All categories

3 years ago

Classify each of the following hydro carbons as alkane,alkene or alkyne

Physics

1 answer:

timofeeve [1]3 years ago

5 0

Answer:

a.)alkene

b.)alkane

d.)alkana

about c..I think there is something wrong with the question

You might be interested in

Which things determine an element's electronegativity? Check all that apply. A. The size of the atom of that element B. Whether

KengaRu [80]

D. The number of electrons

6 0

3 years ago

a motorcycle starts from rest covers 200 meter distance in 6 second calculate final velocity and acceleration

Ivanshal [37]

Explanation:

s = ut + 1/2 a t^2

200 = 0 * 6 + 1/2 * a * (6)^2

200 = 1/2 * a * 36

200 = 18 a

a = 200/18

a= 11.1m/sec^2

v = u + at

v = 0 + 11.1 * 6

v = 66.6m/s

hope it helps you

3 0

3 years ago

Using the law of conservation of angular momentum, estimate how fast a collapsed stellar core would spin if its initial spin rat

Nataly_w [17]

Answer:

$\omega_{f} = 1000000\,\frac{rev}{day}$

Explanation:

The law of conservation of angular momentum states that angular momentum remains constant when there is no external moment or forces applied to the system. Let assume that star can be modelled as an sphere, then:

$\frac{2}{5}\cdot M\cdot R_{o}^{2} \cdot \omega_{o} = \frac{2}{5}\cdot M\cdot R_{f}^{2} \cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \omega_{o}\cdot (\frac{R_{o}}{R_{f}})^{2}$

$\omega_{f} = (1\,\frac{rev}{day} )\cdot (\frac{10000\,km}{10\,km} )^{2}$

$\omega_{f} = 1000000\,\frac{rev}{day}$

3 0

3 years ago

You attach a 2.90 kg mass to a horizontal spring that is fixed at one end. You pull the mass until the spring is stretched by 0.

Murljashka [212]

Answer:

Explanation:

The spring is stretched by .5 m and then released that means its amplitude of oscillation A is 0.5 m .

A = 0.5 m

After the release at one extreme point , the mass comes to rest again at another extreme point after half the time period ie

T / 2 = .3 s

T = 0.6 s

Angular velocity

ω = $\frac{2\times \pi}{T}$

ω = $\frac{2\times \pi}{0.6}$

ω = 10.45

Maximum velocity = ω A

ω and A are angular velocity and amplitude of oscillation.

Maximum velocity = 10.45 x .5

= 5.23 m /s

7 0

3 years ago

You are in charge of designing a movie stunt for a new Fast and Furious film. In this scene a stunt person has to jump from a br

blondinia [14]

Answer:

103.5 meters

Explanation:

Given that a stunt person has to jump from a bridge and land on a boat in the water 22.5 m below. The boat is cruising at a constant velocity of 48.3 m/s towards the bridge. The stunt person will jump up at 6.45 m/s as they leave the bridge.

The time the person will jump to a certain spot under the bridge can be calculated by using the formula below:

h = Ut + 1/2gt^2

since the person will fall under gravity, g = 9.8 m/s^2

Also, let assume that the person jump from rest, then, U = 0

Substitute h, U and g into the formula above

22.5 = 1/2 * 9.8 * t^2

22.5 = 4.9t^2

t^2 = 22.5/4.9

t^2 = 4.59

t = $\sqrt{4.592}$

t = 2.143 seconds

From definition of speed,

speed = distance /time

Given that the boat is cruising at a constant velocity of 48.3 m/s towards the bridge, substitute the speed and the time to get the distance.

48.3 = distance / 2.143

distance = 48.3 * 2.143

distance = 103.5 m

Therefore, the boat should be 103.5m away from the bridge at the moment the stunt person jumps?

3 0

3 years ago