Erbium-165 has a half-life of 10.4 hours. If you start with 1,000 grams of
erbium-165, how much time will it take to have 125 grams of erbium-165 left
in the sample?
A. 41.6 hours
OB. 20.8 hours
C. 31.2 hours
D. 10.4 hours

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

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

Answer:

a

Explanation:

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

The distance and speed are as follows:

Speed of a body is the ratio of the distance covered by a body and the time it takes to cover that distance,

The number of seconds in a year = 365.25 * 24 * 3600 = 31557600 second

a) Distance covered in 51 seconds = 5.5/31557600 * 51

Distance = 8.89 * 10⁻⁶ cm

b) 5.5 cm = 3.4175 * 10⁻⁵ miles

Sped in miles per million years = 3.42 * 10⁻⁵ miles/1 * 10⁻⁶ million years

Speed = 34.2 miles per million years.

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Answer:

1680 seconds

Explanation:

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

(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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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 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.

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 ]

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

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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Answer:

Explanation:

missing info .....

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]

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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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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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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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:

672 000 J      or 672 kJ

Explanation:

change in the thermal energy = mass x specific heat capacity x temperature change

     =   2 kg * 4200J/kg-C *80 C =672000 J

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 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:

12.9m/s^2

Explanation:

As, a=(v^2)/r

=(58^3)/260

=12.9

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

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Answer:

My explanation and answer are down here↓:

Explanation:

p→q is false only when p is true and q is false.

∴p→(q∨r) is false when p is true and (q∨r) is false, and

q∨r false when both q,r are false.

Hence T,F,F.

The distance x will the ball land after flies off with a horizontal initial velocity  is 3.0635 m.

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

A small metal ball with a mass of m = 62.0 g is attached to a string of length l = 1.85 m. It is held at an angle of θ = 48.5° with respect to the vertical.

The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 3.76 m, and then later it hits the ground.

The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.

Kinetic energy  = Potential energy

1/2 mv² =mgh₁

The velocity at the bottom, when the height h = 5m, is

v= √2gh₁...................(1)

The vertical height h₁ = l- lcosθ

h₁ = l- lcosθ

h₁ = 1.85 - 1.85cos48.5°

h₁ =0.6241 m

Putting the values in equation (1), we get

v = √2x 9.81 x0.6241

v = 3.499 m/s

The horizontal distance traveled is

x = vt

x = v x √2h/g

Plug the values, we get

x =  3.499 x √2x3.76 / 9.81

x = 3.0635 m

Thus, the horizontal distance ball travels is  3.0635 m.

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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

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.

(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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