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3 years ago

8

What are the possible effects of global warming?

1 answer:

11Alexandr11 [23.1K]3 years ago

6 0

A possible effect is the risk of Icecaps melted

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A car moves with a speed of 30 metres per second calculate the distance travelled in 30 seconds

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30x30=900

The answer is 900 meters after 30 seconds

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Suppose you take a piece of hard wax from an unlit candle. After you roll the wax between your fingers for a while, it becomes s

romanna [79]

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An amorphous solid does not have a definite melting point; instead, it melts gradually over a range of temperatures, because the bonds do not break all at once. This means an amorphous solid will melt into a soft, malleable state (think candle wax or molten glass) before turning completely into a liquid.

Explanation:

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May I get braineist pls?

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3 years ago

Please help. 25 POINTS, PLEASE ANSWER PROMPTLY

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3 years ago

Suppose that a 102.5 kg football player running at 8.5 m/s catches a 0.47 kg ball moving at a speed of 22.5 m/s with his feet of

nadya68 [22]

Answer:

a) $v=8.564\ m.s^{-1}$

b) $\Delta KE=45.76\ J$

c) $v=8.358\ m.s^{-1}$

d) $\Delta KE=225.24\ J$

Explanation:

Given:

mass of the player, $m_p=102.5\ kg$

mass of the ball, $m_b=0.47\ kg$

initial velocity of the player, $v_p=8.5\ m.s^{-1}$

initial velocity of the ball, $v_b=22.5\ m.s^{-1}$

a)

<u>Case:</u> When the player and the ball are moving in the same direction.

$m_t.v=m_p.v_p+m_b.v_b$

where:

$m_t=$ total mass after the player catches the ball

v = final velocity of the system

$v=\frac{102.5\times 8.5+0.47\times 22.5}{(102.5+0.47)}$

$v=8.564\ m.s^{-1}$

b)

Initial kinetic energy of the system:

$KE_i=\frac{1}{2} [m_p.v_p^2+m_b.v_b^2]$

$KE_i=\frac{1}{2} [102.5\times 8.5^2+0.47\times 22.5^2]$

$KE_i=3821.78\ J$

Final kinetic energy of the system:

$KE_f=\frac{1}{2} m_t.v^2$

$KE_f=\frac{1}{2}\times 102.97\times 8.564^2$

$KE_f=3776.02\ J$

∴Change in kinetic energy

$\Delta KE=KE_i-KE_f$

$\Delta KE=3821.78-3776.02$

$\Delta KE=45.76\ J$

c)

<u>Case:</u> When the player and the ball are moving in the opposite direction.

$m_t.v=m_p.v_p-m_b.v_b$

$v=\frac{102.5\times 8.5-0.47\times 22.5}{(102.5+0.47)}$

$v=8.358\ m.s^{-1}$

d)

Final kinetic energy in this case:

$KE_f=\frac{1}{2} m_t.v^2$

$KE_f=0.5\times 102.97\times 8.358^2$

$KE_f=3596.54\ J$

∴Change in kinetic energy:

$\Delta KE=KE_i-KE_f$

$\Delta KE=3821.78-3596.54$

$\Delta KE=225.24\ J$

3 0

3 years ago