Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.

Answer:

The car travels the distance of 225m before coming to rest.

Explanation:

Here,

v = 45m/s

t = 5s

d = v × t

Therefore,

d = 45 × 5

= 225m

Answer:

The mass of each object is 2kg and 3 kg.

Explanation:

Given that,

Gravitational force,[tex]F=1\times 10^{-8}\ N[/tex]

The distance between masses, d = 20 cm = 0.2 m

The total mass of the two objects, M + m = 5 kg

M = 5-m

The formula for the gravitational force is :

[tex]F=G\dfrac{Mm}{d^2}\\\\1\times 10^{-8}=6.67\times 10^{-11}\times \dfrac{(5-m)m}{(0.2)^2}\\\\\frac{1\times10^{-8}}{6.67\times10^{-11}}=\frac{(5-x)x}{(0.2)^{2}}\\\\\frac{1\times10^{-8}}{6.67\times10^{-11}}\cdot(0.2)^{2}\\\\5.99=(5-x)x\\\\x=2\ kg\ and\ 3 \ kg[/tex]

So, the mass of each object is 2kg and 3 kg.

Answer:

The phases of the cardiac action potential correspond to the surface ECG (ECG) (Figure). The P wave reflects atrial depolarization (phase 0), the PR interval reflects the conduction velocity through the AV node, the QRS complex the ventricular depolarization and QT interval the duration potential ventricular action.

Answer:

2) The bending of a light wave as it passes at an angle from one medium to another .

Hope it is helpful to you ☺️

Answer:

Gravity between you and the sun

Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.

Hope this helps!!!!

Answer:

electric current

Explanation:

The answer is electric current

Full Question:

What is the temperature of a system in thermal equilibrium with another system made up of water and steam at one atmosphere of pressure?

A) 0°F

B) 273 K

C) 0 K

D) 100°C

E) 273°C

Answer:

The correction Option is D) 100°C

Explanation:

The temperature above is referred to as the critical point.

it is the highest temperature and pressure at which water (which has three phases - liquid, solid, and gas) can exist in vapor/liquid equilibrium. If the temperature goes higher than 100 degrees celsius, it cannot remain is liquid form regardless of what the pressure is at that point.

There is also a condition under which water can exist in its three forms: that is  

- Ice (solid)

- Liquid (fluid)

- Gas (vapor)

That state is called triple point. The conditions necessary for that to occur are:

Cheers

Cheers

Answer:

The maximum speed of Bolt for the 100 m race is 14.66 m/s

Explanation:

Given;

initial distance covered by Bolt, d = 200 m

time of this motion, t = 19.3 s

The second distance covered by Bolt, = 100 m

Assuming Bolt maintained the same acceleration for both races.

His acceleration can be determined from the 200 m race.

d = ut + ¹/₂at²

where;

u is his initial velocity = 0

d =  ¹/₂at²

[tex]at^2 = 2d\\\\a = \frac{2d}{t^2} \\\\a = \frac{2\times 200}{19.3^2} \\\\a = 1.074 \ m/s^2[/tex]

Let the final or maximum velocity for the 100 m race = v

v² = u² + 2ad₂

v² =  2 x 1.074 x 100

v² = 214.8

v = √214.8

v = 14.66 m/s

The maximum speed of Bolt for the 100 m race is 14.66 m/s

Answer:

Explanation:

Given:

volume of urine discharged, [tex]V=400~mL=0.4~L=4\times 10^{-4}~m^3[/tex]

time taken for the discharge, [tex]t=30~s[/tex]

diameter of cylindrical urethra, [tex]d=4\times10^{-3}~m[/tex]

length of cylindrical urethra, [tex]l=0.2~m[/tex]

density of urine, [tex]\rho=1000~kg/m^3[/tex]

a)

we have volume flow rate Q:

[tex]Q=A.v[/tex] & [tex]Q=\frac{V}{t}[/tex]

where:

[tex]A=[/tex] cross-sectional area of urethra

[tex]v=[/tex] velocity of flow

[tex]A.v=\frac{V}{t}[/tex]

[tex]\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}[/tex]

[tex]v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}[/tex]

[tex]v=1.06~m/s[/tex]

b)

The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:

[tex]P=\rho.g.l[/tex]

[tex]P=1000\times 9.8\times 0.2[/tex]

[tex]P=1960~Pa[/tex]

Answer:

Answer

Correct option is

A

5×10

−6

tesla

I=5A

x=0.2m

Magnetic field at a distance 0.2 m away from the wire.

