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2 years ago

What is the average speed in the last 1s? With Explanation Please

Physics

1 answer:

I am Lyosha [343]2 years ago

5 0

$this \: is \: my \: attachment \: answer \: hope \: its \: helpful \: to \: you$

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Torque can be best described as which of the following? Give an example of both a force and a torque and explain why in a couple

kicyunya [14]

Answer: rotational force

Explanation:

Torque is the twisting force which cause rotation and the axis of rotation is the point at which the object rotates.

Torque is a rotational force as it leads to the rotation of an object about an axis. Force simply means a pull or push. When an unbalanced ball acts on a force, the ball, the ball will be moved towards the linear motion.

Then, the unbalanced force that is acting in the ball produces torque which causes the ball's rotational motion.

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3 years ago

About how long have wastewater treatment plants been in existence? about how long have wastewater treatment plants been in exist

ella [17]

Water treatment plants have been in existence for almost 220 years

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3 years ago

A force of 210 N is applied to an object and the object accelerates at 14m/s2. Determine the mass of the object in kg.

lisabon 2012 [21]

Answer:

15 kg

Explanation:

F = m*a

F/a = m

210/14 = 15 kg

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3 years ago

Some compounds, like carbon dioxide, are gaseous at room temperature, while others are solid at room temperature. Why is this?

Vesnalui [34]

The answer of a & b are force of cohesion and force of adhesion
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3 years ago

Read 2 more answers

A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert

ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

$F_y=F-mg=8.5-0.46\times 9.8=3.992\ N$

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

$a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2$

Vertical displacement of rocket is same as the height of loop. So, $y=20\ m$

There is no initial velocity in the vertical direction. So, $u_y=0\ m/s$

Now, applying equation of motion in vertical direction. we have:

$y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s$

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

$x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m$

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0

3 years ago