Suppose the resistance of a copper wire is 1.05 N at 30°C.
Find the temperature at which the resistance becomes 1.50
Ω.

The appliance is run with a 12 V car battery and the value of the electric current will be 0.3A.

The pace at which electrons travel through a conductor is known as electric current. The ampere is the SI unit for electric current.

From ohm's law;

Case 1;

V=IR

120 V = 3 A × R

R = 40 ohm

For the same appliances, the value of the resistance is the same;

Case 2;

V=IR

12 V =  I × 40 ohm

I = 0.3 A

Hence the value of the current for case 2 will be 0.3 A.

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Answer:

the 1 gram wasp

Explanation:

To start off with this problem, write down every piece of information and do neccessary conversions.

mass of bee = 2 grams = 0.002 kg

speed of bee = 1 m/s

mass of wasp = 1 gram = 0.001 kg

speed of wasp = 2 m/s

now, we will use the kinetic energy formula and compare the answers

KE BEE = 0.5 (0.002 kg)(1 m/s)^2 = 0.001 Joules

KE WASP = 0.5(0.001 kg)(2 m/s)^2 = 0.002 Joules

0.002 J > 0.001 J

The speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.

According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state how momentum gets conserved.

Unit conversion;

1 km/sec = 1000 m/sec

Given data;

Spaceprobe speed  = 1.795 km/s = 1795 m /sec

Probe mass = 635.0 kg

Fuel mass = 4092.0 kg

Expelled propellent velocity = 4.161 km/s = 41461 m/sec

From the momentum conservation principle;

[tex]\rm P_i = P_f \\\\ (m_p+m_f)v_i = m_pV - m_fv_p \\\\ V = \frac{(635+4092)1795+4092 \times 41461}{635} \\\\ V = 280540.7 \ m/sec \\\\ V = 28.05 m/sec[/tex]

Hence, the speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.

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Answer:

Discrimination is the worst thing in the world you can't even do a thing so you will do such a physical thing or do a mathematics problem ok done it's ok

At 11.3°C the rod will just come into contact

Coefficient of linear expansion of aluminium [tex]\alpha_{Al}[/tex] = [tex]23*10^-^6[/tex] °[tex]C[/tex]

Coefficient of linear expansion of Brass [tex]\alpha_{B} =19*10^-^6[/tex] °[tex]C[/tex]

[tex]\alpha = \Delta{L}/Lo(T2-T1)[/tex]

For aluminium

[tex]\alpha_{Al} = \Delta{L}/Lo(T-0)[/tex]

[tex]\Delta{L_{1}} = (23*10^-^6*50*T)[/tex]

For Brass

[tex]\alphaB = \Delta{L_{2}/Lo(T-0)[/tex]

[tex]\Delta{L_{1} +\Delta{L_{2} =0.024cm(23*10^-^6*50*T)+(19*10-6*50*T) =0.024[/tex]

T =11.43 °C

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That depends on where YOU are when you measure it.

If you're motionless in the laboratory, then you measure the particles flying apart at 1.6c .

If you're riding on one of the particles, you measure the other one flying away from you at less than c

Answer:

The force between two objects is calculated through the equation,

                       F = Gm₁m₂/d²

where m₁ and m₂ are the masses of the objects. In this case, an unknown mass and Earth. d is the distance between them and G is the universal gravitation constant.

In the second case, if the force is to become 2.5 times the original and all the variables are constant except d then,

                      2.5F = Gm₁m₂ / (D²)

                               D = 0.623d

Subsituting the known value of d,

                               D = 0.623(6.9 x 10^8) = 4.298 x 10^8 m

The magnitude of the electric field at x =4m on the x axis at this time 1 N/C.

Electric field at a given distance is calculated as follows;

E = kq/r²

E₂ = (9 x 10⁹ x q)/(2²)

E₂ = 2.25 x 10⁹q

E₂ + E₀ = 0

2.25 x 10⁹q + 4 = 0

2.25 x 10⁹q = - 4

q = -4 / (2.25 x 10⁹)

q = -1.78 x 10⁻⁹

E₄ = (9 x 10⁹ x (-1.78 x 10⁻⁹) ) / (4²)

E₄ = - 1 N/C

|E₄| = 1 N/C

Thus, the magnitude of the electric field at x =4m on the x axis at this time 1 N/C.

The complete question is below:

Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, the resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?

