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4 years ago

7

WHOEVER GET IT RIGHT GETS 50 POINTS

1 answer:

AVprozaik [17]4 years ago

3 0

Answer:

ANSWER BELOW I

I

V

Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in

m/s2

. For example, for Earth, 445 N = 45.4 × 9.8

m/s2

:Notice that the x-axis values will be gravity in

m/s2

, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).

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A child of mass 46.2 kg sits on the edge of a merry-go-round with radius 1.9 m and moment of inertia 130.09 kg m2 . The merrygo-

Vedmedyk [2.9K]

Answer:

The angular velocity is $w_f = 4.503 \ rad/s$

Explanation:

From the question we are told that

The mass of the child is $m_c = 46.2 \ kg$

The radius of the merry go round is $r = 1.9 \ m$

The moment of inertia of the merry go round is $I_m = 130.09 \ kg \cdot m^2$

The angular velocity of the merry-go round is $w = 2.4 \ rad/s$

The position of the child from the center of the merry-go-round is $x = 0.779 \ m$

According to the law of angular momentum conservation

The initial angular momentum = final angular momentum

So

$L_i = L_f$

=> $I_i w_i = I_fw_f$

Now $I_i$ is the initial moment of inertia of the system which is mathematically represented as

$I_i = I_m + I_{b_1}$

Where $I_{b_i}$ is the initial moment of inertia of the boy which is mathematically evaluated as

$I_{b_i} = m_c * r$

substituting values

$I_{b_i} = 46.2 * 1.9^2$

$I_{b_i} = 166.8 \ kg \cdot m^2$

Thus

$I_i =130.09 + 166.8$

$I_i = 296.9 \ kg \cdot m^2$

Thus

$I_i * w_i =L_i= 296.9 * 2.4$

$L_i = 712.5 \ kg \cdot m^2/s$

Now

$I_f = I_m + I_{b_f }$

Where $I_{b_f}$ is the final moment of inertia of the boy which is mathematically evaluated as

$I_{b_f} = m_c * x$

substituting values

$I_{b_f} = 46.2 * 0.779^2$

$I_{b_f} = 28.03 kg \cdot m^2$

Thus

$I_f = 130.09 + 28.03$

$I_f = 158.12 \ kg \ m^2$

Thus

$L_f = 158.12 * w_f$

Hence

$712.5 = 158.12 * w_f$

$w_f = 4.503 \ rad/s$

8 0

3 years ago

(a) What is the cost of heating a hot tub containing 1440 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take h

BaLLatris [955]

Answer:

a) $E = 6.024\,USD$ , For m kilograms, it is 4184m J., 3600000 joules, b) $i = 88.200\,A$

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

$Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})$

Where:

$m_{w}$ - Mass of water, measured in kilograms.

$c_{w}$ - Specific heat of water, measured in $\frac{J}{kg\cdot ^{\circ}C}$ .

$T_{f}$ , $T_{i}$ - Initial and final temperatures, measured in $^{\circ}C$ .

Then,

$Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)$

$Q_{needed} = 180748800\,J$

The energy needed in kilowatt-hours is:

$Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)$

$Q_{needed} = 50.208\,kWh$

The electric energy required to heat up the water is:

$E = \frac{50.208\,kWh}{0.75}$

$E = 66.944\,kWh$

Lastly, the cost of heating a hot tub is: (USD - US dollars)

$E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)$

$E = 6.024\,USD$

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

$i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}$

$i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}$

$i = 88.200\,A$

7 0

3 years ago

when striking, the pike, a predatory fish, can accelerate from rest to a speed of 4.0 m/sm/s in 0.15 ss

Alina [70]

The distance covered will be 0.301m.

firstly we will new to calculate acceleration (a).

we can do so using newtons first law of motion.

v=u + a × t

where

a is the acceleration

t is the time

from here we get

a=(v-u)/t

a=(4-0)/0.15

a=26.67m/s²

using newtons second law of motion

s=u.t+1/2at²

where s is distance travelled in t seconds.

we get,

s=0.301m

learn more about newtons law of motion here:

brainly.com/question/129361

#SPJ4

6 0

2 years ago

In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e

Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

$\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}$

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

$\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}$

or

$\eta=1-\frac{(Q_1-Q_3)}{Q_1}}$

or

$\eta=1-\frac{(Q_3)}{Q_1}}$

also,

$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}$

or

$\frac{T_3}{T_1}=\frac{Q_3}{Q_1}$

thus,

$\eta=1-\frac{(T_3)}{T_1}}$

thus,

$\eta=1-\frac{(240\ K)}{500\ K}}$

or

$\eta=0.52$

or

Efficiency = 52%

8 0

4 years ago

Explica virgula din structura"Vesela verde câmpie acu-i trista,vestezita. "...dau coroana...la limba romană. Plsss

ozzi

This would be a Goal Act Plan, But in translation, Planul Act Gol

8 0

3 years ago