Answer:
the radius of the bigger loop is 5 cm.
Explanation:
Given;
current in the smaller loop, I₁ = 12 A
current in the larger loop, I₂ = 20 A
radius of the smaller loop, r₁ = 3 cm
let the radius of the larger loop, = r₂
Apply Biot-Savart's law to determine the magnetic field at the center of the circular loops.
[tex]B= \frac{\mu_0 I}{2r}[/tex]
The magnetic field at the center of the smaller loop;
[tex]B_1 = \frac{\mu_0 I_1}{2 r_1}[/tex]
The magnetic field at the center of the bigger loop;
[tex]B_2 = \frac{\mu_0 I_2}{2 r_2}[/tex]
If the magnetic field at the center is zero, then B₁ = B₂
[tex]B_1 = B_2 = \frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{2 r_2} \\\\\frac{I_1}{ r_1} = \frac{ I_2}{r_2} \\\\r_2 = \frac{I_2 r_1}{ I_1} = \frac{(20 \ A) \times (3.0 \ cm)}{12 \ A} = 5 \ cm[/tex]
Therefore, the radius of the bigger loop is 5 cm.
For a spring-mass oscillator if you double the mass but keep the stiffness the same, by what numerical factor does the pena original period was and the new period is DT, what is b7 It is useful to write out the expression for the period and ask yours you doubled the mass.
b = _____
If, instead, you double the spring stiffness but keep the mass the same, what is the factor b?
b = _____
If, instead, you double the mass and also double the spring stiffness, what is the factor b?
b = _____
If, instead, you double the amplitude (keeping the original mass and spring stiffness), what is the factor b?
b = _____
Answer:
ygguguguhhihhihijijijjojojinjbgy
Two long, straight wires are separated by 0.120 m. The wires carry currents of 11 A in opposite directions, as the drawing indicates. Find the magnitude of the net magnetic field.
Answer:
The magnitude of the magnetic field is 1.83 x [tex]10^{-5}[/tex] T.
Explanation:
The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.
In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.
The magnetic field (B) between the given wires can be determined by:
B = [tex]\frac{U_{o}I }{2\pi r}[/tex]
where: I is the current, r is the distance between the wires and [tex]U_{0}[/tex] is the magnetic field constant.
But, I = 11 A, r = 0.12 m and [tex]U_{0}[/tex] = 4[tex]\pi[/tex] x [tex]10^{-7}[/tex] Tm/A
So that;
B = [tex]\frac{4\pi *10^{-7}*11 }{2\pi *0.12}[/tex]
= 1.8333 x [tex]10^{-5}[/tex]
B = 1.83 x [tex]10^{-5}[/tex] T
How to calculate voltage U1 ?
Please help!
Answer:
he is a baby art and design
Definition of distance in physics
Answer:
Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
Explanation:
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88
is the correct answer
why do atom absorb photon since it makes it unstable??
[tex]\textsf{When an electron is hit by a }[/tex] [tex]\textsf{photon of light, it absorbs the quanta}[/tex] [tex]\textsf{of energy the photon was carrying}[/tex] [tex]\textsf{and moves to a higher energya}[/tex] [tex]\textsf{ state. Electrons therefore have to }[/tex] [tex]\textsf{jump around within the atom as }[/tex] [tex]\textsf{they either gain or lose energy. }[/tex]
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. Electrons therefore have to jump around within the atom as they either gain or lose energy.
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A 2.5 kg block slides along a frictionless surface at 1.5 m/s.A second block, sliding at a faster 4.1 m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/s. What was the mass of the second block?