B=

2πx

μ

0

I

=

2π×0.2

4π×10

−7

×5

=10×5×10

−7

=5×10

−6

tesla

Answer:

1.61

Explanation:

According to Oxford dictionary, refractive index is, ''the ratio of the velocity of light in a vacuum to its velocity in a specified medium.''

If the clear transparent solid disappears when dipped into the liquid, it means that the index of refraction of the solid and liquid are equal.

Hence, when a transparent solid is immersed in a liquid having the same refractive index, there is no refraction at the boundary between the two media. As long as there is no refraction between the two media, the solid can not be seen because the solid and liquid will appear to the eye as one material.

South

Explanation:

The magnetic force F on a point charge moving with a velocity v in the presence of a magnetic field B is given by

[tex]\vec{\textbf{F}} = q\vec{\textbf{v}}\textbf{×}\vec{\textbf{B}}[/tex]

and according to the right-hand rule, an upward magnetic force on a proton moving westward is only possible if the magnetic field is directed southward.

Answer:

the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm

Explanation:

Given the data in the question;

Diameter of bright central maxima;

⇒ 2 × ( 1.22 × (λD/d) ) ⇒ 2.44( λD/d )

where D is the distance from the the droplet to the screen ( 30 cm )

d is the diameter of the droplet

λ is the wavelength of light ( 690 nm  = 690 × 10⁻⁷ cm )

since the central maximum of the pattern is 0.24 cm in diameter,

we substitute

0.24 cm = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / d )

solve for d

d = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / 0.24 cm

d = 0.0050508 cm² / 0.24 cm

d = 0.021045 cm or 2.1 × 10⁻² cm

Therefore, the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm

Answer:

586 K

Explanation:

Let P is the initial pressure.

Initial temperature, T₁ = 20°C = 293 K

Final pressure, P₂ = 2P

We need to find the final temperature of the gas.

The relation between the pressure and the temperature is as follows

[tex]P\propto T\\\\or\\\\\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}[/tex]

Put all the values,

[tex]\dfrac{P}{2P}=\dfrac{293}{T_2}\\\\\dfrac{1}{2}=\dfrac{293}{T_2}\\\\T_2=2\times 293\\\\T_2=586\ K[/tex]

So, the final temperature of the gas is 586 K.

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]

             a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

The total net force acting on the balloon will be 24498 Newtons

Given that

Volume of the balloon = 2210 cubic meter

Density of the air inside the balloon = 1.13  kg/m3

Here force on the balloon will be equal to the weight of the air displaced by balloon

[tex]F= mass of air displaced\times gravity[/tex]

[tex]F= Density \times volume \times gravity[/tex]

[tex]F=1.13 \times 2210 \times 9.81[/tex]

[tex]F=24498 N[/tex]

The total net force acting on the balloon will be 24498 Newtons

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Answer:

Resistance is inversly proportional to the current.

V=I.R.

P=V.I

Answer:

Option (c)

Explanation:

1.The transformer core is assembled from steel sheets to:

(a) Reduced power loss due to eddy current

(b) Reduced power loss due to hysteresis

(c) Reduced power loss due to current flowing through the winding

(d) Reduce all types of power loss.

A transformer is a device which converts the low voltage into high and vice  versa.

There are two types of a transformer.

Step up: It is used to convert low voltage into high.

Step down It is used to convert high voltage into high.

It depends on the number of turns in primary and the secondary coil.

The core of the transformer is laminated and it is in the form of sheets.

