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2 years ago

What is 0.40 as a fraction?

2 answers:

8 0

Please remember to post in the right category next time :)

.40 as a fraction is 4/10 or 2/5
the 4 is in the tenths place, so put it over a 10 to make it a fraction.
then, simplify the fraction by dividing the numerator and denominator by 2, and you get 2/5

eimsori [14]2 years ago

3 0

It is 4/10 or 2/5.
0.40 or 0.4 is the same as 4/10
the simplified fraction is 2/5

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8. Two forces of 10 N and 30 N are applied to a 10 kg box. Find (1) the box’s acceleration when both forces point due east and (

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(1) acceleration, a = 4 m/ $s^{2}$ (2) acceleration of 10 N, $a_{1}$ = 1 m/ $s^{2}$ and acceleration of 30 N, $a_{2}$ = 3 m/ $s^{2}$

Explanation:

Here, the acceleration of the object could be found using the equation derived in the second law of motion. The equation is given as, F = ma where m is the acceleration of the object, m is the mass of the object and F is the applied on the object.
Let $a_{1}$ be the acceleration for force 10 N, to find acceleration rearrange the equation to a = $\frac{F}{m}$ . When we substitute 10 N force and 10 kg mass of the box in the equation. We will get $a_{1}$ = 1 m/ $s^{2}$

Let $a_{2}$ be the acceleration for force 30 N, to find acceleration rearrange the equation to F = $\frac{F}{m}$ . When we substitute 30 N force and 10 kg mass of the box in the equation. We will get $a_{2}$ = 3 m/ $s^{2}$
To find the combined, just add the force and substitute in the above equation. Hence, a = 4 m/ $s^{2}$

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2 years ago

A physics professor is pushed up a ramp inclined upward at an angle 33.0Â° above the horizontal as he sits in his desk chair tha

Andru [333]

Answer:

2.51 m/s

Explanation:

Parameters given:

Angle, A = 33°

Mass, m = 90kg

Inclined distance, D = 2m

Force, F = 600N

Initial speed, u = 2.3m/s

From the relationship between work and kinetic energy, we know that:

Work done = change in kinetic energy

W = 0.5m(v² - u²)

We also know that work done is tẹ product of force and distance, hence, net work done will be the sum of the total work done by the force from the students and gravity.

Hence,

W = F*D*cosA - w*D*sinA

w = m*9.8 = weight

=> W = 600*2*cos33 - 90*9.8*2*sin33

W = 45.7J

=> 45.7 = 0.5*m*(v² - u²)

45.7 = 0.5*90*(v² - 2.3²)

45.7 = 45(v² - 5.29)

=> v² - 5.29 = 1.016

v² = 6.306

v = 2.51 m/s

The final velocity is 2.51 m/s

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3 years ago

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What type of system is best used to observe conservation of mass because all of the mass stays in one place?

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Answer:

trump

Explanation:

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2 years ago

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When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera

Phantasy [73]

Answer:

253.54 nm

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

$\lambda$ = Wavelength

V = Voltage = 4.9 V

The loss in kinetic energy = Gain in potential energy (energy conservation)

$E=eV\\\Rightarrow E=1.6\times 10^{-19}\times 4.9\\\Rightarrow E=7.84\times 10^{-19}\ J$

Energy is given by

$E=\dfrac{hc}{\lambda}\\\Rightarrow \lambda=\dfrac{hc}{E}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^{8}}{7.84\times 10^{-19}}\\\Rightarrow \lambda=2.53546\times 10^{-7}=253.54\ nm$

The wavelength is 253.54 nm

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3 years ago

The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in 1851. Consider a l

Ket [755]

It is proved that when v<<c , then speed of the light measured in the laboratory frame is , u = c/n + v -v/n^2 .

Given ,

The motion of a transparent medium influences the speed of light .

The water moves with speed v in a horizontal pipe .

Assume that the light travels in the same direction as the water moves .

The speed of the light with respect to the water is c/n

Where n = 1.33 is the refractive index of water .

Let us assume ,

u' be the speed of light in water , in the frame moving with the water .

u' is related to the refractive index of water ,n as :

u'=c/n

where , c is the speed of light .

let , u be the speed of light in water in the lab frame .

Now , u and u' are related as : u = (u'+ v )/(1+ u'v/c^2)

Here v is the speed of water in the horizontal pipe .

we know the value of u' , so by substituting the value , we will get ,

u= (c/n+ v)/(1+cv/nc^2)

u= c/n(1+ nv/c)/(1+v/nc)

(b) We have , v<<c

v/c<<1 .

so , (1+v/nc )^-1 = (1-v/nc)

Now substituting this , we will get ,

u = c/n(1+nv/c) (1-v/nc)

u≈c/n(1+ nv/c-v/cn)

u≈c/n + v - v/n^2

Hence , it is proved that when v<<c , then speed of the light measured in the laboratory frame is , u = c/n + v -v/n^2 .

Learn more about speed here :

brainly.com/question/13943409

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Disclaimer : incomplete question , here is the complete question .

Question: The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in1851. Consider a light beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n=1.33 is the index of refraction of water.(a) Use the velocity transformation equation to show that the speed of the light measured in the laboratory frame isu = c/n (1 + nv/c / 1+ v/nc) . (b) show that for v<<c , the expression from part (a) becomes , to a good approximation , u ≈ c/n + v - v/n^2 .

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1 year ago