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4 years ago

8

A car travels 30.25 kilometers due west and then drives 7.50 kilometers up a hill that inclines 15.0 degrees. What is the vertic

al displacement of the car?
A. 0.60 Kilometers
B. 1.94 Kilometers
C. 3.25 Kilometers
D. 4.35 Kilometers

2 answers:

Naddik [55]4 years ago

8 0

B we use soh sin15.0=o/7.50=1.91

Yuliya22 [10]4 years ago

8 0

The correct answer to the question is : B) 1.94 Km.

EXPLANATION:

As per the question, the car travels 30.25 Km due west.

The car drives 7.50 km up a hill.

The angle of inclination of the hill with the ground is $\theta=\ 15^{0}$ .

We are asked to calculate the vertical displacement of the car.

Resolving 7.50 Km into horizontal and vertical components, we get-

(a) $7.50cos\theta$ which is the horizontal component.

(b) $7.50 sin\theta$ which is the vertical component.

The vertical displacement [ y ] is calculated as -

y = $7.50sin\theta$

= 7.50sin15

= 1.9411.. m.

= 1.94 m.

Hence, the vertical displacement is 1.94 m.

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Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is $R_z = 4.4 \Omega$

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The rate at which internal energy increase in the battery is $Z_1 = 32 W$

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The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

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Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

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Substituting values

$R_T = 2.6 +R_z$

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$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = \frac{E}{R_T}$

Substituting values

$4 = \frac{28}{R_z + 2.6}$

Making $R_z$ the subject of the formula

So $R_z = \frac{28 - 10.4}{4}$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

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The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

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The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

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