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3 years ago

6. A stomp rocket takes 1.5 seconds to reach its maximum height

Physics

1 answer:

anyanavicka [17]3 years ago

4 0

Answer:

a) 15 m/s

b) 11.25 m

Explanation:

a) when the rocket is approaching to the maximum height , its velocity =0 (assume g= 10 m/s² downwards)

applying motion equations upwards

V = U +at

0 = U + (-10)*1.5

U = 15 m/s (initial velocity)

b) let maximum height be S and V =0 (velocity at highest point)

V² = U² + 2aS

0 = 15² + 2 * (-10)*S

S = 11.25 m

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A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur

morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

$\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\$

$\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\$

Substitute the given parameters and solve for the angular speed;

$\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s$

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0

2 years ago

Someone please help !!

rodikova [14]

What is the Investigation about!

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3 years ago

A cube-shaped piece of copper has sides of 4cm each and it's density is

iris [78.8K]

Answer:

$64 cm^3$

Explanation:

<u>Density </u>

The density of a substance is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

$\displaystyle \rho=\frac{m}{V}$

We have a cube-shaped piece of copper of 4 cm of side length. The volume of the piece is:

$V=(4\ cm)^3=64\ cm^3$

Surprisingly, no other magnitude is required, thus the answer is:

$\mathbf{64 cm^3}$

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3 years ago

What are the disadvantages of using alcohol instead of mercury in a thermometer ?

vichka [17]

Answer:

alcohol thermometers are used rather than Mercury thermometers in very cold regions because alcohol has a lower freezing point than Mercury.

Explanation:

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3 years ago

Earth is 149.6 million meters from the Sun and takes 365 days to make one complete revolution around the Sun. Mars is 227.9 mill

luda_lava [24]

Answer:

$\frac{a_{r,earth}}{a_{r,mars}} = 2.325$

Explanation:

The distance of Earth from the Sun is $149.6\times 10^{9}\,m$ and of Mars from the Sun is $227.9\times 10^{9}\,m$ . Let assume that both planets have circular orbits. The centripetal accelaration can be found by using the following expression: