Which is an example of data?<br> A. Variables<br> B. Theories<br> C. Measurements<br> D. Laws

A typical cell phone charger is rated to transfer a maximum of 1.0 Coulomb of charge per second. Calculate the maximum number of

Alexandra [31]

Answer:

The maximum no. of electrons- $2.25\times 10^{22}$

Solution:

As per the question:

Maximum rate of transfer of charge, I = 1.0 C/s

Time, t = 1.0 h = 3600 s

Rate of transfer of charge is current, I

Also,

$I = \frac{Q}{t}$

Q = ne

where

n = no. of electrons

Q = charge in coulomb

I = current

Thus

Q = It

Thus the charge flow in 1. 0 h:

$Q = 1.0\times 3600 = 3600\ C$

Maximum number of electrons, n is given by:

$n = \frac{Q}{e}$

where

e = charge on an electron = $1.6\times 10^{- 19}\ C$

Thus

$n = \frac{3600}{1.6\times 10^{- 19}} = 2.25\times 10^{22}$

3 0

4 years ago

Which object has more heat energy , a hot cup or a cold cup? why

NARA [144]

a hot cup because if a hot cup is hotter it's been in the sun too long

8 0

3 years ago

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A ship maneuvers to within 2.50 x 103 m of anisland's 1.80 x 103 m high mountain peak and fires aprojectile at an enemy ship 6.1

const2013 [10]

Answer:

Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m

Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m

Explanation:

Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.

H = (v₀² Sin²θ)/2g

v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s

θ = 75°, g = 9.8 m/s²

H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m

Range of projectile

R = v₀² (sin2θ)/g

R = 250² (sin2×75)/9.8

R = 250² (sin 150)/9.8 = 3188.8 m

Height of mountain = 1.80 × 10³ = 1800 m

Maximum height of projectile = 2975.2 m

Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m

Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m

Range of projectile = 3188.8 m

Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m

8 0

4 years ago

A. A light wave moves through glass (n=1.5) at an angle of 25°. What angle will it have when it moves from the glass into air (n

torisob [31]

PART A)

By Snell's law we know that

$n_1sin i = n_2 sin r$

here we know that

$n_1 = 1.5$

$i = 25 degree$

$n_2 = 1$

now from above equation we have

$1.5 sin25 = 1 sin r$

$r = 39.3 degree$

so it will refract by angle 39.3 degree

PART B)

Here as we can see that image formed on the other side of lens

So it is a real and inverted image

Also we can see that size of image is lesser than the size of object here

Here we can use concave mirror to form same type of real and inverted image

PART C)

As per the mirror formula we know that

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{d_i} + \frac{1}{60} = \frac{1}{20}$

$d_i = 30 cm$

so image will form at 30 cm from mirror

it is virtual image and smaller in size

3 0

4 years ago

Bill steps off a 3.0-m-high diving board and drops to the water below. At the same time, Ted jumps upward with a speed of 4.2 m/

Pani-rosa [81]

Answer:

equation of motion for Bill is

$y(t) = 4.9t^2$

equation of motion for Ted is

$y(t) = 2 + (-4.2)(t) + 4.9t^2$

Explanation:

Taking downward position positive and upward position negative

g = 9.8 m/s^2

equation of motion for Bill is

$y(t) = y_0 +v_0 t +\frac{1}{2}gt^2$

$y(t) = 0 + 0(t) +\frac{1}{2}gt^2$

$y(t) = \frac{1}{2}\times (9.8t)^2$

$y(t) = 4.9t^2$

equation of motion for Ted is

$y_0 = 2m -1m = 2m$

$y_0 = -4.2 m/s$

$y(t) = y_0 +v_0 t +\frac{1}{2}gt^2$

$y(t) = 2 + (-4.2)(t) +\frac{1}{2}gt^2$

$y(t) = 2 + (-4.2)(t) +\frac{1}{2}\times (9.8t)^2$

$y(t) = 2 + (-4.2)(t) + 4.9t^2$

8 0

4 years ago

Read 2 more answers

Which is an example of data? A. Variables B. Theories C. Measurements D. Laws