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3 years ago

8

Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna

l resistance. Find the current I1 through the first resistor and the potential difference ΔV2 between the ends of the second resistor.]

1 answer:

son4ous [18]3 years ago

4 0

Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

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Andrew gathered several different species of seed plants. Which three characteristics do all the plants most likely share? have

BigorU [14]

Answer:

c

Explanation:

no te tengo porque darte explicaciones

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2 years ago

Read 2 more answers

(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us

konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

$R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m$

(b) The maximum height of the projectile in meters;

$H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m$

(c) The speed with which the projectile hits the ground is;

$v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s$

5 0

3 years ago

A vector quantity must include both magnitude and direction. Which measurement is a vector quantity? A) the rain accumulation at

bogdanovich [222]

Answer:

B)The motion of water in an ocean current

Explanation:

With respect to measurements, a vector has both a magnitude and a direction. The first three examples (maximum height of a hill, air temperature, and rain accumulation) are magnitudes only. The fourth example (motion of water in an ocean current) is a vector, because it has a magnitude (speed) and a direction (with the current).

4 0

3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

3 years ago

The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.

Sholpan [36]

The frequency of rotation of Mars is 0.0000113 Hertz.

<u>Given the following data:</u>

Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

<u>Conversion:</u>

1 day = 24 hours

24 hours to minutes = $60$ × $24$ = $1440$ minutes

$1440 + 37 = 1477 \; minutes$

1 minute = 60 seconds

1477 minute = X seconds

Cross-multiplying, we have:

$X = 60$ × $1477$

X = 88620 seconds

Now, we can find the frequency of rotation of Mars by using the formula:

$Frequency = \frac{1}{Period}\\\\Frequency = \frac{1}{88620}$

<em>Frequency </em><em>of rotation</em> = <em>0.0000113 Hertz</em>

Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.

Read more: brainly.com/question/14708169

8 0

3 years ago