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We need Net resistance

solve from right words

[tex]\\ \rm\Rrightarrow R_1=20+10=30\Omega[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{1}{R_2}=\dfrac{1}{30}+\dfrac{1}{50}=\dfrac{5+3}{150}[/tex]

[tex]\\ \rm\Rrightarrow R_2=\dfrac{150}{8}=18.75\Omega[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{1}{R_3}=\dfrac{1}{30}+\dfrac{1}{40}=\dfrac{4+3}{120}[/tex]

[tex]\\ \rm\Rrightarrow R_3=\dfrac{120}{7}=17.14\Omega[/tex]

[tex]\\ \rm\Rrightarrow R_{net}=17.14+18.75=35.89\Omega[/tex]

Use ohm's law

[tex]\\ \rm\Rrightarrow I=\dfrac{V}{R}[/tex]

[tex]\\ \rm\Rrightarrow I=\dfrac{9}{35.89}[/tex]

[tex]\\ \rm\Rrightarrow I\approx 0.25A[/tex]

Answer:

12.9m/s^2

Explanation:

As, a=(v^2)/r

=(58^3)/260

=12.9

The field exerts 6.5 x 10⁸ Newton of force on EACH COULOMB of charge in the field.

We're putting 1.6 x 10⁻¹⁹ Coulomb of charge into the field.

The force on it will be

      (1.6 x 10⁻¹⁹ Coulomb) x  (6.5 x 10⁻⁸ Newton/Coulomb).

That's 1.04 x 10⁻¹⁰ Newton.

Statement B  is true about the energy level of gas, Gases have more energy than liquids and solids.

A gas is a sample of matter that adopts the shape of the container in which it is housed and develops a uniform density inside the container.

Liquid molecules are more energetic than solid molecules. Further heating will cause the molecules to move so quickly that they won't stick together at all. The gas molecules have the highest energy content.

Gases have more energy than liquids and solids.

Hence statement B  is true about the energy level of gas.

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By using Newton's Second law of motion, the largest mass that can be lifted by this input force is equal to 2,244 kg.

Based on the information provided, we can logically deduce that the output to input force ratio is equal to 22:1. This ultimately implies that, the input force is equal to 1000 N, which would result in an output force of 22,000 N.

Next, we would calculate the largest mass by using Newton's Second law of motion:

Mass = Force/Acceleration

Mass = 22,000/9.8

Mass = 2,244 kg.

Note: Acceleration due to gravity is equal to 9.8 m/s².

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(a) The sign of the unknown charge is negative.

(b) The magnitude of the unknown charge is 2.18 μC.

The charge is the physical quantity which defines the object's electric field. The charged objects creates the electric field around it and attracts or repels another objects coming into that field.

Given is a charged ball of mass m = 0.265 kg and unknown charge q is hanging by a light thread from a ceiling. A fixed charge Q = +5.00 μC on an insulated stand is brought close to the unknown charge. As a result, the unknown charge hangs at an angle 0 = 38.0° to the vertical as shown in the diagram below. The distance between the two charges is r = 22.0 cm.

Tension in the string has two components, Tsinθ and Tcosθ

(a) The fixed charge Q will attract the movable charge q attached to string. The charge Q is positive. So, the charges must of opposite sign.

Thus, the sign of unknown charge is negative.

(b) From the equilibrium of forces, we get

Electrostatic force Fe = kQq/r²

Fe =  Tsinθ .............(1)

and

weight force mg = Tcosθ.............(2)

Dividing both the equations, we get

mg/Fe = tanθ

mg / (kQq/r²) =  tanθ

q = mgr²tanθ / kQ

Substitute the values from the question, we have

q = 0.265 x 9.81 x (0.22)² tan38 / (9x10⁹ x 5 x 10⁻⁶)

q = 2.18 x 10⁻⁶ C

or q = 2.18 μC

Thus, the magnitude of charge is  2.18 μC.

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Answer:

The rod is stretched because there is a larger gravitational force on the mass closest to mass M.

F = G M m / r^2 - G M m / (r + l)^2       the difference of the forces on m's

F = G M m [1 / r^2 - 1 / (r + l)^2]

[1 / r^2 - 1 / (r + l)^2] = (r^2 + 2 r l + l^2 - r^2 / [r^2 (r + l)^2]

= (2 r l + l^2) / [r^2 (r + l)^2]

if l is small compared to r = (2 r l ) / [r^2 (r + l)^2]

= (2 r l ) / [r^4 + 2 r^3 l ]

When a ball is thrown up by doing work, the velocity of the ball will be 2.6 m/s.

