Ask question

Ask question

All categories

Yakvenalex [24]

4 years ago

13

How do electrons differ from protons and neutrons? A. They do not move. B. They are larger. C. They are not in a fixed position.

D. They are located in the nucleus.

1 answer:

Debora [2.8K]4 years ago

5 0

the answer i would go with is C

You might be interested in

Train car A is at rest when it is hit by train car B. The two cars, which have the same mass, are stuck together and move off af

k0ka [10]

The final velocity is half of train car B's initial velocity.

4 0

3 years ago

Read 2 more answers

Why is density used to determine the identity of matter?

dedylja [7]

a substance's density is the same at a certain pressure and temperature, and the density of one substance is usually different than another substance.

4 0

3 years ago

A 0.18 kg baseball moving at +21.80 m/s is slowed to a stop by a catcher who exerts a constant force of -360 N.

Pachacha [2.7K]

Answer:

hey so this website called quiz-let helps you it will give u the answer for every question i use it sometimes when im confused on a test.

7 0

3 years ago

Read 2 more answers

Preston goes on a camel safari. There, he travels 5 km north, then 3 km east.

sammy [17]

7777Answer:

Explanation:

6 0

3 years ago

Read 2 more answers

A proton with mass 1.67×10⁻²⁷kg is propelled at an initial speed of 3.00×10⁵m/s directly toward a uranium nucleus 5.00 m away. T

aleksandr82 [10.1K]

Answer:

$2.4\times 10^5 m/s$

Explanation:

We are given that

Mass of proton= $m_p=1.67\times 10^{-27}kg$

Initial speed of proton= $v_i=3.00\times 10^5 m/s$

$x_i=-5m$

$\alpha=2.12\times 10^{-26}Nm^2$

$F=-\frac{\alpha}{x^2}$

Where x is the separation between the two objects.

We have to find the speed of the proton when it is $8.00\times 10^{-10} m$ from the uranium nucleus.

It means we have to find $v_f \;if x_f=-8.00\times 10^{-10} m$

By work energy theorem

W= $\Delta K.E=\int_{x_i}^{x_f} Fdx$

Substitute the values

$W=\Delta K.E=\int_{x_i}^{x_f} -\frac{\alpha}{x^2}dx$

$W=-\int_{-5}^{-8\times 10^{-10}} \alpha x^{-2}dx$

$W=\alpha[x^{-1}]^{-8\times 10^{-10}}_{-5}$

$W=2.12\times 10^{-26}\times(-\frac{1}{8\times 10^{-10}}+\frac{1}{5})$

$W=-2.65\times 10^{-17}J$

$\frac{1}{2}m(v^2_f-v^2_i)=-2.65\times 10^{-17}$

$\frac{1}{2}\cdot 1.67\times 10^{-27}}(v^2_f-(3\times 10^5)^2)=-2.65\times 10^{-17}$

$v^2_f-9\times 10^{10}=\frac{2\cdot (-2.65)\times 10^{-17}}{1.67\times 10^{-27}}$

$V^2_f-9\times 10^{10}=-3.17\times 10^{10}$

$v^2_f=-3.17\times 10^{10}+9\times 10^{10}$

$v^2_f=5.83\times 10^{10}$

$v_f=\sqrt{5.83\times 10^{10}}=2.4\times 10^5 m/s$

Hence, the speed of proton when it is $8\times 10^{-10} m$ from the uranium nucleus= $v_f=2.4\times 10^5 m/s$

6 0

3 years ago