"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a function of time for a sudden temperature increase from 273 K to 473 K. Before the temperature increase, the instrument output was a steady 109.2 mV. Plot the response y(t) as a function of time. What are the units for y(t)? Find the 90% rise time for y(t90) and the error fraction, Γ(t90).

Answer:

10 seconds

Explanation:

because the cat is moving one m/s slower than the dog, the dog has a relative speed of 1 m/s. 10 meters would take 10 seconds for the dog to cover

Answer:

b. varies inversely with the square of the distance from the center of Earth.

Explanation:

Comparing the Newton's law of universal gravitation and second law of motion;

from Newton's second law of motion,

F = ma ............. 1

from New ton's law of universal gravitation,

F = [tex]\frac{GMm}{r^{2} }[/tex] ........... 2

Equating 1 and 2, we have;

mg =  [tex]\frac{GMm}{r^{2} }[/tex]

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

Therefore, the acceleration due to gravity near Earth, g, is inversely proportional to the square of the distance from the center of Earth.

this process is called parellelogram method of resolving vectors.

Answer:

0.25 m/s

Explanation:

Average velocity = change in displacement / change in time

v = (1.5 m − 2.5 m) / (6 s − 2 s)

v = -0.25 m/s

The average speed is 0.25 m/s.

Answer:

0.25 m/s

Explanation:

khan academy

Answer:

PFA

:-)

Explanation:

Answer:

5.3333 sec

Explanation:

initial speed: u = 108km/hr or 30 m/s

final speed: v = 0m/s

distance travelled: s = 80m

time the car took to stop: = t sec

[tex]v^{2} - u^{2}[/tex] = 2as,

a = ([tex]v^{2} - u^{2}[/tex])/2s

a = (0-900)/160

a = -5.625 [tex]ms^{-2}[/tex]

v = u + at,

t = (v - u)/a

t= (0 - 30)/(-5.625)

t = 5.3333 sec

Answer:

10N

Explanation:

Applying the Hooke law:

F = kx

F: Force

k: stiffness coefficient

x: stretched distance

F = 10N/m x 1m = 10N

Answer:

3504 m/s

Explanation:

Let x be the horizontal component of distance

y - vertical component of distance

t-time

ax- horizontal component of acceleration

ay-Vertical component of acceleration

Vx-horizontal component of velocity

Vy-Vertical component of velocity

horizontally: x = V_x ×t + ½×a_x×t²  

plugging the values we get

23500× cos 55º = 1350×cos25.0º × 10.20 + ½×a_x× (10.20)²  

⇒ax = 19.2 m/s²  

Moreover,

V'x = V_x + a_x×t = 1350×cos25.0º + 19.2×10.20= 1419 m/s  

similarly in vertical direction:

y = V_y×t + ½×a_y×t²  

23500×sin55º = 1350×sin25.0º×10.20s + ½×a_y×(10.20)²  

⇒a_y = 258 m/s²  

Also,

V'y = V_y + a_y×t = 1350×sin25.0º + 258×10.20 = 3204 m/s  

Therefore

V = √(V'x² + V'y²) = 3504 m/s  

therefore,  magnitude of final velocity of missile=3504 m/s

THANKS  

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Brainliest

Answer:

c

Explanation:

The marginal cost curve image  has been attached from which we can clearly, indicate that

ATC = average total cost

AFC = average fixed cost

AVC = average variable cost.

From the graph we can indicate that the marginal cost curve

(c) Lies above the AVC curve when the AVC curve slopes downward.

Answer:

The correct option is;

3. The total mechanical energy is constant as the block moves back and forth

Explanation:

The total mechanical energy is the sum of the potential and kinetic energies of the system

For a system that is isolated from the effects of external forces, but being acted upon by the internal conservative forces within the system, the total mechanical energy is constant

For a black and spring system, we have total mechanical energy, E = 1/2×K×A².

Where;

K = Constant

A = The amplitude of motion

Therefore, where there is no loss to friction, with A, remaining constant, the total mechanical energy will be constant.

Explanation:

To find the average of these numbers, we just have to add the three numbers together and divide by 3.

Answer:1m

Explanation:

2n=0.4m

5n=?

5n×0.4/2n=1m

Answer:

66.86m

Explanation:

Velocity of ball thrown, u = 23.5 m/s

Initial height of the ball above the water, H = 11.5 m

Angle of projection, θ = 39.5°

Vertical components of veloclty = usinθ

Horizontal components of veloclty = ucosθ

The soft ball hits the water after time 't'

Considering the second equation of motion

S = ut + 1/2at^2........ 1

But since the ball went through motion under gravity ( free fall ) rather than linear motion, then equation 1 can be rewritten as:

H = ut +/- 1/2gt^2

H = - 11.5m

U = usinθ

θ = 39.5°

a = -g = -9.8m/s^2

- 11.5m = 23.5(sin39.5°)t + 1/2(-9.8)t^2

-11.5m = 23.5(0.6360)t - 4.9t^2

-11.5m = 14.946t - 4.9t^2

4.9t^2 -14.946t-11.5m = 0

Since the ball drifted horizontally

D = (Ucosθ)t

Where θ = 39.5°

U = 23.5m/s t=

Alternatively,

horizontal component of the velocity is 23.5 cos 39.5º = 18.1331 m/s

now how long does it take the ball to raise to a peak and fall to the water.

