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2 years ago

7

What type of collision is being described in each statement.

1 answer:

Ostrovityanka [42]2 years ago

7 0

Answer:

inelastic, since the girl moves in the same direction as the thrown ball

Explanation:

yess this ok

UwU

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3 years ago

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

pav-90 [236]

Answer: $F_{net}=3.383\times 10^{-7}\ N$

Explanation:

Given

Mass of first object $m_1=225\ kg$

Mass of second object $m_2=525\ kg$

Distance between them $d=3.8\ m$

$m_3=61\ kg$ object is placed between them

So force exerted by $m_1$ on $m_3$

$F_{13}=\frac{Gm_1m_3}{1.9^2}$

$F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}$

$F_{13}=2.5374141274×10^{−7}\ N$

Force exerted by $m_2\ on\ m_3$

$F_{23}=\frac{Gm_2m_3}{1.9^2}$

$F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}$

$F_{23}=5.920632964\times 10^{-7}\ N$

So net force on $m_3$ is

$F_{net}=F_{23}-F_{13}$

$F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}$

$F_{net}=3.383\times 10^{-7}\ N$

i.e. net force is towards $m_2$

(b)For net force to be zero on $m_3$ , suppose

So force exerted by $m_1$ and $m_2$ must be equal

$F_{13}=F_{23}$

$\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}$

$\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}$

$\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}$

$\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}$

$\Rightarrow 3.8-x=1.52752x$

$\Rightarrow 3.8=2.52x$

$\Rightarrow x=1.507\ m$

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3 years ago

An object that is accelerating may be

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It must be either speeding up, or slowing down, or turning. There are no other possibilities.

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3 years ago

What are the uses of X-rays.

Arturiano [62]

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3 years ago

How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl

Vinvika [58]

Answer:

B) $\frac{2}{3}$

Explanation:

The electric force between charges can be determined by;

F = $\frac{kq_{1} q_{2} }{r^{2} }$

Where: F is the force, k is the Coulomb's constant, $q_{1}$ is the value of the first charge, $q_{2}$ is the value of the second charge, r is the distance between the centers of the charges.

Let the original charge be represented by q, so that;

$q_{1}$ = 2q

$q_{2}$ = $\frac{q}{3}$

So that,

F = $q_{1}$ $q_{2}$ x $\frac{k}{r^{2} }$

= 2q x $\frac{q}{3}$ x $\frac{k}{r^{2} }$

= $\frac{2q^{2} }{3}$ x $\frac{k}{r^{2} }$

= $\frac{2}{3}$ x $\frac{kq}{r^{2} }$

F = $\frac{2}{3}$ x $\frac{kq}{r^{2} }$

The electric force between the given charges would change by $\frac{2}{3}$ .

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3 years ago