3 years ago

Can a twig float on water?

1 answer:

5 0

Yes a twig will float on water

1 g/cm³- density of water

roughly 0.50–0.75 g cm-3-density of twig

because a twig is less dense than water the twig will float

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You place a large pebble in a slingshot, pull the elastic band back to your chin, and release it, launching the pebble horizonta

sladkih [1.3K]

Answer:

The larger pebble has 25 times more mass.

Explanation:

To solve the exercise it is necessary to apply the work and energy conservation equations.

For the case described, the work done must be preserved and must be the same, that is,

$W = 0$

By definition work linked to the conservation of kinetic energy would be given by

$\Delta KE = W = 0$

$\Delta KE = 0$

$KE_{larger}-KE_{smaller} = 0$

$KE_{larger}=KE_{smaller}$

$\frac{1}{2}m_lv_l^2 = \frac{1}{2}m_sv_s^2$

$m_lv_l^2 = m_sv_s^2$

The ratio between the mass and the velocity would be,

$\frac{m_l}{m_s}=\frac{v_s^2}{v_l^2}$

$\frac{m_l}{m_s} = \frac{500^2}{100^2}$

$\frac{m_l}{m_s} = 25$

Therefore the answer is: The larger pebble has 25 times more mass.

4 0

3 years ago

Why don't you notice your gravitational force on other objects?

mixas84 [53]

The answer is B tell me if I am wrong.

4 0

3 years ago

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What force would be required to accelerate a 1,100 kg car to 0.5 m/s2?

Klio2033 [76]

Force =mass x acceleration =1100 x0.5 =550

4 0

3 years ago

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

Lubov Fominskaja [6]

Answer:

The moment of inertia is $I=\frac{M}{12} a^{2}$

Explanation:

The moment of inertia is equal:

$I=\int\limits^a_b {r^{2} } \, dm$

If r is $-\frac{a}{2}$

and $dm=\frac{M}{a} dr$

$I=\int\limits^a_b {r^{2}\frac{M}{a} } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}$

$I=\frac{M}{a} \int\limits^a_b {r^{2} } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\ I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}$

7 0

4 years ago

a basketball is tossed upwards with a speed of 5.0 m/s What is the maximum height reached by the basketball from its release poi

den301095 [7]

Answer:

1.275 m

Explanation:

Let the maximum height reached be h.

Here initial velocity, u = 5 m/s

Final velocity, V = 0

Use third equation of motion

V^2 = u^2 + 2 g h

0 = 25 - 2 × 9.8 × h

h = 25/19.6 = 1.275 m

3 0

4 years ago

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