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4 years ago

10

Why are adaptations important to organisms?

1 answer:

Serga [27]4 years ago

5 0

Adaptations can help organisms reproduce and continue living in different conditions

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A 1k ball is thrown into the air the air with an initial velocity of 30 m/s how much kinetic energy does the ball have

Artemon [7]

Ek (Kinetic Energy) = 1/2m*v^2

0.5x1x30^2
0.5x30^2
= 450J (joules)

5 0

3 years ago

if the separation between two masses is doubled, does the gravitational force between them increase or decrease? By what factor?

docker41 [41]

Answer:

increase by a factor of 4.

Explanation:

If the separation distance between two objects is doubled (increased by a factor of 2), then the force of gravitational attraction is decreased by a factor of 4 (2 raised to the second power).

3 0

4 years ago

Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls

Dmitrij [34]

Answer:

$(a) 189.23 N$ , $(b) 47.31 N$ and $(c) 141.92 N$ .

Explanation:

Three balls are shown in figure having charge $q=1.45 \mu C$ . The middle ball, $B$ , is positively charged having charge $+q$ , and the remaining two outside balls, $A$ and $C$ , are negatively charged having charged $-q$ as shown.

$AC=20 cm$ and $AB=BC=10$ cm as B is the mid-point of AC.

Let $d_1=AC=20\times 10^{-3}m$ and $d_2=AB=BC=10\times 10^{-3}m$

From Coulomb's law, the magnitude of the force, $F$ , between two point charges having magnitudes $q_1 \& q_2$ , separated by distance, $d$ , is

$F=\frac {1}{4\pi\epsilon_0}\frac {q_1q_2}{d^2}\;\cdots (i)$

where, $\epsilon_0$ is the permittivity of free space and

$\frac {1}{4\pi\epsilon_0}=9\times 10^9$ in SI units.

This force is repulsive for the same nature of charges and attractive for the different nature of charges.

Now, Using equations(i),

(a) The magnitude of attraction force between balls A and B is

$F_{AB}=F_{BC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_2)^2}$

$\Rightarrow F_{AB}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(10\times 10^{-3}\right)^2}$

$\Rightarrow F_{AB}=189.23 N$

(a) The magnitude of the repulsive force between balls A and C is

$F_{AC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_1)^2}$

$\Rightarrow F_{AC}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(20\times 10^{-3}\right)^2}$

$\Rightarrow F_{AC}=47.31 N$

(c) The magnitude of the net force, $F_{net}$ , on the outside of the ball is,

$F_{net}=189.23-47.31 N$

$\Rightarrow F_{net}=141.92 N$

4 0

3 years ago

A ball is released at the top of a ramp at t = 0. Which is the speed of the ball at t = 4?

Kruka [31]

Answer:

For uniform acceleration

a = (v2 - v1) /t = 2.5 acceleration is constant 2.5

v2 = 2.5 * 4 v1 = 0 at t = 0

v2 = 2.5 m/s^2 * 4 s = 10 m/s

5 0

3 years ago

When light reflects on a mirror, does it undergo a phase change?

Lynna [10]

The light does not undergo a phase change.

Also, the surface of a mirror is a rigid boundary.

5 0

3 years ago