Answer:
1.5kg
Explanation:
Given data
mass m1= 2.5kg
mass m2=??
velocity of mass one v1= 1.5m/s
velocity of mass two v2= 4.1m/s
common velocity after impact v= 2.5m/s
Let us apply the formula for the conservation of linear momentum for inelastic collision
The expression is given as
m1v1+ m2v2= v(m1+m2)
substitute
2.5*1.5+ m2*4.1= 2.5(2.5+m2)
3.75+4.1m2= 6.25+2.5m2
collect like terms
3.75-6.25= 2.5m2-4.1m2
-2.5= -1.6m2
divide both sides by -1.6
m2= -2.5/-1.6
m2= 1.5 kg
Hence the second mass is 1.5kg
find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4
The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:
Q(t) = Aeσ[tex]T^{4}[/tex]
where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.
To determine the rate of energy radiated by the man in the given question;
[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ
But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.
So that;
[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]
= 3.8556 x [tex]10^{-8}[/tex]
= 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex]
Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
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State TRUE or FALSE.
1. We use muscular force to lift a bucket of water.
2. A bow uses mechanical force of the bow string to shoot an arrow.
3. The force of friction enables us to walk on earth.
4. Plants use solar energy to make their food.
5. The energy stored inside the earth is called atomic energy
Answer:
1. True
2. False
3. True
4. True
5. True
Answer:
that is pure falsereeeeeeeee
Explanation:
An object moving with a constant
acceleration changes its velocity from
10ms' to 20 ms' in five seconds. What is the
distance travelled in five seconds
Answer:
Acceleration:
[tex]{ \tt{a = \frac{v - u}{t} }} \\ { \tt{a = \frac{20 - 10}{5} }} \\ { \tt{a = 2 \: m {s}^{ - 2} }}[/tex]
From third equation:
[tex]{ \bf{ {v}^{2} = {u}^{2} + 2as}} \\ { \tt{s = \frac{ {20}^{2} - {10}^{2} }{2 \times 2} }} \\ = { \tt{s = 75 \: m}}[/tex]
Answer:
Formula = m/s
Explanation:
The answer is 10 m / 5 seconds = 2 meters distance
The answer is 20 m / 5 seconds = 4 meters distance
If you dive underwater, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum has an area of about 70 mm217 * 10-5 m22, and it can sustain a force of about 7 N without rupturing. If your body had no means of balancing the extra pressure (which, in reality, it does), what would be the maximum depth you could dive without rupturing your eardrum
Answer:
[tex]h=10m[/tex]
Explanation:
From the question we are told that:
Area [tex]a=70 x 10^{-6}[/tex]
Force [tex]F=7N[/tex]
Generally the equation for Pressure is mathematically given by
Pressure = Force/Area
[tex]P=\frac{F}{A}[/tex]
[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]
[tex]P= 1*10^{5} Pa[/tex]
Generally the equation for Pressure is also mathematically given by
[tex]P=hpg[/tex]
Therefore
[tex]h=\frac{P}{hg}[/tex]
[tex]h=\frac{10000}{1000*9.8}[/tex]
[tex]h=10m[/tex]
Một bơm cánh gạt thủy lực có lưu lượng thực là 20lít/phút, tạo áp suất 230 bar, tốc độ bơm là 1400 vòng/phút. Biết công suất đầu vào là 10kW và hiệu suất cơ là 88%.
a) Tính hiệu suất thể tích của bơm
b) Tính thể tích riêng của bơm (cm/vòng). Câu 3 (2,5đ): Thiết kế hệ thống truyền động khí nén để điều khiển 02 xylanh tác động đơn, sử dụng 02 van đảo chiều 3/2 tác động bằng nút nhấn, 02 van tiết lưu - một chiều. Trình bày nguyên lý làm việc của của hệ thống.
Answer:
English please?
Explanation:
Explain in english?
19 point please please answer right need help
Block on an incline
A block of mass m1 = 3.9 kg on a smooth inclined plane of angle 38is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 2.6 kg hanging vertically. Take the positive direction up the incline and use 9.81
m/s2 for g.
What is the tension in the cord to the nearest whole number?
