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3 years ago

15

A 7.1 resistor and a 4.8 resistor are connected in series with a battery. the potential difference across the 4.8 resistor is me

asured as 12 v. find the potential difference across the battery. v

1 answer:

Taya2010 [7]3 years ago

4 0

For resistors in series in a DC circuit, the potential difference is proportional to the resistance.

The voltage V across the 7.1 ohm resistor can be found from the proportion
V/7.1 = 12/4.8
V = 7.1*12/4.8 = 17.75

The potential difference at the battery terminals is the sum of the potential differences across the two series resistors.
Vbatt = 17.75 V + 12 V
Vbatt = 29.75 V

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How many signifficant figures do the following number have 258

Romashka [77]

Answer:

<u>Facts about 258</u>

<u>Sig Figs</u>

3

<u>258</u>

<u>Decimals</u>

0

<u>Scientific Notation</u>

2.58 × 102

<u>E-Notation</u>

2.58e+2

<u>Words</u>

two hundred fifty-eight

Explanation:

258 Rounded to Fewer Sig Figs

2 260 2.6 × 102

1 300 3 × 102

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2 years ago

Suppose we have a space ship orbiting the earth. What must happen in order for the spaceship to leave orbit and fall back toward

sdas [7]

Answer:

#2) The spaceship's forward motion must be slowed down so the earth's gravitational pull on it will be stronger than the ship's forward motion.

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2 years ago

Read 2 more answers

A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc

liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s, a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

v₀ₓ = 10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

v_y = $v_{oy}$ - g t

$v_{y}^2 = v_{oy}^2 - 2 g (y - y_o)$

at the highest point of the trajectory the vertical speed must be zero

v_y = 0

therefore the velocity of the body is

v = (10 i ^ + 0j ^) m / s

the acceleration is

a = (0 i ^ - g j⁾

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2 years ago

Calculate the wave speed i the wavelength =64m and the wave period = 4s

Marina86 [1]

To do this you would take 64 and divide it by 4.
64/4= 16.
Your answer is 16.

8 0

3 years ago

Given three vectors A = 24i + 33j, B = 55i - 12j and C = 2i + 43j (a) Find the magnitude of each vector. (b) Write an expression

slega [8]

Answer:

(a) , . and .

(b) $\vec A - \vec C=22 \hat i -10 \hat j$ .

(c) $|\vec A - \vec B|=63.13$ and the direction $\theta =$ 124.56°.

Explanation:

Given that,

,

and

$\vec {C}=2 \hat i +43 \hat j$

(a) The magnitude of a vector is the square root of the sum of the square of all the components of the vector, i.e. for a ,.

So, the magnitude of the is

$|\vec A|=\sqrt {24^2+ 33^2}$

$\Rightarrow |\vec A|=\sqrt {1665}$

.

The magnitude of the is

$|\vec B|=\sqrt {55^2+ (-12)^2}$

$\Rightarrow |\vec B|=\sqrt {3169}$

.

And, the magnitude of the is

$|\vec C|=\sqrt {2^2+ 43^2}$

$\Rightarrow |\vec C|=\sqrt {1853}$

.

(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is

$\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)$

$\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j$

$\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j$

(c) The expression of is

$\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)$

$\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j$

$\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)$

The magnitude of is

$|\vec A - \vec B|=\sqrt {(-31)^2+55^2}$

$\Rightarrow |\vec A - \vec B|=\sqrt {3986}$

$\Rightarrow |\vec A - \vec B|=63.13$

Now, if a vector $\vec V= -\alpha \hat i +\beta \hat j$ in 3rd quadrant having direction $\theta$ with respect to $\hat i$ direction, than

in the anti-clockwise direction.

Here, from equation (i), for the vector $\vec A - \vec C$ , $\alpha=31$ and $\beta=45$ .

$\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)$

180°-55.44° [as \pi radian= 180°]

124.56° in the anti-clockwise direction.

(d) Vector diagrams for $\vec A +\vec B$ and $\vec A - \vec B$ has been shown

in the figure(b) and figure(c) recpectively.

Vector $\vec A - \vec B$ is in 3rd quadrant as calculated in part (c).

While Vector $\vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)$

$\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j$ , which is in 1st quadrant as both the components are position has been shown in figure(b).

6 0

3 years ago