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3 years ago

A 0.5 kg ball travelling at a velocity of 1 m/s collides with a 1.5 kg ball at rest, calculate the

Physics

1 answer:

irakobra [83]3 years ago

5 0

Answer:

0.25m/s

Explanation:

Mass, m1 = 0.5kg

m2= 1.5kg

Velocity before collision for mass 1, u1 = 1m/s

Velocity before collision for mass 2, u2 = 0m/s because it is at rest.

Velocity after collision, V = ?

From law of conservation of linear momentum

Sum of momentum before collision = sum of momentum after collision.

Remember momentum = mass x velocity

Therefore,

m1u1 +m2u2 = (m1 + m2)V

0.5x1 + 1.5x0 = (1.5+0.5)V

0.5+0 = 2V

2V = 0.5

Divide both sides by 2

V = 0.5/2

V = 0.25m/s

I hope this was helpful, Please mark as brainliest

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Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

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3 years ago

Standard Heat of Formation: The enthalpy change for the formation of 1 mol of a substance in its standard state from its constit

ArbitrLikvidat [17]

Answer:

True

Explanation:

Standard heat of formation is the heat change that deals with the formation of 1mole at standard rates and states of the given reactants . Standard heat of formation is the difference between the enthalpy change of reactants and products.

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3 years ago

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The distance between two successive peaks on adjacent waves is its

yKpoI14uk [10]

Answer:

Wavelength

Explanation:

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2 years ago

Calculate the force at sea level that a boy of mass 50 kg exerts on a chair in which he is sitting.

kogti [31]

As per the question the mass of the boy is 50 kg.

The boy sits on a chair.

We are asked to calculate the force exerted by the boy on the chair at sea level.

The force exerted by boy on the chair while sitting on it is nothing else except the force of gravity of earth i.e the weight of the body .The direction of that force is vertically downward.

At sea level the acceleration due to gravity g = 9.8 m/s^2

Hence the weight of the boy $w = mg$ [m is the mass of the body]

we have m = 50 kg.

Hence w = 50 kg ×9.8 m/s^2

=490 N kg m/s^2

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3 years ago

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