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2 years ago

7

A metal ball with a charge of -8 C is brought in contact with a similar metal ball that had a charge of +2 C. How much charge wi

ll they have after they touch ?

1 answer:

laila [671]2 years ago

5 0

I think the answer is -6

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i got the right answer for this question (pretty sure i guessed or just asked me teacher the answer?) but can someone explain ho

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A guessing answer the best answer but you have had to subtract the answer by the equation that it was giving u

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3 years ago

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How much potential energy does a 1kg mess have 10m off the ground?

777dan777 [17]

Potential Energy = mass x gravitational acceleration x height
potential Energy = 1 x 9.8 x 10 = 98 joules

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3 years ago

Calculate the distance traveled by a projectile as a function of launch angle. Compare the distances for two projectiles launche

DaniilM [7]

Answer:

$R = x_{max} = \frac{v^2\sin(2\theta)}{g}\\\frac{R_1}{R_2} = \frac{\sin(2\theta_1}{\sin(2\theta_2}$

Explanation:

Using kinematics equations:

$\Delta x = v_{0x}t\\\Delta y = -\frac{1}{2}gt^2+v_{0y}t$

Use $\Delta y = 0$ due to condition of distance traveled.

Solving second equation for time, there are two solutions. t=0 and

$t=\frac{2v_{0y}}{g}$

Use the expression in the first equation to have

$R = \frac{2v^2 \cos\theta\sin\theta}{g}$

Using trigonometric identities, you have the answer of the distance.

By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.

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3 years ago

44.2 cm + 0.123 cm = cm

trapecia [35]

The answer is 44.323 cm

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3 years ago

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A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago