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Semenov [28]
3 years ago
10

Bob wants to measure the length of the red rectangle above. The ruler has 1 cm tick marks. Bob wants other scientists to have a

sense of the likely size of measurement errors. (The uncertainty.) Which reading makes the most sense for him to report? *
Physics
1 answer:
klasskru [66]3 years ago
8 0

Answer:

1 cm ± 0.05 cm

Explanation:

A ruler is readable to ±0.05 cm. This implies that any measurement taken using the ruler could be uncertain by 0.05 cm above or below the true value.

Hence, ±0.05 cm  is called the uncertainty or the precision of the ruler.

We obtained this from the fact that the meter rule is graduated in units of centimetres (cm). This implies that the smallest scale division is 1 mm. Thus, the uncertainty of the meter rule is given by; Δx = smallest increment/2 = 1mm/2 = 0.5mm = 0.05cm.

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After a surface interval my pressure group is D. I want to do a repetitive dive to 38 feet for 32 minutes. What will be my new p
ExtremeBDS [4]

Answer:

New pressure group is N

Explanation:

The Repetitive Dive table has to be checked, the residual nitrogen for the maximum depth of 38 feet as a "D" diver, then add that to the actual dive time to get the total nitrogen time. The total nitrogen time and the maximum depth of 38 feet would put you on a pressure group N.

5 0
3 years ago
Read 2 more answers
Which statements accurately describe mechanical waves? Check all that apply.
kotegsom [21]

Answer:

A

Explanation:

E. An ocean wave moving through water is an example of a mechanical wave

e.g sound waves wave on a rope or string

and Ans a is also correct

4 0
2 years ago
To reduce inductive reactance, what devices are normally placed on transmission or distribution lines?
UkoKoshka [18]

Answer : Capacitors

Explanation : Capacitors are normally  placed on transmission or distribution lines when to reduce inductive reactance.

This is because it enhances electromechanical and voltage stability , limit voltage dips at network nodes and reduces the power loss.

So, we can say that inductive reactance normally replace by the capacitors.



4 0
3 years ago
The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad
kenny6666 [7]

Answer:

a)  v = 2,9992 10⁸ m / s , b)  Eo = 375 V / m ,  B = 1.25 10⁻⁶ T,

c)     λ = 3,157 10⁻⁷ m,   f = 9.50 10¹⁴ Hz ,  T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

        E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

        v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

        k = 2π / λ

        λ = 2π / k

        λ = 2π / 1.99 10⁷

        λ = 3,157 10⁻⁷ m

        w = 2π f

        f = w / 2 π

        f = 5.97 10¹⁵ / 2π

        f = 9.50 10¹⁴ Hz

the wave speed is

        v = 3,157 10⁻⁷   9.50 10¹⁴

        v = 2,9992 10⁸ m / s

b) The electric field is

           Eo = 375 V / m

to find the magnetic field we use

           E / B = c

           B = E / c

            B = 375 / 2,9992 10⁸

            B = 1.25 10⁻⁶ T

c) The period is

           T = 1 / f

            T = 1 / 9.50 10¹⁴

            T = 1.05 10⁻¹⁵ s

the wavelength value is

          λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

this wavelength corresponds to the ultraviolet

5 0
4 years ago
Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot
Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² =  2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ =  m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁  = 4.427 - 2v₂  ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂  = 8.854

v₂ = 8.854 / 3

v₂  = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁  = - 1.477 m/s

v₁  = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

mgh_{max} = \frac{1}{2}mv_{max}^2

(a) The maximum height achieved by the first ball (v₁  = 1.477 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max}  = 0.11 \ m

(b) The maximum height achieved by the second ball (v₂  = 2.95 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max}  = 0.44 \ m

6 0
3 years ago
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