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4 years ago

A 15.0-kilogram mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the

1 answer:

7 0

The answer is 570 J. The kinetic energy has the formula of 1/2mV². The total work in this process W= 1/2m(V2²-V1²) = 1/2 * 15.0 * (11.5²-7.50²) = 570 J.

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If the period of a pendulum were 2.4 s on the moon, which has an acceleration due to gravity of 1.67 m/s^2, then how long would

guajiro [1.7K]

<h2>Reffer the attachment </h2>

Mark as brainlist ❤️❤️

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3 years ago

Two 20.0 g ice cubes at − 20.0 ∘ C are placed into 285 g of water at 25.0 ∘ C. Assuming no energy is transferred to or from the

Lelechka [254]

Answer:

Ft = 17.48°C

Explanation:

Ft is the final temperature. However, ice absorbs heat during two process of melting and cooling and as such, there is no loss of heat to or from the surrounding hence by conservation of energy.

Therefore,

Heat absorbed by water of 20g = heat rejected by water of 265g.

So; M(ice)[C(ice) [(ΔT) + LH(ice) + C(water)(ΔT)] = C(water) M(water) (ΔT)

So, 20[(2.108) [0 - (-20)] + 333.5 + 4.187(Ft - 0)]] = (285)(4.187) (25 - Ft)

To get;

7513 + 83.74 Ft = 29832.4 - 1193.3 Ft

So factorizing, we get;

83.74 Ft + 1193.3 Ft = 29832.4 - 7513

So; 1277.04 Ft = 22319.4

So; Ft = 22319.4/1277.04 = 17.48°C

7 0

3 years ago

Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black

oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

$Q = FA\sigma\Delta T^4$

Where,

F =View Factor

A = Cross sectional Area

$\sigma =$ Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

$L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K$

The view factor between two coaxial parallel disks would be

$\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33$

$\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75$

Then the view factor between base to top surface of the cylinder becomes $F_{12} = 0.26$ . From the summation rule

$F_{13} = 1-0.26$

$F_{13} = 0.74$

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

$\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}$

$\dot{Q_3} = 2\dot{Q_{13}}$

$\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)$

$\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)$

$\dot{Q_3} = 780.76W$

Therefore the rate heat radiation is 780.76W

5 0

3 years ago

a force of 50 newtons pulls a rope attached to a 150 newton sled across a horizontal surface at a constant velocity of 5 meters

lakkis [162]

Answer:

I don't know the answer I hope you find it tho good luck##

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3 years ago

The following are possible ways to express the quantity 0.391 (Give ALL correct answers, i.e., B, AC, BCD...) Note: 3.45E-8 is a

laiz [17]

To answer, evaluate the power of 10 in the given choices. If it is positve, move the decimal n places to the right. If it is negative, move the decimal n corresponding places to the left. From all the choices given, only the choices D, E, and F will give us the correct answer.

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3 years ago