What may have been the first step in the formation of the solar system?

A 873-kg (1930-lb) dragster, starting from rest completes a 401.4-m (0.2509-mile) run in 4.945 s. If the car had a constant acce

Delvig [45]

To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

$x=x_0+v_0t+at^2$

Where

$x_0 =$ Initial position

$v_0 =$ Initial velocity

a = Acceleration

t= time

And the velocity can be expressed as,

$v_f = v_0 + at$

Where,

$v_f = Final velocity$

For our case we have that there is neither initial position nor initial velocity, then

$x= at^2$

With our values we have $x = 401.4m, t=4.945s$ , rearranging to find a,

$a=\frac{x}{t^2}$

$a = \frac{ 401.4}{4.945^2}$

$a = 16.41m/s^2$

Therefore the final velocity would be

$v_f = v_0 + at$

$v_f = 0 + (16.41)(4.945)$

$v_f = 81.14m/s$

Therefore the final velocity is 81.14m/s

8 0

3 years ago

in a class where the number of girls is 36% of the total number,there are 48 boys.how many students are there in the class?

Pavlova-9 [17]

Answer:

There are 75 people in the class. The number of boys is 48 and the number of girls is 27. The percentage of girls is 36% of 75.

Explanation:

3 0

3 years ago

Choose a substance you're familiar with. what are its physical and chemical properties? How would you measure its density? What

Gekata [30.6K]

you can look at magnesium, it can react with oxygen to form oxides. (chemical) it is malleable and a solid at room temperature. (physical)

to measure its density, the mass and volume can be worked out and from this density too. look up the equation, it is quite easy :)
physical changes -- it can be melted, and oxidized

the chemical changes of oxidation magnesium looses electrons to form oxides, this is a chemical reaction- chemical change..--- use to get the density use (rho) or density D = M/V

6 0

2 years ago

What is the electric potential, i.e. the voltage, 0.30 m from a point charge of 6.4 x 10-C?

Gwar [14]

Answer:

V = 192 kV

Explanation:

Given that,

Charge, $q=6.4\times 10^{-6}\ C$

Distance, r = 0.3 m

We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

$V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6.4\times 10^{-6}}{0.3}\\\\V=192000\ V\\\\V=192\ kV$

So, the required electric potential is 192 kV.

3 0

2 years ago

What is a free electron? Can someone help?

Aleksandr [31]

A free electron is one which has become detached from a covalent bond between two atoms and is able to move around from atom to atom and possibly take part in electric current flow.

3 0

3 years ago