What do glucose starch and cellulose have in common

Ple can someone help me to answer this question

ra1l [238]

The length of the uniform meter stick is 50 cm and the mass of the uniform meter stick is 50 g.

<h3 /><h3>A sketch of the uniform stick and the mass</h3>

The sketch of the uniform stick and the mass will be two based on the given statement.

<h3>when the uniform stick balances on knife at 10 cm from one end;</h3>

-------------------------------------------------------

↓ 10cm Δ (L - 10 cm) ↓

200g M

<h3>When the knife edge is moved 5 cm further</h3>

|---------------15 cm-----------------|

-----------------------------------------------------------------------

↓ 8.75 cm Δ (L - 15 cm) ↓

200g M

From the first diagram, apply principle of moment;

200(10) = M(L - 10) ------- (1)

From the second diagram, apply principle of moment;

200(8.75) = M(L - 15) ------- (2)

From equation (1); M = (2000) / (L - 10)

From equation (2); M = (1750) / (L - 15)

Solve (1) and (2);

(2000) / (L - 10) = (1750) / (L - 15)

1750(L - 10) = 2000(L - 15)

L - 10 = 2000/1750(L - 15)

L - 10 = 1.143(L - 15)

L - 10 = 1.143L - 17.14

17.14 - 10 = 1.143L - L

7.14 = 0.143L

L = 7.14/0.143

L = 50 cm

<h3>Mass of the uniform stick</h3>

M = (2000) / (50 - 10)

M = 50 g

Thus, the length of the uniform meter stick is 50 cm and the mass of the uniform meter stick is 50 g.

Learn more about length of meter stick here: brainly.com/question/17225736

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8 0

2 years ago

a runner makes one lap around a 200m track in 25s, what is the runners (a) average speed and (b) average velocity

TEA [102]

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5 0

3 years ago

A starter pistol is fired 150m away from the spectators who hear the gun 0.5s after they see it fired. How fast does sound trave

Natali [406]

Answer:

300m/s

Explanation:

speed = distance/time

150/0.5

8 0

3 years ago

Wireless Internet networks, including many used in homes, often make use of high-frequency radio waves. High-frequency waves are

lutik1710 [3]

The answer is d , small disruptions

4 0

3 years ago

Read 2 more answers

A ball is gently dropped from a height of 30m. If its velocity increase uniform at the rate of 10m/s with what velocity will it

jekas [21]

Explanation:

The velocity of the ball with strikes the ground.
The time of the ball after which it strikes the ground.

$\red{\frak{Given}} \begin{cases} & \sf {A\ ball\ is\ gently\ dropped\ from\ a\ height\ of\ {\pmb{\sf{30\ m}}}.} \\ & \sf {Its\ velocity\ increase\ at\ the\ rate\ of\ {\pmb{\sf{10\ m/s^2}}}.} \end{cases}$

We know that,

The ball is dropped from the height of 30 m as stated in the question. Therefore, it will have a initial velocity of 0 m/s. The distance travelled by the ball will be 30 metres as the distance travelled by the ball is equal to the height of the tower and that is 30 metres.

So, by using the third equation of motion, we will find out the final velocity of the ball.

$\rule{200}{3}$

Therefore,

$\sf \dashrightarrow {2as\ =\ v^2\ -\ u^2} \\ \\ \\ \sf \dashrightarrow {2 \times 10\ m/s^2 \times 30\ m\ =\ v^2\ -\ (0\ m/s)^2} \\ \\ \\ \sf \dashrightarrow {v^2\ =\ 600\ m^2/s^2} \\ \\ \\ \sf \dashrightarrow {v\ =\ \sqrt{600\ m^2s^{-2}}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{Final\ velocity\ (v)\ =\ 24.49\ m/s \approx 24.5\ m/s}}}}_{\sf \blue{\tiny{Required\ velocity\ of\ ball}}}}$

∴ Hence, the required final velocity of the ball with which it strikes the ground is 24.5 m/s. Since, we know that the initial velocity of the ball is 0 m/s and the acceleration of the ball is 10 m/s². So, by using the first equation of motion, we will find out the time of the ball after which it stikes the ground.

$\rule{200}{3}$

Therefore,

$\sf \dashrightarrow {v\ =\ u\ +\ at} \\ \\ \\ \sf \dashrightarrow {t\ =\ \dfrac{v\ -\ u}{a}} \\ \\ \\ \sf \dashrightarrow {t\ =\ \dfrac{24.5\ m/s}{10\ m/s^2}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{Time\ (t)\ =\ 2.45\ s}}}}_{\sf \blue{\tiny{Required\ time}}}}$

∴ Hence, the required time of the ball after which it strikes the ground is 2.45 seconds.

3 0

3 years ago