If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?

Answer:

option A

Explanation:

simple harmonic motion

Answer:

random motion I think not sure

Answer:

The  distance of the proton is  [tex]r_p =10 \ cm[/tex]

Explanation:

Generally the force experience by the electron is mathematically represented as

         [tex]F_e = \frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e}[/tex]

 Where  [tex]\lambda _e[/tex] is the charge density of the charge wire before it is doubled

Also the force experience by the proton is mathematically represented as

         [tex]F_p = \frac{q * \lambda_p }{ 2 \pi * \epsilon_o * r_p}[/tex]

Given that the charge density is doubled i.e [tex]\lambda_p = 2 \lambda_e[/tex] and that the the force are  equal then

      [tex]\frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e} = \frac{q * 2 \lambda_e }{ 2 \pi * \epsilon_o * r_p}[/tex]

      [tex]\frac{ \lambda_e }{ r_e} = \frac{ 2 \lambda_e }{ r_p}[/tex]

      [tex]r_p * \lambda_e =2 \lambda_e * r_e[/tex]

       [tex]r_p =2 r_e[/tex]

Now given from the question that  [tex]r_e[/tex] the distance of the electron from the charged wire is  5 cm

 Then  

        [tex]r_p =2 (5)[/tex]

         [tex]r_p =10 \ cm[/tex]

Answer:

E. None of these statements are true.

Explanation:

We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.

for A, the tension on the rope does not depend on if both teams pull are in equilibrium.

for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.

for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.

for D,  just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.

we go with option E.

Answer

76200meters

Explanation:

we know that 1km=1000meters

to convert km into meters we we divide km by meters

=76.2/1000

=76200meters

Answer:

The inner and outer surfaces of a 0.5-cm thick 2-m by 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass

Explanation:

Complete question:

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of  955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.

Answer:

The initial speed of the block is 1.422 m/s

Explanation:

Given;

mass of the block, m = 2.0 kg

force constant of the spring, K = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

Apply Hook's law to determine applied force on the spring;

F = Kx

F = (955 N/m)(0.046 m)

F = 43.93 N

Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;

F = ma

a = F / m

a = 43.93 / 2

a = 21.965 m/s²

Apply kinematic equation to determine the initial speed of the block;

v² = u² + 2ax

where;

v is the final speed of the block = 0

u is the initial speed of the block

x is the distance traveled by the block = compression of the spring

a is the block deceleration = -21.965 m/s²

0 = u² + 2(-21.965 )(0.046)

0 = u²  - 2.021

u² =  2.021

u = √2.021

u = 1.422 m/s

Therefore, the initial speed of the block is 1.422 m/s

Answer:

PlROCA

Explanation:

Answer:

The answer is 312.5j

Explanation:

The kinetic energy (KE):

KE=1/2*m*v^2

M= mass of the object

v= velocity of the object

We have;

m=25g

v=5m/s

KE=1/2*25g*5^2m/s

KE =312.5j

Explanation:

Distance = speed × time

d = (18.0 m/s) (1 hr × 3600 s/hr)

d = 64,800 m

d = (18.0 m/s) (1 min × 60 s/min)

d = 1080 m

d = (18.0 m/s) (1 s)

d = 18.0 m

Answer: her rotational kinetic energy increases

Answer:

Explanation:

Charge on an electron (q) = 1.6 * 10 ^ -19 C

Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec

We know that, Force exerted on moving particle moving through a magnetic field :

[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]

1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B

B =  0.08573 T

The image in the attachment describes the situation of the fishing rod.

Answer: F = 10.8 N

Explanation: The image shows a fishing rod attached to an axis. To stay in equilibrium, Torque must be equal for the force of magnitude 18N and for the unknow force.

Torque (τ) is a measure of a force's tendency to cause rotation and, in physics, defined as:

τ = F.r.sin(θ)

F is the force acting on the object;

r is distance between where the torque is measured to where the force is applied;

θ is the angle between F and r;

For the fishing rod:

[tex]\tau_{1} = \tau_{2}[/tex]

[tex]F_{1}.r_{1}.sin(\theta) = F_{2}.r_{2}.sin(\theta)[/tex]

Assuming part (1) is related to unknown force:

[tex]F = \frac{F_{2}.r_{2}.sin(\theta}{r_{1}.sin(\theta) }[/tex]

Replacing the corresponding values:

[tex]F = \frac{18*1.8*sin(30)}{3*sin(30)}[/tex]

[tex]F = \frac{18*1.8}{3}[/tex]

F = 10.8

The fishing line exert on the the rod a force of 10.8N.

b. solve for Vo

[tex] vf ^{2} = vo^{2} + 2a(xf - xo) \\ vf ^{2} = vo ^{2} + 2axf - 2axo\\ vo ^{2} = vf ^{2} - 2axf + 2axo \\ vo = \sqrt{vf ^{2} - 2axf + 2axo } \\ vo = - \sqrt{vf^{2} - 2axf + 2axo } [/tex]

I hope I helped you ^_^

Answer:

B = 1.33 10⁻⁸ T , the magnetic field must be in the y + direction

Explanation:

In an electromagnetic wave the electric and magnetic fields are in phase

         c = E / B

         B = E / c

let's calculate

          B = 4.00 / 3 10⁸

          B = 1.33 10⁻⁸ T

To determine the direction we use that the electric and magnetic fields and the speed of the wave are perpendicular.

 If the wave advances in the + Z direction and the electric field is in the + x direction, the magnetic field must be in the y + direction

Answer: d₂ = 170 mGya

Explanation:

the relationship between absonbed 'd' and exposure 'E' is given as;

D(Gv) = F . x (AS/xB)

F is a conversion coefficient depending on medium

so we can simply write

d₁/d₂ = x₁/x₂

Given that;

our x₁ = 60 mAs, x₂ = 120 mAs,  d₁ = 85 mGya, d₂ = ?

from the given formula,

d₂ = (x₂d₁ / x₁)

now we substitute

d₂ = (120 × 85) / 60

d₂ = 170 mGya

∴ if 120 mAa is used,  the new exposure will be 170 mGya

Answer:

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Answer:

The  angle is  [tex]\theta = 36.24 ^o[/tex]

Explanation:

From the question we are told that

    The  mass is  [tex]m = 0.6 \ kg[/tex]

     The radius is  [tex]r = 1.1 \ m[/tex]

     The speed is  [tex]v = 3.57 \ m /s[/tex]

According to  the law of energy conservation

  The  potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e

      [tex]m * g * h = \frac{1}{2} * m * v^2[/tex]

 =>    [tex]h = \frac{1}{2 g } * v^2[/tex]

Here h is the vertical distance traveled by the mass  which is also mathematically represented as

      [tex]h = r * sin (\theta )[/tex]

So

     [tex]\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2][/tex]

substituting values

     [tex]\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2][/tex]

     [tex]\theta = 36.24 ^o[/tex]

Answer:

[tex] \boxed{\sf Kinetic \ energy \ (KE) = 85 \ J} [/tex]

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

[tex] \boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}[/tex]

Substituting values of m & v in the equation:

[tex] \sf \implies KE = \frac{1}{2} \times 6.8 \times {5}^{2} [/tex]

[tex] \sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 3.4 \times 25 [/tex]

[tex] \sf \implies KE =3.4 \times 25 [/tex]

[tex] \sf \implies KE = 85 \: J[/tex]

The kinetic energy of the object reported to two significant figures is: 85 Joules.

Given the following data:

To find the kinetic energy of the object:

Kinetic energy refers to an energy that is possessed by a physical object or body due to its motion.

Mathematically, kinetic energy is calculated by using the formula;

[tex]K.E = \frac{1}{2} MV^2[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]K.E = \frac{1}{2}[/tex] × [tex]6.8[/tex] × [tex]5^2[/tex]

[tex]K.E = 3.4[/tex] × [tex]25[/tex]

Kinetic energy = 85 Joules.

Therefore, the kinetic energy of the object is 85 Joules.

Read more: https://brainly.com/question/23153766

Answer:

[tex]\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi} [/tex]

[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} } [/tex]

[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} } [/tex]

[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 } [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} } [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} } [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21} [/tex]

[tex]\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21} [/tex]

[tex]\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}[/tex]

Answer:

ΔE = GMm/24R

Explanation:

centripetal acceleration a = V^2 / R = 2T/mr

T= kinetic energy

m= mass of satellite, r= radius of earth

= gravitational acceleration = GM / r^2

Now, solving for the kinetic energy:

T = GMm / 2r = -1/2 U,

where U is the potential energy

So the total energy is:

E = T+U = -GMm / 2r

Now we want to find the energy difference as r goes from one orbital radius to another:

ΔE = GMm/2 (1/R_1 - 1/R_2)

So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R

ΔE = GMm/2R (1/3 - 1/4)

ΔE = GMm/24R

Answer:

R = 0.992 Ω

Explanation:

En esta pregunta, dada la información que contiene, debemos calcular la resistencia de la varilla de grafito.

Matemáticamente,

Resistencia = (resistividad * longitud) / Área De la pregunta;

Resistividad = 3,5 * 10 ^ -5 Ωm

longitud = 170 cm = 1,7 m

Área = 60 mm ^ 2 = 60/1000000 = 6 * 10 ^ -5 m ^ 2

Conectando estos valores a la ecuación anterior, tenemos;

Resistencia = (3.5 * 10 ^ -5 * 1.7) / (6 * 10 ^ -5) =

(3.5 * 1.7) / 6 = 0.992 Ω

Answer:

The intensity is  [tex]I_1 = 263.35 \ W/m^2[/tex]

Explanation:

From the question we are told that

    The intensity of the beam is  [tex]I = 285\ W/m^2[/tex]

    The  angle is [tex]\theta = 16^o[/tex]

The  intensity of the light that emerges from the polarizer is mathematically represented by Malus' law as

        [tex]I_1 = I * cos^2 (\theta )[/tex]

substituting values

        [tex]I_1 = 285 * [cos(16)]^2[/tex]

substituting  values

        [tex]I_1 = 285 * [cos(16)]^2[/tex]

        [tex]I_1 = 263.35 \ W/m^2[/tex]

Answer:

According to different sources Edwin Hubble observed red light of galaxies directly proportional to the distance of the galaxy from earth.

