Which should not be a part of scientific inquiry?

den301095 [7]

Answer:

O a force that opposes motion

5 0

3 years ago

An electron is released from rest in a weak electric field given by E =-2.30 x 10-10 N/Cj. After the electron has traveled a ver

Goshia [24]

Explanation:

It is given that,

An electron is released from rest in a weak electric field of, $E=2.3\times 10^{-10}\ N/C$

Vertical distance covered, $s=1\ \mu m=10^{-6}\ m$

We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

$v^2-u^2=2as$

$v^2=2as$ .............(1)

Electric force is $F_e$ and force of gravity is $F_g$ . As both forces are acting in downward direction. So, total force is:

$F=mg+qE$

$F=9.1\times 10^{-31}\times 9.8+1.6\times 10^{-19}\times 2.3\times 10^{-10}$

$F=4.57\times 10^{-29}\ N$

Acceleration of the electron, $a=\dfrac{F}{m}$

$a=\dfrac{4.57\times 10^{-29}\ N}{9.1\times 10^{-31}\ kg}$

$a=50.21\ m/s^2$

Put the value of a in equation (1) as :

$v=\sqrt{2as}$

$v=\sqrt{2\times 50.21\times 10^{-6}}$

v = 0.010 m/s

So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.

6 0

3 years ago

You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity le

BigorU [14]

Answer:

1) the new power coming from the amplifier is 19.02 W

2) The distance away from the amplifier now is 5.50 m

3) u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

Explanation:

Lets say that I am at a distance "u" from the TV,

Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB

SO

S(indB) = 10log (I₁/1₀)

we substitute

125 = 10(I₁/10⁻¹²)

12.5 = log (I₁/10⁻¹²)

10^12.5 = I₁/10^-12

I₁ = 10^12.5 × 10^-12

I₁ = 10^0.5 W/m²

Now I₂ will be intensity of sound when corresponding sound level is 107 dB

107 = 10log(I₂/10⁻²)

10.7 = log(I₂/10⁻¹²)

10^10.7 = I₂ / 10^-12

I₂ = 10^10.7 × 10^-12

I₂ = 10^-1.3 W/m²

Now since we know that

I = P/4πu² ⇒ p = 4πu²I

THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂

Therefore

P₁/P₂ = I₁/I₂

WE substitute

P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)

P₂ = 19.02 W

the new power coming from the amplifier is 19.02 W

2)

P₁ = 4πu²I₁

u =√(p₁/4πI₁)

u = √(1200/4π × 10^0.5)

u = 5.50 m

The distance away from the amplifier now is 5.50 m

3)

Let I₃ be the intensity corresponding to required sound level 85 dB

85 = 10log(I₃/10⁻¹²)

8.5 = log (I₃/10⁻¹²)

10^8.5 = I₃ / 10^-12

I₃ = 10^8.5 × 10^-12

I₃ = 10^-3.5 w/m²

Now, I ∝ 1/u²

so I₂/I₃ = u₁²/u²

u₁ = √(I₂/I₃) × u

u₁ = √(10^-1.3 / 10^-3.5) × 5.50

u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

8 0

3 years ago

Which of the following instrument measure time most accurately

valkas [14]

C. Quarts watches is the correct answer

8 0

3 years ago

If a train is travelling 200km/hour eastward for 1800 seconds how far does it travel?

Olenka [21]

Answer:

Distance, d = 99990 meters

Explanation:

It is given that,

Speed of the train, v = 200 km/h = 55.55 m/s

Time taken, t = 1800 s

Let d is the distance covered by the train. We know that the speed of an object is given by total distance covered divided by total time taken. Mathematically, it is given by :

$v=\dfrac{d}{t}$

$d=v\times t$

$d=55.55\times 1800$

d = 99990 m

So, the distance covered by the train is 99990 meters. Hence, this is the required solution.

5 0

3 years ago