vertical component of velocity = 23.5 sin 39.5º = 14.947m/s

time to reach peak t = v/g = 11.947/9.8 = 1.5252 sec

peak reached above cliff top is

h = ½gt² = ½(9.8)(1.5252)²

= ½×22.797

= 11.3985m

now the ball has to fall 11.3985+ 11.5 = 22.8985m

time to fall from that height is

t = √(2h/g) = √(2• 22.8986/9.8) = 2.1617 sec

add up the two times to get time it is in the air, 2.1617 + 1.5252 = 3.6869

now haw far does the ball travel horizontally in that time

d = vt = 18.1331 ×3.6869= 66.856m

= 66.86m

Answer:

The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.

Answer:

First, as you may know, the light travels at a given velocity.

In vaccum, this velocity is c = 3x10^8 m/s.

And we know that:

distance = velocity*time

Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.

This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)

Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.

The astronomers could know what was happening inside galaxies way back then by the fact that;

they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find

Astronomers Measure the wavelength of the light that is stretched, so the light is seen as 'shifted' towards the red part of the spectrum by using spectroscopy. This measure is also called redshift.

This invokes a ray of light through a triangular prism that splits the light into various components known as spectrum.

The way the astronomers could use this concept to know what was happening in the galaxies before is by examining the spectra of galaxies that have the highest redshifts.

Read more at; https://brainly.com/question/15995216

Answer:

The correct option is energy levels

Explanation:

Rutherford's model of an atom suggests that an atom has a tiny positively charged central mass (now called the nucleus) which is surrounded by electrons (negatively charged) in a cloud-like manner.

Bohr's model went a bit further than the Rutherford's model in describing an atom by suggesting that the electrons which surrounds in the nucleus travel in fixed circular orbits. This description by Bohr was able to describe the energy levels of orbitals which assumes that smallest orbitals have the lowest energy while the largest orbitals have the highest energy.

Answer:

energy levels

hope this helped!

Explanation:

Answer:

585×10⁸ m

Explanation:

Distance = rate × time

d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)

d = 585×10⁸ m

Answer; 4.4cm

Explanation: There is a ruler upon which the cylinder start from 1.4cm and reaches 10.2cm

distance traveled =10.2-1.4=8.8

since this cylinder is small so the linear distance can be approximately taked as rotational distance(as in case of point charge) so

2x2πxr =8.8

so the circumference will be 2πr=4.4cm

Answer: A. 2.7 x 10^2 joules

Explanation: I’m sorry for the guy above me!

Answer:

They are:

1) change position

2) distract yourself

3) Get fresh air

4) Face the direction you are going.

5) Drink water.

6) Play music.

7) Put your eyes on horizon.  

Explanation:

Hope it helps.

Answer:

The nail exerts an equal force on the hammer in the opposite direction.

Explanation:

The Newtons third law states that there is an equal an opposite reaction for every action. When hammer pushes the nail, the nail will push the hammer back in opposite direction. When the hammer hits a nail then nail will exert the equal and opposite force to the hammer. These both objects will exert force on each other in opposite directions.

Hello, I am BrotherEye

Answer:

Answers are

1. "The nail exerts an equal force on the hammer in the opposite direction."

2. "500 N"

3. "The iron piece exerts a force of 1 N on the magnet in the opposite direction."

4. "When mass moves closer to the point of rotation, rotational inertia decreases."

5. "The skater spins slower because his rotational inertia has increased."

Explanation:

Answer:

E= 600 W

Explanation:

Given that

m = 2 kg

Speed ,  v= 10 m/s

Force , F= 60 N

Given that box is moving with constant velocity, it means that friction force will be 60 N.

f = 60 N

Therefore total energy generated

E= f x v

E= 60 x 10 = 600 W

E= 600 W

Thus the answer will be 600 W.

Answer:

Explanation:

The velocity ratio of  a wheel and axle is the ratio of the radius (R) of the wheel to the radius (r) of the axle. It is expressed as;

VR = R/r

Since radius = diameter/2

VR = (D/2)/(d/2)

VR = D/d

D is the diameter of the wheel and 'd' is the diameter of the axle.

Given VR = 3 and d = 5cm

3 = D/5

D = 15 cm

If the diameter of the wheel is 15cm, the radius of the wheel will be 15/2 = 7.5cm.

b) Workdone by the load = Load * distance moved by load

Given load = 60kg

Distance moved by load = 2π*radius of axle

Distance moved by load  = 2π(0.025) = 0.157

workdone by load = 60* 0.157 = 9.42J

Effort = Workdone by load/distance moved by the wheel

Effort = 9.42/2π(0.075)

Effort = 9.42/0.471

Effort = 20kg

Hence the effort applied is 20kg

c) MA = Load/Effort

MA = 60/20

MA = 3

d) Efficiency = MA/VR * 100%

Efficiency = 3/3 * 100%

Efficiency = 100%

Answer:

Rounded to three significant figures:

(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].