It states that the Work done in moving a body is equal to the change in kinetic energy of the body

Kinetic energy = 1/2 mv²

Given is a ball of mass m = 0.750 kg and the work done on ball W = 2.50 J

The ball is initially at rest. So, initial velocity is zero. Then, change in kinetic energy will be

W= ΔK.E = K.Ef - K.Ei

According to work energy theorem, work done is

W = 2.5J  = 1/2 x 0.750 x (v)² -0

v =2.6 m/s

Thus, the velocity of the ball is 2.6 m/s

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(a) Let [tex]v[/tex] be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

[tex]a_{\rm rad} = \dfrac{v^2}R[/tex]

where [tex]R[/tex] is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

[tex]F = (1.500\,\mathrm{kg}) a_{\rm rad}[/tex]

so that

[tex](1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N[/tex]

Solve for [tex]v[/tex] :

[tex]v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}[/tex]

(b) The net force equation in part (a) leads us to the relation

[tex]F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}[/tex]

so that [tex]v[/tex] is directly proportional to the square root of [tex]R[/tex]. As the radius [tex]R[/tex] increases, the maximum linear speed [tex]v[/tex] will also increase, so the cord is less likely to break if we keep up the same speed.

Answer:

b. B

Explanation:

Picture B has the smallest peaks among all which henceforth makes the wavelength i.e. distance between two adjacent crests or troughs, small.

Answer:b

Explanation:Wavelength can be calculated using the following formula: wavelength = wave velocity/frequency. Wavelength usually is expressed in units of meters. The symbol for wavelength is the Greek lambda λ, so λ = v/f.

Answer:

a

Explanation:

Answer:

Approximately [tex]3.14\; {\rm rad \cdot s^{-1}}[/tex].

Explanation:

Fact: the angular velocity [tex]\omega[/tex] of a simple harmonic oscillator is the ratio between the maximum velocity [tex]v_{\text{max}}[/tex] and the maximum displacement [tex]x_\text{max}[/tex] of this oscillator. In other words:

[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}}\end{aligned}[/tex].

Derivation of the previous equation:

Let [tex]A[/tex] denote the amplitude of this oscillation, and let [tex]\omega[/tex] denote the angular velocity.

The displacement of the oscillator at time [tex]t[/tex] would be:

[tex]x(t) = A\, \sin(\omega\, t)[/tex].

The maximum displacement of this oscillator would be [tex]x_\text{max} = A[/tex].

The velocity of this oscillator at time [tex]t[/tex] is the derivative of displacement with respect to time:

[tex]\begin{aligned} v(t) &= \frac{d}{d t}\, [x(t)] \\ &= \frac{d}{d t} [A\, \sin(\omega\, t)] \\ &= A\, \omega\, \cos(\omega\, t)\end{aligned}[/tex].

The maximum velocity of this oscillator would be [tex]v_\text{max} = A\, \omega[/tex].

Notice that dividing [tex]v_\text{max} = A\, \omega[/tex] by [tex]x_\text{max} = A[/tex] would give:

[tex]\displaystyle \frac{v_\text{max}}{x_\text{max}} = \frac{A\, \omega}{A} = \omega[/tex].

It is given that [tex]v_\text{max} = 2.76\; {\rm m\cdot s^{-1}}[/tex] while [tex]x_\text{max} = 0.0880\; {\rm m}[/tex]. Therefore:

[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}} \\ &= \frac{2.76\; {\rm m\cdot s^{-1}}}{0.0880\; {\rm m}} \\ &\approx 3.14\; {\rm s^{-1}}\end{aligned}[/tex].

(Radians per second.)

The force, in newtons, exerted by each of the 10 braces is 2.135 x 10⁴ N.

The force is defined as the shear stress or pressure applied per unit area.

F = P/A

Given is a 17.0-m-high and 11.0-m-long wall and its bracing under construction are shown in the figure(attached). The force is exerted by each of the 10 braces, if a strong wind exerts a horizontal force of 655 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths.

Considering the pivot at the base of wall.

