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8 days ago

7

If you start with the number 3.0 and move the decimal point one unit to the left, you wind up with 0.30. If you move the decimal

point a total of four units to the left, the result is the same as that of dividing the original number by _________.

1 answer:

Keith_Richards [2.2K]8 days ago

4 0

Answer:

10000

Explanation:

When shifting the decimal point left, the number effectively gets divided by 10 for each movement.

Conversely, shifting it to the right corresponds to multiplying the number by 10 for every unit moved.

If the decimal point is moved four units to the left, it becomes 0.0003, which is equivalent to dividing 3.0 by 10000

$\Rightarrow \frac{3.0}{10000} = 0.0003$

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A household refrigerator runs one-fourth of the time and removes heat from the food compartment at an average rate of 800 kJ/h.

Ostrovityanka [2204]

Answer:

1454.54 kJ/h or 0.404 kW

Explanation:

Provided:

The heat being removed by the refrigerator is 800 kJ/h.

The refrigerator's coefficient of performance (COP) is 2.2.

Since the refrigerator operates for one-fourth of the time,

it effectively removes heat, Q = $4\times 800 kJ/h$ = 3200 kJ/h.

Next, the power consumed by the refrigerator during its operation is calculated as:

$\frac{Q}{COP}$

= $\frac{3200 kJ/h}{2.2}$ .

= 1454.54 kJ/h

Thus, the fridge uses 1454.54 kJ/h

or 0.404 kW (given 1 kW = 3600 kJ/h).

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An astronaut is standing on the surface of a planet that has a mass of 6.42×1023 kg and a radius of 3397 km. The astronaut fires

serg [2593]

The final calculated height is 22.2 km. By converting 3397 km to meters, we achieve 3397000m. Using the gravitational constant, we compute the gravitational acceleration on the planet in accordance with Newton's law of gravitation, where M and R represent the planet's mass and radius, respectively. As the bullet ascends to its peak height, its kinetic energy transitions to potential energy. We account for the bullet mass and its initial velocity of 406 m/s, simplifying both sides of the equation to ultimately express the height reached as 22.2 km.

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6 days ago

Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to

kicyunya [2264]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance for a metal can be calculated by

R = ρ L / A

Where ρ indicates the resistivity of aluminum, L is the resistance's length, and A indicates the cross-sectional area

We use this formula for both configurations

For small face measurements (W x W)

The length is

L = W

Area

A = W W = W²

R₁ = ρ W / W² = ρ / W

For larger face measurements (D x L)

Length L = D= 2W

Area A = W L

R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

From this, we find the relation to be

R₂ / R₁ = 2W²/L

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1 day ago

An object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in

Maru [2355]

a.) 10 Hz b.) 0.1 s c.) 187.4 m/s d.) -412.6 m/s²

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3 days ago

The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

ValentinkaMS [2425]

Answer:

The force magnitude is $F_{net}= 1.837 *10^4N$

and it is directed at 57.98° from the horizontal in a counterclockwise manner.

Explanation:

The problem states that

At t = 0, $\theta = 20^o$

The angular rate of increase is $w = 2 \ ^o/s$

Converting to revolutions per second gives us $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The rope length is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , Tension T of the rope is 18 kN.

The weight of the para-sailor is $M_p = 75kg$

In analyzing the question, we observe that the equation for length can be represented as a linear displacement equation.

The derivative of displacement results in velocity.

Hence,

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

signifies the velocity, and further differentiation yields acceleration.

Therefore,

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now considering the moment when the rope forms a 30° angle with the water,

typically angular velocity is expressed as

$w = \frac{\Delta \theta}{\Delta t}$

where $\theta$ represents the angular displacement.

Next, evaluating the interval from $20^o \ to \ 30^o$ gives us

$2 = \frac{30 -20 }{t -0}$

making t the focal point.

$t = \frac{10}{2}$

$= 5s$

At this time, the displacement measures

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity computes to

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

Whereas linear acceleration calculates as

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally, radial acceleration is given by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Simultaneously, angular acceleration can be represented as

$\alpha_t = r \theta'' + 2 r' \theta '$

Then $\theta '' = \frac{d (0.0349)}{dt} = 0$

Thus,

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The resultant acceleration is mathematically denoted as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the acceleration's direction is mathematically expressed as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The y-axis force acting on the para-sailor is mathematically shown as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The x-axis force acting on the para-sailor is represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The overall force is calculated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The directional force is evaluated as

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

7 0

22 days ago