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Cloud [144]
3 years ago
10

The number 0.00325 × 10-8 cm can be expressed in millimeters as A) 3.25 × 10-11 mm. B) 3.25 × 10-10 mm. C) 3.25 × 10-12 mm. D) 3

.25 × 10-9 mm.
Physics
1 answer:
krok68 [10]3 years ago
6 0

Answer:

Option B. 3.25×10¯¹⁰ mm.

Explanation:

Measurement (cm) = 0.00325×10⁻⁸ cm

Measurement (mm) =?

The measurement in mm can be obtained as follow:

1 cm = 10 mm

Therefore,

0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm

0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm

Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.

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c) those caused by parasites

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m
erastovalidia [21]

Answer:

a)  W = 46.8 J  and b)   v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

    W = ΔK = k_{f} -K₀

a) work is the scalar product of force by distance

    W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

    W = F d cos θ

    W = 39.0 1.20 cos 0

    W = 46.8 J

b) zero initial kinetic language because the package is stopped

    W -W_{fr} = k_{f} -K₀

    W - fr d= ½ m v² - 0

    W - μ N d = ½ m v

   on the horizontal surface using Newton's second law

     N-W = 0

     N = W = mg

 

     W - μ mg d = ½ m v

    v² = (W -μ mg d) 2/m  

    v = √(W -μ mg d) 2/m

    v = √[(46.8 -  0.30 4.30 9.8 1.20) 2/4.3 ]

    v = √(31.63 2/4.3)

    v = 3.84 m/s

8 0
3 years ago
A 2.7-kg ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 45.0 m high cliff. At the instant the ball i
scoray [572]

Answer:

The distance traveled by the woman is 34.1m

Explanation:

Given

The initial height of the cliff

yo = 45m final, positition y = 0m bottom of the cliff

y = yo + ut -1/2gt²

u = 20.0m/s initial speed

g = 9.80m/s²

0 = 45.0 + 20×t –1/2×9.8×t²

0 = 45 +20t –4.9t²

Solving quadratically or by using a calculator,

t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s

So this is the total time it takes for the ball to reach the ground from the height it was thrown.

The distance traveled by the woman is

s = vt

Given the speed of the woman v = 6.00m/s

Therefore

s = 6.00×5.69 = 34.14m

Approximately 34.1m to 3 significant figures.

3 0
3 years ago
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