By using such type of core, the power loss due to the windings is reduced.

option (c) .

Answer:

n.m maby

Explanation:

i think or its kg m/s

Answer:

answer (B) & (C)

Explanation:

kg • /N • s

Answer: 19.15 meters on the waves away from the survivor.

Explanation:

Answer:

254 m

Explanation:

Applying,

v = λf............... Equation 1

Where v = velocity of radio wave, λ = wave length, f = frequency

make λ the subject of the equation

λ = v/f............ Equation 2

From the question,

Given: f = 1180 kHz = 1180000 Hz

Constant: v = 3×10⁸ m/s

Substitite into equation 2

λ  = 3×10⁸/1180000

λ  = 2.54×10²

λ  = 254 m

Answer:

1) [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex], 2) [tex]T_{B} \approx 1.137\cdot T_{A}[/tex], where [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex].

Explanation:

1) Pendulum A is a simple pendulum, whose period ([tex]T_{A}[/tex]) is determined by the following formula:

[tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex] (1)

Where:

[tex]l[/tex] - Length of the massless bar.

[tex]g[/tex] - Gravitational acceleration.

2) Pendulum B is a physical pendulum, whose period ([tex]T_{B}[/tex]) is determined by the following formula:

[tex]T_{B} = 2\pi \cdot \sqrt{\frac{I_{O}}{m\cdot g\cdot l} }[/tex] (2)

Where:

[tex]m[/tex] - Total mass of the pendulum.

[tex]g[/tex] - Gravitational acceleration.

[tex]l[/tex] - Length of the uniform bar.

[tex]I_{O}[/tex] - Moment of inertia of the pendulum with respect to its suspension axis.

The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:

[tex]I_{O} = \frac{1}{2} \cdot m\cdot l^{2}+\frac{1}{24}\cdot m\cdot l^{2} + \frac{3}{4}\cdot m\cdot l^{2}[/tex]

[tex]I_{O} = \frac{31}{24}\cdot m\cdot l^{2}[/tex] (3)

By applying (3) in (2) we get the following expression:

[tex]T_{B} = 2\pi \cdot \sqrt{\frac{\frac{31}{24}\cdot m \cdot l^{2} }{m\cdot g \cdot l} }[/tex]

[tex]T_{B} = 2\pi \cdot \sqrt{\frac{31\cdot l}{24\cdot g} }[/tex]

[tex]T_{B} = \sqrt{\frac{31}{24} } \cdot \left(2\pi \cdot \sqrt{\frac{l}{g} }\right)[/tex]

[tex]T_{B} \approx 1.137\cdot T_{A}[/tex]

1. The period of pendulum A for small oscillations is  

[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]

2. The period of pendulum B for small oscillations.

[tex]T_B=1.137.T_A[/tex]

Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.

1) Pendulum A is a simple pendulum, whose period () is determined by the following formula:

[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]

Where:

l - Length of the massless bar.

g - Gravitational acceleration.

2) Pendulum B is a physical pendulum, whose period () is determined by the following formula:

[tex]T_A=2\pi\sqrt{\dfrac{I_o}{mgl}}[/tex] .............................2

Where:

m - Total mass of the pendulum.

g - Gravitational acceleration.

l - Length of the uniform bar.

Io- Moment of inertia of the pendulum with respect to its suspension axis.

The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:

[tex]I_o=\dfrac{1}{2}ml^2+\dfrac{1}{24}ml^2+\dfrac{3}{4}ml^2[/tex]

[tex]I_o=\dfrac{31}{24}ml^2[/tex]..................................3

By applying (3) in (2) we get the following expression:

[tex]T_B=2\pi\sqrt{\dfrac{\frac{31}{24}ml^2}{mgl}[/tex]

[tex]T_B=2\pi\sqrt{\dfrac{31l}{24g}}[/tex]

[tex]T_B=\sqrt{\dfrac{31}{24}}. (2\pi\sqrt{\dfrac{l}{g}})[/tex]

[tex]TB=1.137.T_A[/tex]

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Answer:

6 bits

Explanation:

The quality of digitized signal can be improved by reducing quantizing error. This is done by increasing the number of amplitude levels, thereby minimizing the difference between the levels and hence producing a smoother signal.

