A theory is a way of explaining observations. true or false.

Which of the following might be measured in

gulaghasi [49]

Answer:

The correct answer to this question is D. The area of a desk top.

Explanation:

Square meters, or m^2, are the units that represent area, or the product of two lengths. This is because the length and the width both have the units meters, so when multiplied to calculate area, the resulting units are square meters or m2. Thus, the answer for this question is D.

The length of a pendulum string would be measured in meters.

The volume of an atom would be measured in m^3 (volume is the product of the length, width, and height, giving it a cubic power). However, the volume would probably be measured in a much smaller unit because atoms are very small.

The mass of a book would likely be measured in grams or kilograms.

4 0

4 years ago

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One degree Celsius indicates the same temperature change as?

VARVARA [1.3K]

If you've ever seen the formula from Celsius to Fahrenheit...

$C=\frac59(F-32)$

As you can see, if you put in some number of Celsuis C your answer is going to be multiplied by 5/9...meaning that 1°C = 5/9°F.

As for Kelvin, it's literally just Celsuis, but about 273.15° more.

B. 5/9°F

6 0

3 years ago

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6.0 m, y = 10.0 m, and has vel

mihalych1998 [28]

Explanation:

Given that,

Position of the particle at t = 0,

$y=(6i+10j)\ m$

Velocity of the particle at t = 0

$u=(1i+6j)\ m$

Acceleration of the particle,

$a=(5i+7j)\ m/s^2$

Solution,

(a) Let v is the velocity at t = 10 s. Using the equation of kinematics as :

$v=u+at$

$v=(1i+6j)+(5i+7j)10$

$v=(51i+76j)\ m/s$

(b) Let y' is the position at t = 1 s. Again using second equation of kinematics as :

$y'=y+ut+\dfrac{1}{2}at^2$

$y'=(6i+10j)+(1i+6j)1+\dfrac{1}{2}\times (5i+7j)1^2$

$y'=\dfrac{19}{2}i+\dfrac{39}{2}j$

(c) Magnitude of y',

$|y'|=\sqrt{(\dfrac{19}{2})^2+(\dfrac{39}{2})^2}$

|y'| = 21.69 meters

Direction of the y',

$tan\theta=\dfrac{y}{x}$

$tan\theta=\dfrac{39/2}{19/2}$

$\theta=64.02^{\circ}$

Hence, this is the required solution.

4 0

3 years ago

Electromagnetic energy travels inselect one:<br> a. long stringsb. short stringsc. burstsd. waves

aleksandrvk [35]

I think it's waves.
electromagnetic waves

6 0

4 years ago

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How can I solve the following?

skad [1K]

Answer:

Part A - 4.084 mJ

Part B - 0.908 mJ

Part C - 8.168 mJ

Explanation:

Part A: How much energy is stored in the capacitor network as shown in (Figure 1)?

Since capacitors C₂ and C₃ are in series, their equivalent capacitance is C',

1/C' = 1/C₂ + 1/C₃ (Since C₁ = C₂ = C₃ = C)

1/C' = 1/C + 1/C

1/C' = 2/C

C' = C/2

Since C' is in parallel with C₁, the equivalent capacitance for the circuit is C" = C₁ + C' = C + C/2 = 3C/2

C" = 3C/2

The energy stored in the circuit, W = 1/2C"V² where C" = equivalent capacitance = 3C/2 and V = voltage = 15.0 V

W = 1/2C"V²

W = 1/2(3C/2)V²

W = 3CV²/4

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/4

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/4

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/4

W = 16335/4 × 10⁻⁶ FV²

W = 4083.75 × 10⁻⁶ J

W = 4.08375 × 10⁻³ J

W = 4.08375 mJ

W ≅ 4.084 mJ

Part B: How much energy would be stored in the capacitor network if the capacitors were all in series?

If the capacitors are connected in series, their equivalent resistance is C'

and 1/C' = 1/C₁ + 1/C₂ + 1/C₃

Since C₁ = C₂ = C₃ = C

1/C' = 1/C + 1/C + 1/C

1/C' = 3/C

C' = C/3

The energy stored in the circuit, W = 1/2C'V² where C' = equivalent capacitance = C/3 and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(C/3)V²

W = CV²/6

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = CV²/6

W = 24.2 × 10⁻⁶ F (15.0 V)²/6

W = 24.2 × 10⁻⁶ F × 225 V²/6

W = 5445/6 × 10⁻⁶ FV²

W = 907.5 × 10⁻⁶ J

W = 0.9075 × 10⁻³ J

W = 0.9075 mJ

W ≅ 0.908 mJ

Part C: How much energy would be stored in the capacitor network if the capacitors were all in parallel?

If the capacitors are connected in parallel, their equivalent resistance is C'

and C' = C₁ + C₂ + C₃

Since C₁ = C₂ = C₃ = C

C' = C + C + C

C' = 3C

The energy stored in the capacitor network, W = 1/2C'V² where C' = equivalent capacitance = 3C and V = voltage = 15.0 V

W = 1/2C'V²

W = 1/2(3C)V²

W = 3CV²/2

since C = 24.2 μF = 24.2 × 10⁻⁶ F

W = 3CV²/2

W = 3 × 24.2 × 10⁻⁶ F (15.0 V)²/2

W = 3 × 24.2 × 10⁻⁶ F × 225 V²/2

W = 16335/2 × 10⁻⁶ FV²

W = 8167.5 × 10⁻⁶ J

W = 8.1675 × 10⁻³ J

W = 8.1675 mJ

W ≅ 8.168 mJ

5 0

3 years ago