(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].

Explanation:

Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.  

Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:

The path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].

The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].

Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].

The dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.

Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.

For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].

Answer:

The distance the car travels is 115500 m in S.I units

Explanation:

Distance d = vt where v = speed of the car and t = time taken to travel

Now v = 99 km/h. We now convert it to S.I units. So

v = 99 km/h = 99 × 1000 m/(1 × 3600 s)

v = 99000 m/3600 s

v = 27.5 m/s

The speed of the car is 27.5 m/s in S.I units

We now convert the time t = 70 minutes to seconds by multiplying it by 60.

So, t = 70 min = 70 × 60 s = 4200 s

The time taken to travel is 4200 s in S.I units

Now the distance, d = vt

d = 27.5 m/s × 4200 s

d = 115500 m

So, the distance the car travels is 115500 m in S.I units

Answer:

B.

Explanation:

as there is a '-' sign before the magnification value, it forms a real image.

so the last two options get cancelled.

as the value of magnification is 0.87 ie. lesser than 1, we can say that the image is smaller.

The type of image is formed by a lens if m = -0.87 would be smaller than the object and the object is real

A lens is a transmissive optical tool that employs refraction to focus or disperse a light beam. The power of the lens is expressed in the dioptre which is the reciprocal of the focal length of the lens.

The image formation through the lens is calculated with the help of the lens formula given as follows

1/f = 1/v - 1/u

where f is the focal length of the lens

u is the distance of the object from the lens

v is the distance of the image formed from the

magnification (m) = size of image /size of the object

As given in the problem image is formed by a lens if m = -0.87

Thus, the negative sign represents that the image formed by the object is real, and the magnitude of the magnification is less than 1 this represents that size of the image is smaller than the size of the object.

Learn more about the lens from here

https://brainly.com/question/766997

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Derived Quantities

Answer:

These are quantities calculated from two or more measurements

Explanation:

They can't me measured directly.

They can only be computed.

They are calculated in PHYSICAL SCIENCE.

hope it helps.

Answer:

Explanation:

Energy of gamma ray = 2.2233 MeV

Let mass of neutron be n  amu  

mass defect of deuteron = 2.013553212 - ( 1.007 276 467 + n ) u .

in terms of energy this mass defect will be equal to energy of gamma ray

1 amu = 931 MeV

931 [ 2.013553212 - ( 1.007 276 467 + n ) ] =  2.2233

( 1.007 276 467 + n ) -  2.013553212  = .00238807733

n = 1.008664822 amu

so mass of neutron = 1.008664822 amu

Answer:

12 i. The work done by Wale = 107.910 kJ

The work done by Lekan = 117.720 kJ

Total work done = 225.36 kJ

ii. Wale's power =  4.3164 kW

Lekan's power = 3.924 kW

Wale has more power and is more powerful than Lekan

13. 313.92 N

Explanation:

i. The work done, W = Force, F × Distance moved by the force, D

The given parameters are

The mass of Wale = 55 kg

The mass of Lekan = 60 kg

The acceleration due to gravity, g =9.81 m/s²

The motion force of Wale and Lekan are;

Motion force of Wale = 9.81 × 55 = 539.55 N

Motion force of Lekan = 9.81 × 60 = 588.6 N

The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ

The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ

107910 + 117720 =225630 J = 225.36 kJ

ii. Power = Work done/time

Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W

Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W

Wale has more power and is more powerful than Lekan

13. The velocity ratio = 5

V. R. = Distance moved by effort/(Distance moved by load)

Efficiency = 80%

Work done by effort = x

Work done by machine = Efficiency × Work done by effort  = 0.8 × x

Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D

Work done by effort = Force × Distance moved = 200×9.81× E

Work done by effort = 1962×E = 1962×E = 1962×5×D

Work done by machine = 1962 × D, when D = 1, we have;

0.8 × 1962×1 = 1569.6 J

Work done by effort = Force × Distance moved

Work done by effort = Force × 5×D = Force × 5 (D = 1)

From the principle of conservation of energy, we have;

Energy is neither created nor destroyed

Therefore

Work done by effort = Force × 5 = 1569.6 J

Force = 1569.6 /5 = 313.92 N.

Explanation:

It is given that,

The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:

[tex]x=4t^3+3t^2-5t+2[/tex]

Where,

x is in meters and t is in sec

We know that,

Velocity,

[tex]v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5[/tex]

(a) i. t = 2 s

[tex]v=12(2)^2+6(2)-5=55\ m/s[/tex]

At t = 4 s

[tex]v=12(4)^2+6(4)-5=211\ m/s[/tex]

(b) Acceleration,

[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6[/tex]

Pu t = 3 s in above equation

So,

[tex]a=24(3)+6\\\\a=78\ m/s^2[/tex]

Hence, (a) (i) v = 55 m/s (ii) v = 211 m/s and (b) 78 m/s²