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5 days ago

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A 1.47-newton baseball is dropped from a height of 10.0 meters and falls through the air to the ground. The kinetic energy of th

e ball is 12.0 joules the instant before the ball strikes the ground. The maximum amount of mechanical energy converted to internal energy during the fall is_________.

2 answers:

Sav [1.1K]5 days ago

8 0

Answer:

The peak mechanical energy transformed into internal energy throughout the descent is 26.7 joules.

Explanation:

Potential Energy (PE) is calculated as the weight of the baseball multiplied by the height, which gives us 1.47N × 10m = 14.7Nm or 14.7 joules.

The kinetic energy (KE) is recorded as 12 joules.

The maximum mechanical energy that converts to internal energy during the fall can be expressed as PE + KE = 14.7 joules + 12 joules = 26.7 joules.

serg [1.1K]5 days ago

7 0

Answer:

ΔEinternal = - 2.7 J

Explanation:

Provided values:

W = m*g = 1.47 N

hinitial = 10.0 m

Kfinal = 12.0 J

The formula we can use is:

ΔE = ΔEinternal

Thus, ΔE can be expressed as (Kfinal + Ufinal) - (Kinitial + Uinitial):

⇒ ΔE = (12.0 J + 0 J) - (0 J + Uinitial) = Kfinal - Uinitial

⇒ ΔE = Kfinal - (W*hfinal) = 12 J - (1.47 N*10.0 m)

⇒ ΔE = 12 J - 14.7 J = - 2.7 J

Therefore, ΔEinternal = - 2.7 J

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1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

Softa [913]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

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12 days ago

Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of t

Softa [913]

Answer:

1/7 kg

Explanation:

Refer to the attached diagram for enhanced clarity regarding the question.

One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.

g denotes the acceleration due to gravity.

Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.

Given M = 1.0 kg and a = 3/4g.

By applying Newton's second law; $\sum fy = ma_y$

For the body with mass m;

T - mg = ma... (1)

For the body with mass M;

Mg - T = Ma... (2)

Combining equations 1 and 2 gives;

+Mg -mg = ma + Ma

Ma-Mg = -mg-ma

M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into this equation leads to;

3/4 g-g = -m(3/4 g+g)

3/4 g-g = -m(7/4 g)

-g/4 = -m(7/4 g)

1/4 = 7m/4

Multiplying gives: 28m = 4

m = 1/7 kg

Hence, the mass of the other box is 1/7 kg

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6 days ago

A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation

Ostrovityanka [942]

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given the data:

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The heat generation formula can be articulated as follows:

q = πr²Lq'

q = π. 0.1². L. 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Applying the energy balance equation,

Energy In = Energy Out

This equates to q, which is 754L

From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C

Additionally, the outer surface temperature records as 51° C

Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C

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An object initially at rest experiences a constant horizontal acceleration due to the action of a resultant force applied for 10

inna [987]

Answer:

a = 18.28 ft/s²

Explanation:

the values provided are:

duration of force application, t= 10 s

Work done = 10 Btu

mass of the object = 15 lb

acceleration, a =? ft/s²

1 Btu = 778.15 ft.lbf

thus, 10 Btu = 7781.5 ft.lbf

$m = \dfrac{15}{32.174}\ slug$

m = 0.466 slug

So,

the work is equivalent to the change in kinetic energy

$W = \dfrac{1}{2} m (v_f^2-v_i^2)$

$7781.5 = \dfrac{1}{2}\times 0.466\times v_f^2$

$v_f = 182.75\ ft/s$

The acceleration of the object is therefore

$a = \dfrac{v_f-v_o}{t}$

$a = \dfrac{182.75-0}{10}$

a = 18.28 ft/s²

the constant acceleration of the object is calculated to be 18.28 ft/s²

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5 days ago

A rocket has a mass 250 x (10^3) slugs on earth. Specify(a) its mass in SI units, and(b)its weight in SI units.If the rocket is

Yuliya22 [1153]

Explanation:

We are given that,

Mass of the rocket, $m=250\times 10^3\ slugs$

(a) The standard unit of mass is kilogram (kg). The conversion between slugs and kilograms is as follows:

1 slug = 14.59 kg

Thus, $250\times 10^3\ slugs=250\times 10^3\times 14.59$

Mass of the rocket, m = 3647500 kg

(b) The weight of the rocket can be expressed as:

W = m g

$W=3647500\times 9.8=35745500\ N$

or

$W=3.5\times 10^7\ N$

(c) If the rocket were on the moon, the gravitational acceleration on the moon is given as $a=5.3\ ft/s^2=1.61\ m/s^2$

Mass refers to the quantity of matter present in an object. Therefore, the mass of the rocket remains constant at 3647500 kg

The weight of the rocket on the moon would be, $W=mg$

$W=3647500\times 1.61$

W = 5872475 N

or

$W=5.8\times 10^6\ N$

Thus, this is the final answer required.

8 0

3 days ago