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2 years ago

One cubic meter (1.0 m3

1 answer:

8 0

Because the masses that you give are for blocks that are 1 cubic meter in volume, they also serve as the densities for the two metals that you are comparing.

mass = density*volume 

volume = (4/3)*pi*r^3 

volume of iron sphere = (4/3)*3.14*0.0201^3 = 3.40*10^-5 m^3 

mass of iron sphere = 7860* 3.40*10^-5 m^3 = 0.27 kg = mass of Aluminum Sphere 

Volume of Al Sphere = 0.27/2700 = 9.90*10^-5 m^3 

Radius = cube root (volume / (4/3) / pi) = 2.87 cm. 

I did this using the MS calculator, and I'm not 100% sure on the numerical answer, but the process is what you need to do to solve the problem. You should double check my answer.

hope this helped :)

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Explanation:

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2 years ago

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2 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

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Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

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2 years ago

There is a 3-kg toy cart moving at 4 m/s. Calculate the kinetic energy of the cart.

m_a_m_a [10]

The kinetic energy of the cart is 24 J.

Explanation:

The acceleration of a given mass from rest to the velocity is known as kinetic energy. It gains energy from acceleration and remains in this state until the speed of the object changes.

The kinetic energy is the given by,

K.E = 1/2 mv^2

Given the mass m = 3 kg, v = 4 m / s.

K.E = 1/2 $\times$ 3 $\times$ (4)^2

K.E = 24 J.

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2 years ago