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3 years ago

What is the kinetic energy f a 25kg object movingat a velocity of 10m/s?

Physics

1 answer:

alex41 [277]3 years ago

5 0

Using K.E=1/2MV^2
answer is 125joules

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A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him

juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal direction

$m_{b}v_{b}\cos\theta= m_{c}v_{c}$

$v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}$

Put the value into the formula

$v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}$

$v_{c}=0.0345\ m/s$

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0

2 years ago

Why in example multiplying the length by 2

dolphi86 [110]

What??????? idk!!!!!

3 0

2 years ago

Which of the following shows the units of angular motion?

Whitepunk [10]

Answer:

The units are (rad/s^2)

4 0

3 years ago

The animation shows a ball which has been kicked upward at an angle. Run the animation to watch the motion of the ball. Click in

sergejj [24]

Answer:

The acceleration of the ball is $a_y = - 0.3672 \ m/s^2$

Explanation:

From the question we are told that

The maximum height the ball reachs is $H_{max} = 42.24 \ m$

The horizontal component of the initial velocity of the ball is $v_{ix} = 5.57 \ m/s$

The vertical component of the initial velocity of the ball is $v_{iy} = = 16.18 m/s$

The vertically motion of the ball can be mathematically represented as

$v_{fy}^2 = v_{iy} ^2 + 2 a_{y} H_{max}$

Here the final velocity at the maximum height is zero so $v_{fy} = 0 \ m/s$

Making the acceleration $a_y$ the subject we have

$a_y = \frac{v_{iy} ^2}{2H_{max}}$

substituting values

$a_y = - \frac{5.57^2}{2* 42.24}$

$a_y = - 0.3672 \ m/s^2$

The negative sign shows that the direction of the acceleration is in the negative y-axis

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3 years ago

How does the law of universal gravitation apply to objects near Earth’s surface?

tekilochka [14]

In other words, the Earth attracts objects near its surface to itself. This universal force also acts between the Earth and the Sun, or any other star and its satellites. Each attracts the other. Sir Isaac Newton defined this attraction mathematically. Hope this helps! Mark brainly please!

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2 years ago