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Y_Kistochka [10]

3 years ago

13

Superman strikes a golf ball on the ground at a 38 degree angle above the horizontal at 147 m/s. What is the maximum height the

golf ball reaches?
255 m

418 m

687 m

1103 m

1 answer:

tino4ka555 [31]3 years ago

5 0

The ball is hit with an initial vertical velocity of

(147 m/s) sin(38º) ≈ 90.5 m/s

Recall that

<em>v </em>² - <em>u </em>² = 2 <em>a</em> ∆<em>y</em>

where <em>u</em> and <em>v</em> are initial and final velocities, respectively; <em>a</em> is acceleration; and ∆<em>y</em> is displacement.

Vertically, the ball is in freefall, so it is only subject to acceleration due to gravity, with magnitude <em>g</em> = 9.80 m/s² in the downward direction. At its maximum height, the ball has zero vertical velocity (<em>v</em> = 0) and the displacement is equal to the maximum height, so

0² - (90.5 m/s)² = 2 (-<em>g</em>) ∆<em>y</em>

∆<em>y</em> = (90.5 m/s)² / (2<em>g</em>)

∆<em>y</em> ≈ 418 m

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You are trying to determine the specific gravity of a solid object that floats in water. If m is the mass of your object, mS is

Alisiya [41]

Answer:

Specific Gravity = m/[m(s)-m(os)]

Explanation:

Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance. By this definition we need to find out the ratio of density of the object of mass m to the density of the surrounding liquid.

m = mass of the object

<u>Weight in air</u>

W (air) = mg, where g is the gravitational acceleration

<u>Weight with submerged with only one mass</u>

m(s)g + Fb = mg + m(b)g, <em>consider this to be equation 1</em>

where Fb is the buoyancy force

Weight with submerged with both masses

m(os)g + Fb’ = mg + m(b)g, <em>consider this to be equation 2</em>

<u>equation 1 – equation 2 would give us</u>

m(s)g – m(os)g = Fb’ – Fb

where Fb = D x V x g, where D is the density of the liquid the object is submerged in, g is the force of gravity and V is the submerged volume of the object

m(s)g – m(os)g = D(l) x V x g

m(s) – m(os) = D(l) x V

we know that Mass = Density x V, which in our case would be, D(b) x V, which also means

V = Mass/D(b), where D(b) is the density of the mass

<u>Substituting V into the above equation we get</u>

m(s) – m(os) = [D(l) x m)/ D(b)]

Rearranging to get the ratio of density of object to the density of liquid

D(b)/D(l) = m/[m(s)-m(os)], where D(b)/D(l) denotes the specific gravity

8 0

3 years ago

enot [183]

Answer:

A. $U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. $U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. $U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

$C=\dfrac{\epsilon A}{d}$

$C$ is the capacitance, $A$ is the common plate area, $d$ is the plate separation and $\epsilon$ is the permittivity of the material between the plates.

For air or free space, $\epsilon$ is $\epsilon_0$ called the permittivity of free space. In general, $\epsilon=\epsilon_r \epsilon_0$ where $\epsilon_r$ is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, $\epsilon_r=1$ .

The energy stored in a capacitor is the average of the product of its charge and voltage.

$U = \dfrac{QV}{2}$

Its charge, $Q$ , is related to its capacitance by $Q=CV$ (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for $U$ ,

$U = \dfrac{CV^2}{2}$

A. Substituting for $C$ in $U$ ,

$U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. When the distance is $3d$ ,

$U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}$

$U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. When the distance is restored but with a dielectric material of dielectric constant, $K$ , inserted, we have

$U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

6 0

3 years ago

Roller coasters accelerates from initial speed of 6.0 M/S2 final speed of 70 M/S over four seconds. What is the acceleration?

choli [55]

Answer:

$a=16\ m/s^2$

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly in time.

The formula to calculate the change of velocities is:

$v_f=v_o+at$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The roller coaster moves from vo=6 m/s to vf=70 m/s in t=4 seconds. To calculate the acceleration, solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{70-6}{4}=\frac{64}{4}$

$\boxed{a=16\ m/s^2}$

3 0

3 years ago

A body weighing 108N moves with speed of 5m/s in a horizontal

poizon [28]

Answer:

i) 5 $m$ / $s^{2}$ ii) 54 $N$ iii) 54 $N$

Explanation:

i) $a = \frac{v^{2}}{r}$ ⇒ $a$ = 5² ÷ 5 = 5 $m/s^{2}$

ii) $m = \frac{W}{g}$ ⇒ $m$ = 108 ÷ 10 = 10.8 $kg$ , $F = ma$ ⇒ $F =$ 10.8 × 5 = 54 $N$

iii) F1 = F2 = 54 $N$

4 0

3 years ago

During fabrication processes, polymeric materials are generally subjected to which of the following conditions? The ambient temp

geniusboy [140]

Answer:

- Elevated temperatures.

- Elevated pressures.

Explanation:

In the manufacture of plastic products are used:

Raw material (pellets) which are the monomers that promote the chemical reaction.

To these are added the charges, in order to reduce the cost of the final product and improve some properties. These charges can be fiberglass, paper, metal structures.

Additives are also added whose mission is to improve or achieve certain properties, such as reducing friction, reducing chemical degradation, increasing electrical conductivity, coloring the product, and all this happens in the presence of a catalyst that is responsible for initiating and accelerating the chemical reaction process.

There are different methods of production of plastics, one of the most frequent is by injection.

Injection is a process that is carried out on machines similar to extrusion machines, in which the spindle, in addition to rotating, has an axial displacement.

In the injection, once the mold is filled, it is separated from the nozzle of the machine, breaking the feed channel. After a certain time, the piece already cooled is demoulded.

High pressures and temperatures are necessary, but parts of good finish and at high production speeds are obtained

3 0

3 years ago