All of the following statements about protons are true except?

A 7.00-kg object accelerates from rest to a final velocity in 55 seconds. If the magnitude of the

Len [333]

Answer:

The final velocity of the object is 330 m/s.

Explanation:

To solve this problem, we first must find the acceleration of the object. We can do this using Newton's Second Law, given by the following equation:

F = ma

If we plug in the values that we are given in the problem, we get:

42 = 7 (a)

To solve for a, we simply divide both sides of the equation by 7.

42/7 = 7a/7

a = 6 m/s^2

Next, we should write out all of the information we have and what we are looking for.

a = 6 m/s^2

v1 = 0 m/s

t = 55 s

v2 = ?

We can use a kinematic equation to solve this problem. We should use:

v2 = v1 + at

If we plug in the values listed above, we should get:

v2 = 0 + (6)(55)

Next, we should solve the problem by performing the multiplication on the right side of the equation.

v2 = 330 m/s

Therefore, the final velocity reached by the object is 330 m/s.

Hope this helps!

7 0

4 years ago

Calculate the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal. Assume that the spe

goblinko [34]

Answer:

the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal is 1835.16 .

Explanation:

We know, wavelength is expressed in terms of Kinetic Energy by :

$\lambda=\dfrac{h}{\sqrt{2mE}}$

Therefore , $E=\dfrac{h^2}{2 \lambda^2 m}$

It is given that both electron and proton have same wavelength.

Therefore,

$E_e=\dfrac{h^2}{2 \lambda^2 m_e}$ .... equation 1.

$E_p=\dfrac{h^2}{2 \lambda^2 m_p}$ .... equation 2.

Now, dividing equation 1 by 2 .

We get ,

$\dfrac{E_e}{E_p}=\dfrac{\dfrac{h^2}{2 \lambda^2 m_e}}{\dfrac{h^2}{2 \lambda^2 m_p}}\\\\\\\dfrac{E_e}{E_p}=\dfrac{m_p}{m_e}$

Putting value of mass of electron = $9.1\times 10^{-31}\ kg$ and mass of proton = $1.67\times 10^{-27}\ kg.$

We get :

$\dfrac{E_e}{E_p}=\dfrac{1.67\times 10^{-27}\ kg}{9.1\times 10^{-31}\ kg}=1835.16$

Hence , this is the required solution.

4 0

4 years ago

Read 2 more answers

In part d, you saw that elements with even atomic numbers tend to be more abundant than neighboring elements with odd atomic num

jolli1 [7]

The atomic procedure clarifies why this is the situation is beginning from carbon ( the nuclear number is 6), the most widely recognized atomic responses include the combination of an extra helium core. I hope the answer will help you.

8 0

3 years ago

4.) mi/hr/s and m/s/s are units for

egoroff_w [7]

Answer:

Acceleration

Explanation:

Acceleration has units of length per time squared.

6 0

3 years ago

A jogger jogs around a circular track with a diameter of 275 m in 14.0 minutes. what was the jogger's average speed in m/s?

Pie

The jogger's average speed 1.03 m/s

<h3>The Speed and the Velocity of a Particle in a Circle</h3>

The speed of a particle is a circle will always be constant while the velocity will not. That is, velocity varies.

Given that a jogger jogs around a circular track with a diameter of 275 m in 14.0 minutes. First convert the minutes to seconds

Given parameters are;

Diameter = 275 m

Radius r = D/2 = 137.5 m

Time t = 14 minutes = 14 x 60s = 840 s

Speed V = 2πr ÷ t

V = ( 2 × π × 137.5 ) ÷ 840

V = 863.9 / 840

V = 1.028 m/s

Therefore, the jogger's average speed 1.03 m/s approximately

Learn more about Circular Motion here: brainly.com/question/20905151

#SPJ1

5 0

2 years ago