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3 years ago

What do significant figures in a measurement include___________________________________.

Physics

1 answer:

yuradex [85]3 years ago

4 0

The significant figures in a measurement include:

All digits measured exactly,

One estimated digit.

<span>Significant figures can help us show the preciseness of a number.</span>

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If two planets orbit a star, but planet B is twice as far from the star as planet A, planet A will receive ____ times the flux t

Tresset [83]

Answer:

The nearest plant (A) receives 4 times more radiation from the farthest plant

Explanation:

The energy emitted by the star is distributed on the surface of a sphere, whereby intensity received is the power emitted between the area of the sphere

I = P / A

P = I A

The area of the sphere is

A = 4π r²

Since the amount of radiation emitted by the star is constant, we can write this expression for the position of the two planets

P = I₁ A₁ = I₂ A₂

I₁ / I₂ = A₂ / A₁

Suppose index 1 corresponds to the nearest planet,

r2 = 2 r₁

I₁ / I₂ = r₁² / r₂²

I₁ / I₂ = r₁² / (2r₁)²

I₁ / I₂ = ¼

4 I₁ = I₂

The nearest plant (A) receives 4 times more radiation from the farthest plant

7 0

3 years ago

a flowerpot falls from a windowsill 25m above the sidewalk how fast is the flowerpot moving when it strikes the ground

Mariana [72]

s = 25m

v = ?

a = 9.81m/s²

v² = u² + 2as

v² = 0 + 2x9.81x25

v² = 490.5

v = √490.5

= 22.15m/s

3 0

3 years ago

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

3 years ago

Read 2 more answers

A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic

xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2$

=2.5m

on the axis of x at x = 2.5

$r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}$

r = 2.5m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T$

(Along z direction)

B)r = 5.00m

Displacement vector is

$\bar{r}=x\hat i+y\hat j+z \hat k\\$

$\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2$

=5.00m

on the axis of x at x = 5.0

$r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}$

r = 5.00m

And unit vector

$\hat r =\frac{\bar{r}}{r}$

$= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\$

Therefore, the magnetic field is as follow

$d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}$

$d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T$

(Along x direction)

7 0

3 years ago

A 600 kg car is at rest, and then accelerates to 5 m/s.

Crank

Answer:

7500J

Explanation:

Given parameters:

Mass of car = 600kg

Velocity = 5m/s

Unknown:

Original kinetic energy = ?

Final kinetic energy = ?

Work used = ?

Solution:

The kinetic energy of a body is the energy due to the motion of a body.

It can be solve mathematically using expression below;

K.E = $\frac{1}{2}$ m v²

where m is mass

v is velocity

original kinetic energy;

The car started at rest and v = 0, therefore K.E = 0

Final kinetic energy;

K.E = $\frac{1}{2}$ x 600 x 5² = 7500J

Work done:

Work done = Final K.E - Initial K.E = 7500 - 0 = 7500J

3 0

3 years ago