All categories

2 years ago

Which could describe the motion of an object?

Physics

2 answers:

Volgvan2 years ago

3 0

mhm mhmhmhmhmmhmhmmh

shutvik [7]2 years ago

3 0

Mhm mhmhmhmmmmmhmhmm

You might be interested in

What is E of a hydrogen atom in the 3p state?

notsponge [240]

Answer:

E=-1.51 eV.

$L=\hbar\sqrt{2}$

Explanation:

The nth level energy of a hydrogen atom is defined by the formula,

$E_{n}=-\frac{13.6}{n^{2} }$

Given in the question, the hydrogen atom is in the 3p state.

Then energy of n=3 state is,

$E_{n}=-\frac{13.6}{(3)^{2} }\\E_{n}=-1.51eV$

Therefore, energy of the hydrogen atom in the 3p state is -1.51 eV.

Now, the value of L can be calculated as,

$L=\hbar\sqrt{l(l+1)}$

For 3p state, l=1

$L=\hbar\sqrt{1(1+1)}\\L=\hbar\sqrt{2}$

Therefore, the value of L of a hydrogen atom in 3p state is $L=\hbar\sqrt{2}$ .

4 0

3 years ago

Which of the following describes an element?

bixtya [17]

Answer:

A substance made of two or more types of atoms

Explanation:

8 0

2 years ago

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road

m is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

$v=85 mph \cdot \frac{1609}{3600}=38.0 m/s$ is the speed

And solving for $\mu$ , we find the coefficient of friction required to keep the car in circular motion:

$\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

2 years ago

A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.

stiks02 [169]

Ignoring air resistance, the bullet's horizontal velocity is constant:

$v_x=v_{0x}=100\,\dfrac{\mathrm m}{\mathrm s}$

In 1.3 seconds, we can expect it to travel

$v_xt=\left(100\,\dfrac{\mathrm m}{\mathrm s}\right)(1.3\,\mathrm s)=130\,\mathrm m$

4 0

3 years ago

Electric charge is distributed over the disk x2 + y2 ≤ 4 so that the charge density at (x, y) is rho(x, y) = 4x + 4y + 4x2 + 4y2

maw [93]

Answer:

Q=185.84C

Explanation:

We have to take into account the integral

$Q=\int \rho dV$

In this case we have a superficial density in coordinate system.

Hence, we have for R: x2 + y2 ≤ 4

$Q=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\rho dydx$

but, for symmetry:

$Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C$

HOPE THIS HELPS!!

8 0

2 years ago