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3 years ago

1.Round off the following measurement to the three significant figures. 1.296 g.

Physics

1 answer:

Kobotan [32]3 years ago

5 0

Answer:

<h2>1) 1.30g</h2><h2>2) 3 sf</h2><h2>3) 5.66 grams</h2>

Explanation:

1) Significant figure means the amount of figure we are interested in. Note that 0 is insignificant. For example 30 is a one significant figure i. e the only figure of interest is 3.

Given the figure 1.296 g, as we can see, there are initially for figures in the value. This means we are to eliminate one to make it 3. To do that we will remove the 6 at the back by round it up to 1 and add it to the value of 29 preceding it to give 30. The figure will become 1.30 (to 3sf).

<em>Note that only values greater than 4 are rounded up to 1 and added to the preceding value</em>

2) Given the figure 2020, the number of significant figures present is 3 significant figure. Note that all zeros at the end of a value are considered insignificant hence the only significant value here is 202 (which is 3 sig fig.)

3) For the figure 5.658, the value to 3sf is 5.66 (Note that the value of 8 at the end is rounded up to 1 and added to 65 to make it 66). This is as explained in the first answer above.

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Answer:

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2 years ago

Two automobiles traveling at right angles to each other collide and stick together. Car A has a mass of 1200 kg and had a speed

sergij07 [2.7K]

Answer:

$v_{B0}=15.73 m/s$

Explanation:

We can use the conservation of momentum. The initial momentum is equal to the final momentum:

x-coordinate

$p_{0x}=p_{fx}$

$m_{A}v_{A0}=(m_{A}+m_{B})v_{cx}$

$m_{A}v_{A0}=(m_{A}+m_{B})v_{c}cos(40)$ (1)

y-coordinate

$p_{0y}=p_{fy}$

$m_{B}v_{B0}=(m_{A}+m_{B})v_{cy}$

$m_{B}v_{B0}=(m_{A}+m_{B})v_{c}sin(40)$ (2)

We can divide equations (2) and (1):

$\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=\frac{sin(40)}{cos(40)}$

$\frac{m_{B}v_{B0}}{m_{A}v_{A0}}=tan(40)$

$v_{B0}=\frac{m_{A}v_{A0}}{m_{B}}*tan(40)$

$v_{B0}=\frac{1200*25}{1600}*tan(40)$

$v_{B0}=15.73 m/s$

I hope it helps you!

4 0

3 years ago

Read 2 more answers

A 0.5-kg ball moving at 5 m/s strikes a wall and rebounds in the opposite direction with a speed of 2 m/s. If the impulse occurs

mart [117]

<em>Given that:</em>

mass of the ball (m) = 0.5 Kg ,

ball strikes the wall (v₁) = 5 m/s ,

rebounds in opposite direction (v₂) = 2 m/s,

time duration (t) = 0.01 s,

<em> Determine the force (F) = ?</em>

We know that from Newton's II law,

<em>F = m. a</em> Newtons

(velocity acting in opposite direction, so <em>a = ( (v₁ + v₂)/t</em>

= m × (v₁ + v₂)/t

= 0.5 × (5 + 2)/0.01

= 350 N

<em>The force acting up on the ball is 350 N</em>

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3 years ago

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konstantin123 [22]

Answer:

Once a carnivorous plant has procured an item for dinner, it has to have some way to turn it into fertilizer. What carnivorous plants do is very similar to what humans do with their dinner after they have eaten it. Most carnivorous plants have glands that secrete acids and enzymes to dissolve proteins and other compounds. The plants may also enlist other organisms to help with digestion. The plants then absorb the nutrients made available from the prey.

Drosera releases digestive juices through the glands at the tip of its tentacles and absorbs the nutrients through the tentacles, leaf surface, and sessile glands. In order to do this it bends its tentacles and rolls or bends the leaf to get as many tentacles as possible into contact with the prey for digestion and to make as much leaf surface available for absorption. Its relative Drosophyllum has differently structured, non moving tentacles and doesn't use them directly for digestion. Instead it has specialized glands on the surface of the leaf that release the digestive enzymes (see Carniv. Pl. Newslett. 11(3):66-73 ( PDF ) for drawings and discussion).

The sealed trap of Dionaea does digestion in a way similar to the leaf surface digestion carnivores—upon capture of a prey, digestive enzymes in mucous are released. The advantage of the sealed trap of Dionaea is rain won't wash away the nutrients as digestion proceeds.

The sealed trap carnivores Aldrovanda and Utricularia already have water in their traps so they only need to release enzymes. Utricularia appears to release the enzymes continuously into its traps.

The other carnivorous plants use either a mixed mode of digestive enzymes and partner organisms (Genlisea, Sarracenia, most Nepenthes, Cephalotus, some Heliamphora, Roridula) or other organisms exclusively for digestion (most Heliamphora, some Nepenthes, Darlingtonia). Part of the reason for partnering with other organisms is that the plants actually have little choice in the matter. This could also be a factor for the leaf surface and sealed trap digesters as well. The prey will have gut flora that are quite capable of digesting their host when it dies. In addition, insect larvae, frog tadpoles, and predacious protozoans will or will attempt to take up residence in water-filled traps. The plant releasing digestive enzymes and acids into the traps will help tip the nutrition balance to themselves, but there are limits.

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