How does a push or pull affect motion?

Physics

1 answer:

STatiana [176]2 years ago

8 0

Answer:

The motion of an object acted on by a force depends partly on the strength of the push or pull. The stronger the push or pull, the faster the object will move.

Explanation:

More mass is the correct answer.

3 0

3 years ago

True or false you can use a light microscope to see atoms and fabric

Elza [17]

Visible light has a wavelength of 400-700nm, which means light microscopes cannot see anything smaller than this. Atoms are much smaller than this, so they are not visible to light microscopes.

8 0

3 years ago

Read 2 more answers

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$

So, the speed at point B is $3.78\times 10^5\ m/s$ .

7 0

3 years ago

Which atomic model proposed that electrons move in specific orbits around the nucleus of an atom

BabaBlast [244]

Bohr's atomic model

8 0

3 years ago