Explanation:
We can write Newton's 2nd law as applied to the sliding mass [tex]m_1[/tex] as
[tex]T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)[/tex]
For the hanging mass [tex]m_2,[/tex] we can write NSL as
[tex]T - m_2g = -m_2a\:\:\:\:\:\:\:(2)[/tex]
We need to solve for a first before we can solve the tension T. So combining Eqns(1) & (2), we get
[tex](m_1 + m_2)a = m_2g - m_1g\sin38[/tex]
or
[tex]a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g[/tex]
[tex]\:\:\:\:= 0.30\:\text{m/s}^2[/tex]
Using this value for the acceleration on Eqn(2), we find that the tension T is
[tex]T = m_2(g - a) = (2.6\:\text{kg})(9.51\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:=24.7\:\text{N}[/tex]
An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X is stationary. An observer on Planet X sees rocket 2 travel toward Earth at 0.4c. What is the speed of rocket 1 according to an observer on rocket 2?
Answer:
0.625 c
Explanation:
Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.
In the context,
The relative speed of body 2 with respect to body 1 can be expressed as :
[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]
Speed of rocket 1 with respect to rocket 2 :
[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]
[tex]$u' = \frac{0.7 c}{1.12}$[/tex]
[tex]u'=0.625 c[/tex]
Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c
: A fan is placed on a horizontal track and given a slight push toward an end stop 1.80 meters away. Immediately after the push, the fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2. What is the maximum possible velocity (magnitude) the cart can have after the push so that the cart turns around just before it hits the end-stop
Answer:
The initial velocity is 1.27 m/s.
Explanation:
distance, s = 1.8 m
acceleration, a = - 0.45 m/s^2
final velocity, v = 0
let the initial velocity is u.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times 0.45\times 1.8\\\\u = 1.27 m/s[/tex]
We have that the Initial velocity is mathematically given as
u=1.27m/s
Maximum possible velocity
Question Parameters:
a slight push toward an end stop 1.80 meters away
he fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2
Generally the equation for the third equation of motion is mathematically given as
Vf^2 = Vi^2 + 2ad
Therefore
0=u^2+0.45*1.8
u=1.27m/s
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Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point
Answer:
The right answer is "0.273 m".
Explanation:
Given:
Power (P),
[tex]\frac{1}{f} = 2D[/tex]
Near point,
u = 0.6 m
As we know,
⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]
By substituting the values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]
[tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]
[tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]
[tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]
By applying cross-multiplication, we get
[tex]0.6=2.2 \ v[/tex]
[tex]v = \frac{0.6}{2.2}[/tex]
[tex]S_{near} = 0.273 \ m[/tex]
If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.70 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Answer:
the magnitude of the magnetic field is 1.23 x 10⁸ T.
Explanation:
Given;
magnitude of the electric field, E = 3.7 V/m
The magnitude of the magnetic field is calculated as;
E = cB
where;
B is the magnitude of the magnetic field
c is the speed of light = 3 x 10⁸ m/s
From the above equation, the magnetic field, B, is calculated as;
[tex]B = \frac{E}{c} \\\\B = \frac{3.7 }{3\times 10^8 } \\\\B = 1.23 \times 10^{-8 } \ T[/tex]
Therefore, the magnitude of the magnetic field is 1.23 x 10⁸ T.
Solve numerical problem. Please give me step - step explanation Help me out plz
Answer:
You should multiply 60 kg*9.8 and answer will come.
Hope this will help you.
Answer:
yes she is right you should multiple 60*9.8
have a great day God bless you
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How far horizontally will it travel in 2 seconds?
A. 30 m
B. 90 m
C. 45 m
D. 60 m
Answer:
It will travel Vx * t = 30 m/s * 2 s = 60 m
A motorcyclist start from rest to reaches 6m/s with uniform acceleration for 3s what his acceleration?
