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3 years ago

What is a mixture?

Physics

2 answers:

Kazeer [188]3 years ago

8 0

Potassium and Chlorine is a mixture.

Verizon [17]3 years ago

8 0

Answer:number fourExplanation:

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a rocket initially at rest on the ground lifts off vertically with a constant acceleration of 2.0 x 10^1 meters per second^2. Ho

xenn [34]

Here's the formula for the distance covered by an accelerating body in some amount of time ' T '. This formula is incredibly simple but incredibly useful. It pops up so often in Physics that you really should memorize it:

D = 1/2 a T²

Distance = (1/2)·(acceleration)·(time²)

This question gives us the acceleration and the distance, and we want to find the time.

(9,000 m) = (1/2) (20 m/s²) (time²)

(9,000 m) = (10 m/s²) (time²)

Divide each side by 10 m/s²:

(9,000 m) / (10 m/s²) = (time²)

900 s² = time²

Square root each side:

<em>T = 30 seconds</em>

7 0

3 years ago

A catcher in a baseball game stops a pitched ball that was originally moving at 44 m/s over a distance of 12.5 cm. the mass of t

eduard

Answer:

Force = -1161.6 Newton

Explanation:

Given the following data;

Initial velocity, u = 44m/s

Distance ,s = 12.5cm to m = 12.5/100 = 0.125m

Mass = 0.15kg

To find the acceleration;

We would use the third equation of motion;

V ² = U² + 2as

0² = 44² + 2*a*0.125

0 = 1936 + 0.25a

0.25a = -1936

a = -1936/0.25

Acceleration, a = -7744m/s2

Force = mass * acceleration

Substituting into the equation, we have;

Force = 0.15 * (-7744)

Force = -1161.6 Newton

The value of its force is negative because the glove decreases the velocity of the ball.

4 0

3 years ago

two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t

Tomtit [17]

The charges have opposite sign and magnitude $6 \mu C$

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

$q_1 = q_2 = q$

So we can rewrite the equation as

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

And solving for q:

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C$

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

$q_1 = 6 \mu C\\q_2 = -6 \mu C$