I need help solving my homework problem , can somebody help me please . This class is physics two . My question is “ where the potential difference across ab is 49.5 V “ ?

Answer:

The net torque is 0.98 Nm.

Explanation:

The torque is given by

Torque = force x perpendicular distance

The clock wise torque is taken as negative while the counter clock wise torque is taken as positive.

Take the torques about the fulcrum.

Torque =  1 x 9.8 x 0.4 - 0.5 x 9.8 x 0.6

Torque = 3.92 - 2.94 = 0.98 Nm

Answer:

Explanation:

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Answer:

Option A. The speed of sound through water is 4.3 times faster than sound through air.

Explanation:

To answer the question correctly, we shall determine the speed of the wave in both cases. This is illustrated below:

For Air:

Frequency (fₐ) = 195 Hz

Wavelength (λₐ) = 1.76 m

Velocity (vₐ) =?

vₐ = λₐfₐ

vₐ = 1.76 × 195

vₐ = 343.2 m/s

For Water:

Frequency (fᵥᵥ) = 195 Hz

Wavelength (λᵥᵥ) = 7.6 m

Velocity (vᵥᵥ) =?

vᵥᵥ = λᵥᵥfᵥᵥ

vᵥᵥ = 7.6 × 195

vᵥᵥ = 1482 m/s

Finally, we shall compare the speed in water to that of air. This can be obtained as follow:

Velocity in air (vₐ) = 343.2 m/s

Velocity in water (vᵥᵥ) = 1482 m/s

Water : Air

vᵥᵥ : vₐ =>

vᵥᵥ / vₐ = 1482 / 343.2

vᵥᵥ / vₐ = 4.3

Cross multiply

vᵥᵥ = 4.3 × vₐ

Thus, the speed in water is 4.3 times the speed in air.

Option A gives the correct answer to the question.

Answer:

228

Explanation:

Answer: The given statement is False.

Explanation:

Boyle's law states that pressure is inversely proportional to the volume of the gas at constant temperature and the number of moles.

Mathematically,

[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)

OR

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are the initial pressure and volume of the gas

[tex]P_2\text{ and }V_2[/tex] are the final pressure and volume of the gas

Hence, the given statement is False.

(C)

Explanation:

From Ohm's law,

V = IR

Solving for I,

I = V/R

= (230 V)/(25 ohms)

= 9.2 A

Answer:

Soory

Explanation:

I really dont know but i will send you wait

Answer:

a)  t = 6.46 s, b)  x = 72.98 m, c)    v₁ = 26.75 m / s,   v₂ = 22.61 m / s

Explanation:

This is an exercise in kinematics, let's write the expressions for each person

Hanna

leaves a time t₀o = 1s after Sam's output, both with zero initial velocity and acceleration of a₁ = 4.90 m / s²

          x₁ = 0 + ½ a₁ (t-t₀) ²

          v₁ = 0 + a₁ (t-t₀)

Sam

with an acceleration of a₂ = 3.50 m / s² and with an initial velocity of zero  

         x₂ = 0+ ½ a₂ t²

         v₂ = 0 + a₂ t

a) at the point where the position of the two is found is the same

          x₁ = x₂

         ½ a₁ (t-t₀) ² = ½ a₂ t²

let's solve

         t-t₀ = [tex]\sqrt{\frac{a_2}{a_1} }[/tex]   t

         t (1 - [tex]\sqrt{ \frac{a_2}{a_1} }[/tex]) = t₀

         t = [tex]\frac{t_o}{ 1-\sqrt{ \frac{a_2}{a_1} } }[/tex]

let's calculate

          t = [tex]\frac{ 1}{1- \sqrt{\frac{3.50}{4.90} } }[/tex]

          t = [tex]\frac{1}{1- 0.845}[/tex] 1 / 1- 0.845

          t = 6.46 s

b) the distance traveled is

         x = ½ a₂ t²

         x = ½ 3.5 6.46²

         x = 72.98 m

c) Hanna's speed

         v₁ = 4.9 (6.46 -1)

         v₁ = 26.75 m / s

sam's speed

          v₂ = a2 t

          v₂ = 3.50 6.46²

          v₂ = 22.61 m / s

Answer:

The correct answer is - B ≠ 0, E = 0.