I hope it helps and follow me for more good answer♥️♥️

Answer:

The glove takes 1.02s to return to the pitchers hand.

Explanation:

Given;

initial velocity the pitcher's glove, u = 5 m/s

Apply kinematic equation

s = ut - ¹/₂gt²

where;

g is acceleration due to gravity = 9.8 m/s²

t is the time takes the glove to return to the pitchers hand

s is the displacement of the glove, which will be equal to zero when the glove returns to the pitchers hand. (s = 0)

0 = ut - ¹/₂gt²

ut = ¹/₂gt²

u = ¹/₂gt

gt = 2u

t = (2u) / g

t = (2 x 5) / 9.8

t = 1.02 s

Therefore, the glove takes 1.02s to return to the pitchers hand.

Answer:

c. the work done by the environment on the system equals the changein internal energy.

Explanation:

Adiabatic process:

When the boundary of a system is perfectly insulated, it means that the energy can not flow from the system and into the system ,these system is known as adiabatic system.

When the energy transfer in the system is zero ,then these type of process is known as adiabatic process.

From the first law of thermodynamics

Q= ΔU + W

Q=Heat transfer

ΔU=Change in internal energy

W=Work transfer

In adiabatic process , Q= 0

Therefore

0=ΔU +W

W=- ΔU

Negative sign indicates that ,the work done by the environment.

Therefore the correct option will be (c).

Answer:

Explanation:

Relation between flux and inductance is as follows

φ = Li

where φ is flux associated with induction of inductance L when a current i flows through it

putting the values

3.25 x 10⁻³ x 800 = L x 2.9

L = .9 H

for induced emf in an induction , the relation is

emf induced = L di / dt

Putting the values

7.5 x 10⁻³ = .9 x di / dt

di / dt = 8.33 x 10⁻³ A / s

(a) The self inductance of the solenoid is 0.897 H.

(b) The magnitude of the rate of change of the current is 0.00836 A/s.

The given parameters;

The self inductance of the solenoid is calculated as follows;

[tex]emf = \frac{d\phi}{dt}\\\\emf = \frac{Ldi}{dt} \\\\d\phi = Ldi\\\\\phi = BA\\\\NBA = LI\\\\L = \frac{NBA}{I} \\\\L = \frac{N\phi}{I} \\\\L = \frac{800 \times 3.25\times 10^{-3}}{2.9} \\\\L = 0.897 \ H\\\\[/tex]

The magnitude of the rate of change of the current is calculated as follows;

[tex]emf = L \frac{di}{dt} \\\\\frac{di}{dt} \ = \frac{emf}{L} \\\\\frac{di}{dt} = \frac{7.5 \times 10^{-3}}{0.897} \\\\\frac{di}{dt} = 0.00836 \ A/s[/tex]

Learn more here:https://brainly.com/question/17086348

Answer:

D.

Inverted and reduced

If object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.

When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.

A concave mirror is a name for this sort of mirror. An enlarged image is caused when the concave mirror is positioned too near to the object.

A concave mirror has a focal length of magnitude f. An object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.

Hence option B is correct.

To learn more about the concave mirror refer to the link;

https://brainly.com/question/25937699

Answer: 0.0617

Explanation:

Given: The probability of wet weather on any given day in a city of Punjab : p=15%=0.15

Let X be a binomial variable that represents the number of days having wet weather.

Binomial probability formula : [tex]P(X=x)=^nC_xp^x(1-p)^x[/tex], where n= total outcomes, p = probability of success in each outcomes.

Here, n= 7 ( 1 week = 7 days)

The probability that it will take a week for it three wet weather on 3 separate days:

[tex]P(X=3)^=\ ^7C_3(0.15)^3(1-0.15)^{7-3}\\\\=\dfrac{7!}{3!(7-3)!}(0.15)^3(0.85)^4\\\\=\dfrac{7\times6\times5}{3\times2}\times 0.003375\times0.52200625\approx0.0617[/tex]

Hence, the required probability =0.0617

Answer:

The intensity of light from the first polarizer  is [tex]I_1 = 26 W/m^2[/tex]

Explanation:

  The intensity of the unpolarized light is  [tex]I_o = 52.0 \ W/m^2[/tex]

Generally the intensity of light that emerges from the first polarized light is

            [tex]I_1 = \frac{I_o}{2 }[/tex]

 substituting values

             [tex]I_1 = \frac{52. 0}{2 }[/tex]

             [tex]I_1 = 26 W/m^2[/tex]