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Yakvenalex [24]

3 years ago

13

Which of these statements accurately describes polymers?

1 answer:

alina1380 [7]3 years ago

3 0

Answer:polymer are made of long chains of molecules

Explanation:

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What is the displacement of the car between t=1s and t=4s

tensa zangetsu [6.8K]

Answer:

Option C. 30 m

Explanation:

From the graph given in the question above,

At t = 1 s,

The displacement of the car is 10 m

At t = 4 s

The displacement of the car is 40 m

Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:

Displacement at t = 1 s (d1) = 10 m

Displacement at t= 4 s (d2) = 40

Displacement between t = 1 and t = 4 (ΔD) =?

ΔD = d2 – d1

ΔD = 40 – 10

ΔD = 30 m.

Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.

4 0

3 years ago

7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ba

Dima020 [189]

$\\ \tt \longmapsto \: h = - 4.8 {t}^{2} + 8.2t + 1.8 \\ \\ \tt \longmapsto \: h = - 4.8(1.7) {}^{2} + 8.2 \times 1.7 + 1.8 \\ \\ \tt \longmapsto \: - 4.8 \times 2.89 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 13.8 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 17.06$

6 0

2 years ago

You are given a sample of metal and asked to determine its specific heat. You weight the sample and find that its weight is 28.4

lidiya [134]

Answer:

The samples specific heat is 14.8 J/kg.K

Explanation:

Given that,

Weight = 28.4 N

Suppose, heat energy $E=1.25\times10^{4}\ J$

Temperature = 18°C

We need to calculate the samples specific heat

Using formula of specific heat

$Q=mc\Delta T$

$c=\dfrac{Q}{m\Delta T}$

Where, m = mass

c = specific heat

$\Delta T$ = temperature

Q = heat

Put the value into the formula

$c=\dfrac{1.25\times10^{4}}{\dfrac{28.4}{9.8}\times(18+273)}$

$c=14.8\ J/kg. K$

Hence, The samples specific heat is 14.8 J/kg.K

8 0

3 years ago

In the diagram below letters a b c and d represents four different processes that occurs during the water cycle

lyudmila [28]

For the first one, trees are organisms, so it’s B.

The second one,, the heavy percolation and strong winds come from hurricanes. So it’s hurricanes

5 0

3 years ago

We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination hav

Dmitrij [34]

Answer:

$t=1.5\times 10^{-4}\ s$

Explanation:

Given:

file size to be transmitted, $D=30\ Mb$

transmission rate of data, $\dot D=10\ Mb.s^{-1}$

propagation speed, $v=2\times 10^8\ m.s^{-1}$

distance of data transfer, $s=10000\ km=10^4\ m$

<u>Now the delay in data transfer from source to destination for each 10 Mb:</u>

$t'=\frac{s}{v}$

$t'=\frac{10^4}{2\times 10^8}$

$t'=5\times 10^{-5}\ s$

<u>Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:</u>

So,

$t=3\times t'$

$t=3\times 5\times 10^{-5}$

$t=1.5\times 10^{-4}\ s$

3 0

3 years ago