Answer:
[tex]\boxed {\boxed {\sf 2 \ m/s^2}}[/tex]
Explanation:
Acceleration is the rate of change in velocity with respect to time. It is calculated by dividing the change in velocity by the change in time. The formula is:
[tex]a= \frac{ \Delta v}{\Delta t}[/tex] or [tex]a= \frac{v_f-v_i}{\Delta t}[/tex]
The change in velocity is the difference between the initial velocity and the final velocity. The motorcycle starts at rest, or 0 meters per second and reaches 6 meters per second. The change in time is 3 seconds.
[tex]\bullet \ v_f= 6 \ m/s\\\bullet \ v_i= 0 \ m/s \\\bullet \ \Delta t = 3 \ s[/tex]
Substitute the values into the formula
[tex]a= \frac { 6 \ m/s - 0 m/s}{3 \ s}[/tex]
Solve the numerator.
[tex]a= \frac{6 \ m/s}{3 \ s}[/tex]
[tex]a= 2 \ m/s^2[/tex]
The motorcyclist's acceleration is 2 meters per second squared.
Chase grew up wanting to wear his sister's dresses over his brother's pants and button up shirts. When Chase turns 18, he decides to begin living as woman, though he's still only sexually attracted to women. He decides he doesn't want to undergo surgery. Chase is
Explanation:
she is a woman, i dont understand why youre still using he/him after she comes out
a cell phone is released from the top with the speed of 10ms what is the speed 3s after?
Answer:
30ms
Explanation:
you need to multiple the 10ms by 3s which gives you 30ms
What is an internal resistance?
Explanation:
some thing inside a resistor
Several years ago, there was a solar eclipse visible only from a remote part of Mexico (Los Cabos). To view this event, many thousands of tourists went to that remote area. What effect would this event have on the price of pesos in tersm of U.S. dollars
The price of Pesos is increased, and it takes more to buy a given amount of Pesos
According to the demand and supply economic theory, made popular by Adam Smith in 1776, which provides a price determination model based on the relationship between the amount of a good producers of the good are willing to sell at a given price and the amount of that good that the end users wish to purchase, there is an inverse relationship between price and supply, therefore as the supply decreases, the price increases, and therefore, there is a direct relationship between the demand and price, such that as the demand increases, the price of the goods increases because the ratio of supply to demand is decreased
Given that the arrival of thousands of tourists to the remote Los Cabos, will result in the tourists requiring Pesos, the demand of Pesos will rise, which according to the demand and supply theory, will increase the price of Pesos
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Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be
Answer:
a) [tex]F_g=1.5*10^9Ibf[/tex]
b) [tex]F_t=12490Ibf/ft^2[/tex]
[tex]F_b=0[/tex]
Explanation:
From the question we are told that:
Height [tex]h=200ft[/tex]
Width [tex]w=1200ft[/tex]
a)
Generally the equation for Dam's Hydro static force is mathematically given by
[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]
Where
[tex]\rho=Density\ of\ water[/tex]
[tex]\rho=62.4Ibm/ft^3[/tex]
Therefore
[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]
[tex]F_g=1.5*10^9Ibf[/tex]
b)
Generally the equation for Dam's Force per unit area is mathematically given by
[tex]F=\rho*g*h[/tex]
For Top
[tex]F_t=\rho*g*h[/tex]
[tex]F_t=62.4*32.2*200[/tex]
[tex]F_t=12490Ibf/ft^2[/tex]
For bottom
[tex]Here \\H=0 zero[/tex]
Therefore
[tex]F_b=0[/tex]
The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].
The force per unit area near the top is 86.74 psi.
The force per unit area near the bottom is zero.
Hydrostatic force
The hydrostatic force on the dam is the force exerted on the dam by the column of the water.
[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]
Force per unit area near the topThe force per unit area is the pressure exerted near the top of the dam.
[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]
where;
P is pressure in PSI
ρ is density of water in lb/gal
h is the vertical height in ft
[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]
The pressure near the bottom is zero, become the vertical height is zero.