Also, Sampling frequently (also known as oversampling) can help in improving signal quality.

To get the number of bits, we use:

2ⁿ = amplitude level

where n is the number of bits.

Given an amplitude level of 64, hence:

2ⁿ = 64

2ⁿ = 2⁶

n = 6 bits

Answer:

(7.115 ; 13.405)

Explanation:

Given :

Sample size, n = 22

Mean bill, μ = 10.26

Standard deviation, s = 5.21

To obtain the 99% confidence interval for the mean bill of all orders ;

Mean ± margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 99%, df = n-1, 22 - 1 = 21

Tcritical = 2.831

Margin of Error = 2.831 * (5.21/√22) = 3.145

Confidence interval = 10.26 ± 3.145

Lower boundary = 10.26 - 3.145 = 7.115

Upper boundary = 10.26 + 3.145 = 13.405

Confidence interval :

(7.115 ; 13.405)

Answer:

B) 1500m/s

Explanation:

Ans is 1500m/s

Answer:

(a) Iron > plastic > ice

(b) Same on all

(c) Iron downwards, plastic net force zero, ice upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.

Explanation:

Three spheres have same volume , plastic, ice and iron.

(a) The weight is given by

Weight = mass x gravity = volume x density x gravity

As the density of iron is maximum and the density of ice is least so the order of the weight is

Weight of iron > weight of plastic > weight of ice

(b) Buoyant force is given by

Buoyant force = Volume immersed x density of fluid x g

As they have same volume, density of fluid is same so the buoyant force is same on all the spheres.

(c) Net force is

F = weight - buoyant force  

So, the net force on the iron sphere is downwards

On plastic sphere is zero as the density of plastic sphere is same as water. On ice sphere it is upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.  

Answer:

The correct answer is "6.96 rad/s".

Explanation:

The given values are:

Length,

L = 0.6 m

Mass,

m₁ = 0.5 kg

m₂ = 0.2 kg

Initial velocity,

V = 8 m/s

Now,

The final angular velocity will be:

⇒ [tex]\omega =\frac{6m_1V}{(4m_1+3m_2)L}[/tex]

By substituting the values, we get

⇒     [tex]=\frac{6\times 0.2\times 8}{(4\times 0.2+3\times 0.5)0.6}[/tex]

⇒     [tex]=\frac{9.6}{1.38}[/tex]

⇒     [tex]=6.96 \ rad/s[/tex]

Answer:

The value of x is 2.1 cm from the center of the coil.

Explanation:

Radius, R = 2.7 cm

Number of turns, N = 800

The magnetic field at the axis is half of the magnetic field at the center.

[tex]B_{axis}=\frac{B_{center}}{2}\\\\\frac{\mu o}{4\pi}\times \frac{2 \pi I N R^2}{\left (R^2 + x^2 \right )^{\frac{3}{2}}} = 0.5\frac{\mu o}{4\pi}\times\frac{2\pi N I}{R}\\\\\frac{R^2}{(R^2 + x^2)^\frac{3}{2}} = \frac{1}{2R}\\\\4R^6 = (R^2+x^2)^3\\\\1.6 R^2 = R^2 + x^2\\\\x^2 = 0.6 \times 2.7\times 2.7 \\\\x = 2.1 cm[/tex]

Answer:

g₂ = 11 m/s²

Explanation:

The value of free-fall acceleration on the surface of a planet is given by the following formula:

[tex]g = \frac{Gm}{r^2}[/tex]

where,

g = free-fall acceleration

G = Universal Gravitational Constant

m = mass of the planet

r = radius of planet

FOR PLANET 1:

[tex]g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2[/tex] --------------------- equation (1)

FOR PLANET 2:

[tex]g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\[/tex]

using equation (1):

[tex]g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}[/tex]

g₂ = 11 m/s²