Explanation:

A force q E is exerted by the electric field on the charged particle that accelerates always, that is, it increases the speed of the particle. The particle can never be rotated in a rotation in the circle in an electric field.

The magnitude of the velocity cannot be changed by the magnetic field but changes only the velocity of the direction.

If the particle moves in a circle it means that the speed should begin for a year. The only direct velocity is constant and only the change be b can be achieved. So, B ≠ 0 and E = 0.

Answer:

3

Explanation:

first you find the original component of the force be a cause in 36 degrees then with ka sin 36 find the average interforce by multiplying it with hundred Newton then divide the original force by the mass 25 Kg is equals tto ma

Answer:

See image

Explanation:

Plato

Answer:

This is the answer. hope this help u

Answer: The answer will be R1 = B

Explanation:

Cause Here x component of A = A

X component of 2B       = 2BcosФ

X component of Resultant = 0

I.e. A + 2BcosФ = 0

cosФ = -A/2B

also R1 = √(A² + B² + 2ABcosФ)

            =√ [A² + B² + 2AB(-A/2B)]

            =√A² + B² -A²

R1 = √B²

R1 = B so R1 = B is correct.

Answer:

According to this model, the atom is a sphere of positive charge, and negatively charged electrons are embedded in it to balance the total positive charge.

Explanation:

Hope this helps you

Answer:

The plum pudding model of the atom states that  had negatively-charged electrons embedded within a positively-charged "soup."

Explanation:

Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup." Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus. Based on these results, Rutherford proposed the nuclear model of the atom.

Answer:

Transmits high-frequency (1 to 5 megahertz) sound pulses into your body using a probe. The sound waves travel into your body and hit a boundary between tissues (e.g. between fluid and soft tissue, soft tissue and bone).

Explanation:

Doppler ultrasound works by measuring sound waves that are reflected from moving objects, such as red blood cells.

Answer:

The guy above is pretty good

Explanation:

I'd go with that answer, give them brainliest

Answer:

so in a given orbital there can be 3 electrons.

Explanation:

The Pauli exclusion principle states that all the quantum numbers of an electron cannot be equal, if the spatial part of the wave function is the same, the spin part of the wave function determines how many electrons fit in each orbital.

In the case of having two values, two electrons change. In the case of three allowed values, one electron fits for each value, so in a given orbital there can be 3 electrons.

Answer:

There are no options, so....

Explanation:

Chlorophyll a absorbs its energy from the Violet-Blue and Reddish orange-Red wavelengths, and little from the intermediate (Green-Yellow-Orange) wavelengths.

Answer:

See explanation

Explanation:

The degradation of the drug is a first order process;

Hence;

ln[A] = ln[A]o - kt

Where;

ln[A] = final concentration of the drug

ln[A]o= initial concentration of the drug = 5 gm/100

k= degradation constant = 0.05 day-1

t= time taken

When [A] =[ A]o - 0.5[A]o = 0.5[A]o

ln2.5 = ln5 - 0.05t

ln2.5- ln5 = - 0.05t

t= ln2.5- ln5/-0.05

t= 0.9162 - 1.6094/-0.05

t= 14 days

b) when [A] = [A]o - 0.9[A]o = 0.1[A]o

ln0.5 = ln5 -0.05t

t= ln0.5 - ln5/0.05

t= -0.693 - 1.6094/-0.05

t= 46 days

Answer:

The pond's water level will fall.

Explanation:

Archimedes principle explains that a floating body will displace the amount of water that weighs the same as it, whereas a body resting on the bottom of the water displaces the amount of water that is equal to the body's volume.

When the anchor is in the boat it is in the category of floating body and when it is on the bottom of the pond it is in the second category.

Since anchors are naturally heavy and denser than water, the amount of water displaced when the anchor is in the boat is greater than the amount of water displaced when the anchor is on the bottom of the pond since the way anchors are doesn't make for them to have considerable volume.

When the anchor is dropped to the bottom of the pond, the water level will therefore fall. If the anchor doesn't reach the bottom it is still in the floating object category and there will be no difference to the water level, but once it touches the bottom of the pond, the water level of the pond drops.

