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Delvig [45]
4 years ago
13

state the type of force which provide the needed centripetal force in the following instance :an electron orbiting the nucleus

Physics
1 answer:
Vladimir [108]4 years ago
6 0

i think the answer is electrostatic force hope this helps u stay safe

You might be interested in
Through which media did sound waves travel the fastest and the slowest?
In-s [12.5K]

Answer:

Sound waves travels fastest in solids.

Sound waves travels slowest in gases.

Explanation:

Sound waves travels fastest in solids.

Sound waves travels slowest in gases.

7 0
4 years ago
A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp
Anna71 [15]

\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}

Explanation:

       Natural length of a spring is 0.5\text{ }m. The spring is streched by 0.02\text{ }m. The resultant energy of the spring is 0.5\text{ }J.

       The potential energy of an ideal spring with spring constant k and elongation x is given by \dfrac{1}{2}kx^{2}.

       So, in the current problem, the natural length of the spring is not required to find the spring constant k.

       \text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}

∴ The spring constant of the spring = 2500\text{ }\frac{kg}{s^{2}}

4 0
4 years ago
A block is launched up a frictionless 40° slope with an initial speed v and reaches a maximum vertical height h. The same block
Strike441 [17]

Answer: 1. h

Explanation:

The block would reach exactly the same height from the ground. It would travel a greater distance away from the source, but the height away from the earth would remain the same as you are giving it the same energy each time. Therefore, it will reach the same gravitation potential energy.

Another approach to look at it this is seeing it when the Block moves up the slope, its kinetic energy decreases and the potential energy increases. In both cases, the kinetic energy decreases by same amount, therefore the block rises to same height H.

Try to use the formula;

1/2MV2 = mgh

Where V = √(2gh)

I hope this helps

3 0
3 years ago
What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
if pete (mass =90.00kg) weighs himself and finds that he weighs 30.0 pounds, how far away from the earth is he?
Svet_ta [14]

Here we know that mass of the person is 90 kg

His weight is given as 30 lbf

so here we can convert it into Newton as we know that

1 lbf = 4.45 N

Now from above conversion

30 lbf = 30 \times 4.45 = 133.45 N

now we can use this to find the gravity at this height

mg = 133.45

90\times g = 133.45

g = 1.49 m/s^2

now we know that with height gravity varies as

g' = g(\frac{R}{R+H})^2

1.49 = 9.8(\frac{6.37\times 10^6}{6.37\times 10^6 + h})^2

h = 1.0 \times 10^7 m

so above is the height from the surface of earth

4 0
3 years ago
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