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Anuta_ua [19.1K]

3 years ago

6

What is a motion graph?

1 answer:

givi [52]3 years ago

6 0

<u>Motion graph: </u>

<em>It is defied as </em><em>"A considerable amount of information about the motion like velocity, acceleration". </em><em>and the information obtained by the slope of graphs.</em>

<em>The graphs distance, velocity and acceleration are the function of time, that defines the motion using the motion equations. </em>

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B

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A cat runs 6m in 4seconds what is it speed

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Explanation:

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What is an ore?

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4 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

BlackZzzverrR [31]

Answer:

Vf = 0.0108 m³

Explanation:

Assuming there is no change in internal energy, we can calculate the changed volume by the following formula:

$Energy\ Supplied = Work = W = P \Delta V \\W = P(V_{f}-V_{i})\\$

where,

W = Work = Energy Supplied = 1000 J

P = Pressure = 101325 Pa

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Vi = Initial Volume = 0.001 m³

Therefore,

$1000\ J = (101325\ Pa)(V_{f}-0.001\ m^3)\\\\V_{f}-0.001\ m^3 = \frac{1000\ J}{101325\ Pa}\\\\V_{f} = 0.001\ m^3 + 0.0099\ m^3$

<u>Vf = 0.0108 m³</u>

6 0

3 years ago

Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t

scoundrel [369]

Answer: The correct answer is option (D).

Explanation:

Vertical distance of first water jet = $y_1=50 cm$

Vertical distance of second water jet = $y_2=100 cm$

Horizontal velocity of first water jet = $u_y_ 1=1 m/s$

Horizontal velocity of second water jet = $u_y_2=0.5 m/s$

Vertical distance of first water jet:

$y_1=(u_y_1)t_1+\frac{1}{2}gt_1^2$

$u_y_1=0,(u_y_1)t_1=0$

$t_1=\sqrt{\frac{2y_1}{g}}$ ...(1)

Similarly, Vertical distance of second water jet:

$y_2=(u_y_2)t_2+\frac{1}{2}gt_2^2$

$u_y_2=0,(u_y_2)t_2=0$

$t_2=\sqrt{\frac{2y_2}{g}}$ ...(2)

Horizontal distance of first water jet:

$x_1=(u_x_1)t_1+\frac{1}{2}at_1^2$

There is no acceleration in horizontal direction.

$a=0,at_1^2=0$

$x_1=(u_y_1)t_1$ ...(3)

Horizontal distance of second water jet:

$x_2=(u_x_2)t_2+\frac{1}{2}at_2^2$

$a=0,at_2^2=0$

$x_2=(u_x_2)t_2$ ...(4)

On dividing (3) and (4):

$\frac{x_1}{x_2}=\frac{(u_y_1)t_1}{(u_y_2)t_2}$

$\frac{x_1}{x_2}=\frac{u_y_1}{u_y_2}\times\frac{\sqrt{\frac{2y_1}{g}}}{\sqrt{\frac{2y_2}{g}}}}=\frac{u_y_1}{u_y_2}\sqrt{\frac{y_1}{y_2}}$

(from (1) and (2))

$\frac{x_1}{x_2}=\frac{1 m/s}{0.5 m/s}\sqrt{\frac{50 cm}{100 cm}}=1.41:1$

Hence, the correct answer is option (D).

6 0

4 years ago