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Natasha2012 [34]

3 years ago

11

Captain American throws his shield straight down off a roof with a speed of 50 m/s . What is the acceleration of the shield half

way down to the ground?

1 answer:

Nutka1998 [239]3 years ago

6 0

<em>IF</em> this is happening on <u>Earth</u>, then the acceleration of gravity wherever the captain and the roof are is about 9.8 m/s² downward.

<em>IF</em> his shield is not affected by air resistance, and it's free to behave under the influence of gravity alone, then its acceleration is <em>9.8 m/s² downward</em>, from the moment it leaves his hand until the moment it hits a building, a tree, a car, a bus, an evildoer, or the ground.

Whatever speed it has when it leaves his hand, in whatever direction, makes no difference.

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A jet accelerates at a=3.92 m/s2 from rest until it reaches its takeoff velocity of 360 km/hr. It has to travel for 5 seconds at

Art [367]

Answer:

$\Delta s = 1775.510\,m$

Explanation:

The minimum distance for takeoff is:

$\Delta s = \Delta s_{1} + \Delta s_{2}$

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a} + v_{f}\cdot \Delta t$

$\Delta s = \frac{(100\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}}{2\cdot (3.92\,\frac{m}{s^{2}})}+(100\,\frac{m}{s} )\cdot (5\,s)$

$\Delta s = 1775.510\,m$

7 0

3 years ago

If a water wave completes one cycle in 2 seconds, what is

Novay_Z [31]

The period of a wave is the time it takes the wave to complete one cycle (at a fixed location).

So if a wave completes one cycle in <em>2 seconds</em>, then that is its period.

8 0

3 years ago

The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le

lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = $\sqrt{2gh}$

V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

= 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

= - 0.329 J

Hence, kinetic energy is not conserved.

8 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.

BARSIC [14]

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

$y_{i}=0m$

$v_i=80\frac{m}{s}$

$a=4\frac{m}{s^2}$

$y_{e}=1000m$

$g=9.8\frac{m}{s^2}$

From Kinematics, the position $y$ as a function of time when the engine still works will be:

$y(t)=v_it+\frac{1}{2}at^2$

At what time the altitud will be $y_{e}=1000m$ ?

$v_it+\frac{1}{2}at^2=y_{e}$ ⇒ $\frac{1}{2}at^2+v_it-y_{e}=0$

Using the quadratic formula: $t_1=10s$ .

How much time does it take for the rocket to touch the ground? No the function of position is:

$y(t)=y_{e}+v_et-\frac{1}{2}gt^2$

Where our new initial position is $y_{max}$ , the velocity when the engine breaks is $v_e=v_i+at=120\frac{m}{s}$ and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

$y_{e}+v_et-\frac{1}{2}gt^2=0$

Again, using the quadratic ecuation:

$t_2=24.49s$

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

$t_T=t_1+t_2=34.49s$ .

6 0

4 years ago

Azurite is a mineral that contains 55.1% of copper. How many meter of copper wire with diameter of 0.0113 in can be produced fro

soldier1979 [14.2K]

Answer:

1402.73 m

Explanation:

Mass of Azurite=3.25 lb

Percent of copper in AZurite mineral=55.1%

Diameter of copper wire,d=0.0113 in

Radius of copper wire= $r=\frac{d}{2}=\frac{0.0113}{2}=0.00565 in=\frac{565}{100000}=\frac{565}{100}\times \frac{1}{1000}=5.65\times 10^{-3}in$

$\frac{1}{1000}=10^{-3}$

Density of copper= $\rho=8.96g/cm^3$

1 lb=454 g

3.25 lb= $3.25\times 454=1475.5 g$

Mass of Azurite= $1475.5 g$

Mass of copper= $\frac{55.1}{100}\times 1475.5=813 g$

Density= $\frac{Mass}{volume}$

Using the formula

$8.96=\frac{813}{volume\;of\;copper}$

Volume of copper wire= $\frac{813}{8.96}=90.7cm^3$

Radius of copper wire= $5.65\times 10^{-3}\times 2.54=14.35\times 10^{-3} cm$

1 in=2.54 cm

Volume of copper wire= $\pi r^2 h$

$\pi=3.14$

Using the formula

$90.7=3.14\times (14.35\times 10^{-3})^2\times h$

$h=\frac{90.7}{3.14\times (14.35\times 10^{-3})^2}$

$h=140273 cm$

1 m=100 cm

$h=\frac{140273}{100}=1402.73 m$

Hence, the length of copper wire required=1402.73 m

7 0

3 years ago