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3 years ago

What is one way in which objects transfer momentum? Explain.

Physics

2 answers:

defon3 years ago

7 0

Through vibration when it vibrates
it transfer momentum.

aksik [14]3 years ago

3 0

By colliding with the other object, an object transfer momentum.

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I will mark brainlist

tensa zangetsu [6.8K]

Answer:

False

Explanation:

A wave is a disturbance that transfers energy from one place to another without transferring matter.

6 0

3 years ago

Read 2 more answers

The wavelength of the visible line in the hydrogen spectrum that corresponds to m = 5 in the Balmer equation is: A. 656 nm. B. 4

BaLLatris [955]

Answer:

The wavelength of the visible line in the hydrogen spectrum is 434 nm.

Explanation:

It is given that, the wavelength of the visible line in the hydrogen spectrum that corresponds to n₂ = 5 in the Balmer equation.

For Balmer series, the wave number is given by :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

R is the Rydberg's constant

For Balmer series, n₁ = 2. So,

$\dfrac{1}{\lambda}=1.097\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{5^2})$

$\lambda=4.34\times 10^{-7}\ m$

$\lambda=434\ nm$

So, the wavelength of the visible line in the hydrogen spectrum is 434 nm. Hence, this is the required solution.

6 0

3 years ago

How are gravity and friction similiar

oee [108]

Gravity is the force that attracts to bodies toward each other friction is the resistance from one thing to another

8 0

3 years ago

A driver of a 1900 kg car traveling with 460,000 J of kinetic energy must slam on the brakes in order to avoid hitting a deer in

elena-s [515]

Uhhh about 40 miles probably so round that

7 0

3 years ago

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

IgorLugansk [536]

Answer:1.5

Explanation:

Given

mass of first cart $m_1=6 kg$

initial Velocity $u_1=3 m/s$

mass of second cart $m_2=3 kg$

$u_2=0 m/s$

In the absence of External Force we can conserve momentum

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$

$v=\frac{6\times 3+3\times 0}{6+3}$

$v=2 m/s$

Final kinetic Energy of two masses

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}\cdot (3+6)\cdot (2)^2$

$K.E._2=18 J$

Initial Kinetic Energy

$K.E._1=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2$

$K.E._1=\frac{1}{2}6\times 3^2+0$

$K.E._1=27 J$

$ratio =\frac{K.E._1}{K.E._2}=\frac{27}{18}=1.5$

5 0

3 years ago