special lamp can produce UV radiation. Which two statements
describe the electromagnetic waves emitted by a UV lamp? *
They have a higher frequency than X-rays.
They have the same wave speed as visible light
They have a longer wavelength than microwaves.
They have a lower frequency than gamma rays.
They have a greater wave speed than radio waves.
Answer:
The correct options are:
B) They have the same wave speed as visible light
D) They have a lower frequency than gamma rays.
Explanation:
B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s
D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).
We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.
In which direction does a bag at rest move when a force of 20 newtons is applied from the right?
ОА.
in the direction of the applied force
OB.
in the direction opposite of the direction of the applied force
OC. perpendicular to the direction of the applied force
OD
in a circular motion
Answer:
in the direction of the applied force
Explanation:
A student stretches an elastic band by 0.8 m in 0.5 seconds. The spring constant of the elastic band is 40 N/m. What was the power exerted by the student
Answer:
The power exerted by the student is 51.2 W
Explanation:
Given;
extension of the elastic band, x = 0.8 m
time taken to stretch this distance, t = 0.5 seconds
the spring constant, k = 40 N/m
Apply Hook's law;
F = kx
where;
F is the force applied to the elastic band
k is the spring constant
x is the extension of the elastic band
F = 40 x 0.8
F = 32 N
The power exerted by the student is calculated as;
P = Fv
where;
F is the applied force
v is velocity = d/t
P = F x (d/t)
P = 32 x (0.8 /0.5)
P = 32 x 1.6
P = 51.2 W
Therefore, the power exerted by the student is 51.2 W
Run 2 17. Set # of slits to 2 18. Set Wave Length to 400nm 19. Set Slit width to 1600 nm 20. Set Slit spacing to 5000nm In row 18 21. Record distance to 1st bright fringe 22. Record distance to 2nd bright fringe 23. Record distance to 3rd bright fringe Knowing the screen distance to be 1m 24. Calculated the measured angle to 1st bright fringe 25. Calculated the measured angle to 2nd bright fringe 26. Calculated the measured angle to 3rd bright fringe Using sin(θ)=mλ/d 27. Calculate θ for 1st bright fringe
Answer:
a) m=1, y₁ = 0.08 m , θ₁ = 4.57º , b) m=2, y₂ = 0.16 m , θ₂ = 9.09º , c) m=3, y₃ = 0.24 m , θ₃ = 13.5º
Explanation:
After reading your strange statement, I understand that this is an interference problem, I transcribe the data to have it more clearly. Number of slits 2, distance between slits 5000 nm, wavelength 400 nm, distance to the screen 1 m.
They ask us to calculate the angles for the first, second and third interference, they also ask us to write down the distance from the central maximum.
The expression for constructive interference for two slits is
d sin θ = m λ
where d is the distance between the slits, λ is the wavelength used, m is an integer representing the order of interference
Let's use trigonometry to find the distance from the central maximum
tan θ = y / L
in all interference experiments the angle is small,
tan θ = sin θ / cos θ = sin θ
sint θ = y / L
let's replace
d y / L = m λ
y = m λ L / d
let's calculate
distance to the first maximum m = 1
y₁ = 1 400 10⁻⁹ 1/5000 10⁻⁹
y₁ = 0.08 m
distance to second maximum m = 2
y₂ = 2 400 10⁻⁹ 1/5000 10⁻⁹
y₂ = 0.16 m
distance to the third maximum m = 3
y₃ = 3 400 10⁻⁹ 1/5000 10⁻⁹
y₃ = 0.24 m
with these values we can search for each angle
tan θ = y / L
θ = tan⁻¹ y / L
for m = 1
θ₁ = tan⁻¹ (0.08 / 1)
θ₁ = 4.57º
for m = 2
θ₂ = tan⁻¹ (0.16 / 1)
θ₂ = 9.09º
for m = 3
θ₃ = tan⁻¹ (0.24 / 1)
θ₃ = 13.5º
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a proton and an electron are situated 971 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed.
You find that the forces are attractive and the magnitude of each force is:______
Answer:
The magnitude of each force is 2.45 x 10⁻¹⁶ N
Explanation:
The charge of proton, +q = 1.603 x 10⁻¹⁹ C
The charge of electron, -q = 1.603 x 10⁻¹⁹ C
Distance between the two charges, r = 971 nm = 971 x 10⁻⁹ m
Apply Coulomb's law;
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where;
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
q₁ and q₂ are the charges of proton and electron respectively
F is the magnitude of force between them
Substitute in the given values and solve for F
[tex]F = \frac{(8.99*10^9)(1.603*10^{-19})^2}{(971*10^{-9})^2} \\\\F = 2.45*10^{-16} \ N[/tex]
Therefore, the magnitude of each force is 2.45 x 10⁻¹⁶ N