Answer:

θ = 34.77°

Explanation:

From diffraction equation:

[tex]m\lambda = dSin\theta[/tex]

where,

m = order of diffraction

λ = wavelength of light used

d = slit separation

θ = angle

Therefore, for initial case:

m = 2

λ = 600 nm = 6 x 10⁻⁷ m

d = slit seperation = ?

θ = angle 20°

Therefore,

[tex](2)(6\ x\ 10^{-7}\ m)=d(Sin\ 20^o)\\\\d = \frac{12\ x 10^{-7}\ m}{0.342}\\\\d = 3.5\ x\ 10^{-6}\ m[/tex]

Now, for the second case:

m = 5

λ = 600 nm = 6 x 10⁻⁷ m

d = slit seperation = (1.5)(3.5 x 10⁻⁶ m) = 5.26 x 10⁻⁶ m

θ = angle = ?

Therefore,

[tex](5)(6\ x\ 10^{-6}\ m) = (5.26\ x\ 10^{-6}\ m)Sin\theta\\\\Sin\theta = \frac{(5)(6\ x\ 10^{-7}\ m)}{(5.26\ x\ 10^{-6}\ m)}\\\\\theta = Sin^{-1}(0.5703)[/tex]

θ = 34.77°

Answer:

The absolute pressure is 28.15 psi.

Explanation:

Gauge pressure = 28 inch of Mercury

Absolute pressure, Po = 14.4 psi

The absolute pressure is the sum of the gauge pressure and the absolute pressure.  

gauge pressure = 28 inch = 0.7112 m of Mercury

= 0.7112 x 13.6 x 1000 x 9.8 = 94788.736 Pa

= 13.75 psi

The absolute pressure is

P = 14.4 + 13.75 = 28.15 psi

Answer:

0 J

Explanation:

Since work done W = PΔV where P = pressure and ΔV = change in volume.

Since the volume is constant, ΔV = 0

So, Work done, W = PΔV = P × 0 = 0 J

So, the work done is 0 J.

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

[tex]$f_{m_1}= 65 \ Hz$[/tex] ,  and

[tex]$f_{m_2}= 95 \ Hz$[/tex]

Sampling rate [tex]f_s = \ 245 \ Hz[/tex]

The positive frequencies at the output of the sampling system are :

[tex]$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $[/tex]

When n = 0,

[tex]$f_{o_1}= f_{m_1} = 65 \ Hz,\ \ f_{o_2}= f_{m_2} = 95 \ Hz $[/tex]

when n  = 1,

[tex]$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm f_s $[/tex]

[tex]$f_{o_1}= \pm 65 \pm 245,\ \ f_{o_2}=\pm 95 \pm 245$[/tex]

[tex]$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \ f_{o_2}= 150 \ Hz,340 \ Hz$[/tex]

When n = 2,

[tex]$f_{o_1}= \pm 65 \pm 2(245),\ \ f_{o_2}=\pm 95 \pm 2(245)$[/tex]

[tex]$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \ f_{o_2}= 395 \ Hz,585 \ Hz$[/tex]

Therefore, the first six positive frequencies present in the replicated spectrum are :

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Answer:

Wavelength of light in film (let y = lambda)

y = 500 nm / (4/3) = 375 nm    

There will be a phase change at the air/film interface (not the other side)

S = 4 t       thickness of film = S/4 where S equals 1 wavelength

This is because of the phase change at one surface

375 nm = 4 * t

t = 93.8 nm

V = 99 volts

Explanation:

The voltage drop can be calculated using Ohm's law:

V = IR

= (1.10 A)(90 Ω)

= 99 volts

Answer:

The compression of the spring is 24.6 cm

Explanation:

magnitude of the charge on the left, q₁ = 4.6 x 10⁻⁷ C

magnitude of the charge on the right, q₂ = 7.5 x 10⁻⁷ C

distance between the two charges, r = 3 cm = 0.03 m

spring constant, k = 14 N/m

The attractive force between the two charges is calculated using Coulomb's law;

[tex]F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(4.6\times 10^{-7})(7.5\times 10^{-7})}{(0.03)^2} \\\\F= 3.45 \ N[/tex]

The extension of the spring is calculated as follows;

F = kx

x = F/k

x = 3.45 / 14

x = 0.246 m

x = 24.6 cm

The compression of the spring is 24.6 cm