All categories

3 years ago

In the car experiment you completed what part of the scientific method would the following statement be classified as?

Physics

1 answer:

Deffense [45]3 years ago

6 0

answer:

explanation:

only got 70 followers on the gram

You might be interested in

How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?

aalyn [17]

There is too much information given, it's hard to understand exactly which variables are important in this problem.

7 0

3 years ago

Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH = 1.06

Slav-nsk [51]

Answer:

$1.06085\times 10^{-10}\ m$

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

e = Charge of electron = $1.6\times 10^{-19}\ C$

n = 1 (ground state)

Angular momentum is given by

$L=mvr$

From Bohr's atomic model we have

$L=\dfrac{nh}{2\pi}$

$mvr=\dfrac{nh}{2\pi}\\\Rightarrow v=\dfrac{nh}{2\pi mr}$

The centripetal force will balance the electrostatic force

$\dfrac{ke^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow \dfrac{ke^2}{r}=mv^2\\\Rightarrow \dfrac{ke^2}{r}=m(\dfrac{nh}{2\pi mr})^2\\\Rightarrow r=\dfrac{n^2h^2}{4\pi^2mke^2}\\\Rightarrow r=\dfrac{1^2\times (6.626\times 10^{-34})^2}{4\pi^2 \times 9.11\times 10^{-31}\times 8.99\times 10^{9}\times (1.6\times 10^{-19})^2}\\\Rightarrow r=5.30426\times 10^{-11}\ m$

The diameter is $2\times 5.30426\times 10^{-11}=1.06085\times 10^{-10}\ m$

7 0

3 years ago

A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb

NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

$\sum \tau = F*d$

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

$F*(27-6)= 6*600$

$F = \frac{6*600}{21}$

$F= 171.42 lb$

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0

3 years ago

The Earth orbits the Sun at a speed of 30 km/s. At that speed it completes one path around the Sun every year. Of course, as tha

romanna [79]

Answer:

a. 299,792,458 m/s

Explanation:

Since the speed of light in a vacuum is invariant and has the value of 299,792,458 m/s, we would measure this value of 299,792,458 m/s for the speed of light from the star as it arrives on Earth.

3 0

3 years ago

How does an electromagnet become permanent

love history [14]

<span>A moving electrical charge produces a magnetic field and a moving magnetic field produces an electrical field. An electromagnet works by coiling a bunch of wire and spinning a couple of magnets around that wire at high speeds. When this occurs the magnets induce an electric current in the wire and hence the electricity production. Once the magnets stop spinning, the induced electrical field dissipates and the current stops flowing through the wire.

</span>

6 0

3 years ago