Learn more about hydrostatic pressure here: https://brainly.com/question/11681616
If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.10 V/mV/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Answer:
B = 1.03 10⁻⁸ T
Explanation:
For an electromagnetic wave, the electric and magnetic fields must oscillate in phase so that they remain between them at all times, otherwise the wave will extinguish
This relational is expressed by the relation
E /B = c
B = E / c
let's calculate
B = 3.10 / 3 10⁸
B = 1.03 10⁻⁸ T
How much amount of water can be decomposed
through electrolysis by passing 2 F charge?
Answer:
So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.
A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand.
(a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used?
(b) What other angle gives the same range, and why would it not be used?
(c) How long did this pass take?
Answer:
a) θ = 14.23º, b) θ₂ = 75.77, c) t = 0.6019 s
Explanation:
This is a missile throwing exercise.
a) the reach of the ball is the distance traveled for the same departure height
R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]
sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]
sin 2θ = 7.00 9.8 / 12.0²
2θ = sin⁻¹ (0.476389) = 28.45º
θ = 14.23º
the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º
θ ’= 90 -14.23
θ’= 75.77º
b) the two angles that give the same range are
θ₁ = 14.23
θ₂ = 75.77
the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.
C) the time of the pass can be calculated with the expression
x = v₀ₓ t
t = x / v₀ₓ
t = 7 / 11.63
t = 0.6019 s
In first case a mass M is split into two parts with one part being 1/6.334 th of the original mass. In second case M is split into two equal parts. In both the cases the two parts are separated by same distance. What ratio of the magnitude of the gravitational force in first case to the magnitude of the gravitational force in the second case
Answer:
[tex]F_r=0.132:0.25[/tex]
Explanation:
From the question we are told that:
[tex]M_1=M*\frac{1}{6.334}[/tex]
Therefore
[tex]M_2=M-M*\frac_{1}{6.334}[/tex]
[tex]M_2=M*\frac{5.334}{6.334}[/tex]
Generally the equation for Gravitational force of attraction is mathematically given by
For Unequal split
[tex]F=\frac{GM_1M_2}{d^2}[/tex]
[tex]F=\frac{G(M*\frac_{1}{6.334})(M*\frac{5.334}{6.334})}{d^2}[/tex]
[tex]F=\frac{GM^2}{d^2}*(0.132)[/tex]
For equal split
[tex]F=\frac{GM_1M_2}{d^2}[/tex]
[tex]F=\frac{G(\frac{M}{2})((\frac{M}{2}}{d^2}[/tex]
[tex]F=0.25 \frac{GM^2}{d^2}[/tex]
Therefore the ratio of the gravitational force is
[tex]F_r=0.132:0.25[/tex]
A student wants to start a small business in school. Write down six items that
he/she can sell in school at a profit.
Answer:
packets of pen
packets of pencil
copies
books
bottles
mask
Six items that a student can sell in school at a profit:
- Homemade baked goods
- School supplies
-Drinks
- Healthy snacks
- Personalized accessories
- Stickers
What is a profit?Profit is the difference between the revenue earned by a business or individual and the costs incurred to produce the goods or services sold.
It is an important measure of financial success for companies and is often used to determine the value of a business.
We have,
Here are six items that a student can sell in school at a profit:
Homemade baked goods - cupcakes, cookies, brownies, and other treats can be sold individually or as a pack.
School supplies - items such as pens, pencils, erasers, rulers, notebooks, and binders are always in demand.
Drinks - bottled water, juices, and sodas are popular beverages that students may purchase during the school day.
Healthy snacks - fresh fruit, granola bars, and trail mix are nutritious snacks that many students are interested in buying.
Personalized accessories - items like keychains, bracelets, and bookmarks with unique designs or student names can be popular among peers.
Stickers - fun and colorful stickers can be sold individually or in packs and are often a favorite of younger students.
Thus,
Six items that a student can sell in school at a profit:
- Homemade baked goods
- School supplies
-Drinks
- Healthy snacks
- Personalized accessories
- Stickers
Learn more about profit here:
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