From the equilibrium of forces, we have

r₁ x Fwind = r₁ x Fbsinθ

Put the values, we get

Fb = Fwind /10sin35°.............(1)

The wind force is also given by

Fwind = Horizontal force or pressure x Area

Fwind = F/A x hw

           =655 x 17 x 11

F wind = 122,485 N

From equation (1), we have

Fb = Fwind /10sin35°

Fb =  122,485 /10sin35°

Fb = 21,354.6080 N

0r Fb = 2.135 x 10⁴ N

Thus, the force exerted on each of the 10 brace is 2.135 x 10⁴ N.

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The weight of the object on pluto will be 15.56 N

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

Given data;

[tex]\rm g_e[/tex] is the acceleration due to gravity on earth.

m is the mass of an object

[tex]\rm g_p[/tex] is the acceleration due to gravity on pluto.

[tex]\rm W_e[/tex] is the weight of the object on earth

[tex]\rm W_ p[/tex] is the weight of an object on pluto

The mass of the object on the earth's surface is found as;

We = m × ge

m = W / ge

m = 250 / 9.81

m = 25.51  kg

The weight of the object on pluto.

Wp = m × gp

Wp = 25.51  kg × 0.61  m/s²

Wp =15.56 N.

Hence the weight of the object on pluto will be 15.56 N

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Answer: B.

15.6 newtons

Explanation: edmentum

Answer:

coplanar When all forces are acting in the same 

resultunt the single force and associated torque obtained by combining a system of forces and torques acting on a rigid body via vector addition

Answer:

Explanation:

missing info .....

(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.

(b) The velocity and acceleration at any time t is v =  (4ti + j) m/s and a = a = 4i m/s²

(c)  The average acceleration in the time interval given in part (a)​ is 3.98 m/s².

x = at²i + btj

x = 2t²i + tj

Δv = Δx/Δt

x(2) = 2(2)²i + 2j

x(2) = 8i + 2j

|x(2)| = √(8² + 2²) = 8.246

x(3) =  2(3)²i + 3j

x(3) = 18i + 3j

|x(3)| = √(18² + 3²) = 18.248

Δv = (18.248 - 8.246)/(3 - 2)

Δv =  10 m/s

v = dx/dt

v =  (4ti + j) m/s

a = dv/dt

a = 4i m/s²

v(2) = 4(2)i + j

v(2) = 8i + j

|v(2)| = 8.06 m/s

v(3) = 4(3)i + j

v(3) = 12i + j

|v(3)| = 12.04 m/s

a = (12.04 - 8.06)/(3 - 2)

a = 3.98 m/s²

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We will use the design process to meet the requirements of the competition by the given explained in the main answer.

The initiative was taken by two or more parties operating independently to win the business of a third party by competing to hire contractors on the best terms possible.

The complete question is

"A local businessperson placed an advert in the local newspaper for a competition for girls only from different schools to participate in the competition. The competition is about drawing ideas for a shopping complex for the local community. Girls from the different schools are excited about the competition. you are a group of boys in the local schools who saw the advert and feel it is unfair for you to be excluded from the competition.

consider yourself as one of the learners in the local schools.

Explain why the project should be given to boys also who have better skills for this project than the girls do.

1.2 think about the design process skills, investigates, design, make, evaluate and communicate."

Explain how you will use the design process to meet the requirements of the competition.

[ 1.1 ]Boys should also be given the project because some of them have more artistic talent than girls do when it comes to sketching designs for local shopping centers.

Boys can also present a clear picture of how local communities can expand their shopping options.

For instance, they can encourage local merchants to offer seasonal items because doing so will benefit their businesses.

They can also suggest that retailers encourage customers to bring a friend to participate in competitions, which will be highly regarded.

[1.2] In order to fulfill the needs of the competition, I would make sure to explore how a shopping center for the neighborhood functions, for instance, by making sure I am aware of the strategies one may use to enhance his or her firm in the neighborhood.

In order to win the competition, I would make sure that I was confident in what I was saying and that I was audible enough to be heard during the competition.

Finally, I would make sure I met the competition's conditions by staying on-topic and solely addressing the issues raised in the advertisement

Hence will use the design process to meet the requirements of the competition.

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Answer:

1680 seconds

Explanation:

[tex]28 hrs * \frac{60 s}{1 hr} =1680s[/tex]

Answer:

28

Explanation:

Or from this equation, when we can write the magnification of talents is F times F plus the object distance knowledge, substitute the value to find out the magnification and that equals minus 54 centimeters Over -54 minus 26 centimeters. And we got the magnification of the image produced by The lens is 0.675.

The direction of this vector is 170° from the positive x-axis.

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