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3 years ago

Why are carbon atoms able to form many organic compounds?

Physics

1 answer:

STatiana [176]3 years ago

3 0

Answer:

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Explanation:

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A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

Alexxandr [17]

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: $900\; {\rm J}$ .

Work done when firing this projectile: $2500\; {\rm J}$ .

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity $v_{x}$ of this projectile would stay the same (at $30\; {\rm m\cdot s^{-1}}$ ) throughout the flight.

The vertical velocity $v_{y}$ of this projectile would be $0\; {\rm m\cdot s^{-1}}$ at the vertex (highest point) of its trajectory. (Otherwise, if $v_{y} > 0$ , this projectile would continue moving up and reach an even higher point. If $v_{y} < 0$ , the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be $v = 30\; {\rm m\cdot s^{-1}}\!$ when it is at the top of the trajectory. The kinetic energy $\text{KE}$ of this projectile (mass $m = 2.0\; {\rm kg}$ ) at the vertex of its trajectory would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}$ .

Apply the Pythagorean Theorem to find the initial speed of this projectile:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Hence, the initial kinetic energy $\text{KE}$ of this projectile would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}$ .

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be $2500\; {\rm J}$ .

7 0

2 years ago

In the graph, which two regions show the particle undergoing zero acceleration and negative acceleration respectively?

GuDViN [60]

when velocity and time both are constant and when velocity will decrease the acceleration will be negative

7 0

4 years ago

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

3 years ago

If my mass is 196 lbm and I tackle one of my teammates - while decelerating from a velocity of 6.7 m/s to 0 m/s in 0.5 s, how mu

Olin [163]

Answer:

the force acting on the team mate is 1.19 kN.

Explanation:

given,

mass = 196 lbm

while tackling, the deceleration is from velocity 6.7 m/s to 0 m/s

time taken for deceleration = 0.5 sec

F = mass × acceleration

acceleration = $\dfrac{0-6.7}{0.5}$

= -13.4 m/s²

1 lbs = 0.453 kg

196 lbs = 196 × 0.453 = 88.79 kg

F = 88.79 × 13.4

F = 1189.786 N = 1.19 kN

hence, the force acting on the team mate is 1.19 kN.

8 0

3 years ago

This position-time graph describes an object's motion. Use it to determine the

kirza4 [7]

We're happy that you're asking for the "displacement", because displacement is simply the straight-line distance between the start-point and end-point, and we don't care about any of the motions or gyrations along the way.

From the graph:

-- The location of the object at time-zero, when time begins, is 10 meters.

-- The location of the object after 6.0 seconds is 4 meters.

-- The distance between the start-point and end-point is

(final location) - (initial location)

-- So Displacement = (4 meters) - (10 meters)

<em>Displacement = -6 meters</em>

3 0

3 years ago