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Vera_Pavlovna [14]

3 years ago

15

In this chapter, we treated force as a push or pull, now we say it is part of an interaction. Is a force a push or pull, or part

of an interaction?

1 answer:

mafiozo [28]3 years ago

8 0

Answer:

A force can be described as both.

A force is a push or pull upon an item coming about because of the object's connection with another object.

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A deuteron particle consists of one proton and one neutron and has a mass of 3.34x10^-27 kg. A deuteron particle moving horizont

nikklg [1K]

_dThe radius of curvature of a subatomic particle under a magnetic field is given by the following formula:

$r=\frac{mv}{qB}$

Where:

$\begin{gathered} r=\text{ radius} \\ v=\text{ velocity} \\ q=\text{ charge} \\ B=\text{ magnetic field} \end{gathered}$

We can determine the quotient between the velocity and the charge of the deuteron particle from the formula. First, we divide both sides by the mass:

$\frac{r_d}{m_d}=\frac{v}{q_B_}$

Now, we multiply both sides by the magnetic field "B":

$\frac{Br_d}{m_d}=\frac{v}{q}$

Since the charge of the deuterion is the same as the charge of the proton and the velocity we are considering are the same this means that the quotient between velocity and charge is the same for both particles. Therefore, we can apply the formula for the radius again, this time for the proton:

$r_p=\frac{m_pv}{qB}$

And substitute the quotient between velocity and charge:

$r_p=\frac{m_p}{B}(\frac{Br_d}{m_d})$

Now, we cancel out the magnetic field:

$r_p=\frac{m_pr_d}{m_d}$

Now, we substitute the values:

$r_p=\frac{(1.67\times10^{-27}kg)(0.385m)}{(3.34\times10^{-27}kg)}$

Solving the operations:

$r_p=0.193m=19.3cm$

Therefore, the radius is 19.3 cm.

3 0

2 years ago

Which of the following is an example of an anaerobic exercise?

Nataly [62]

Sitting. Because its no move at all

7 0

3 years ago

3. Infiltration is the process by which rainwater becomes

amid [387]

<em></em>Your answer is :<em>

A.
</em>
Groundwater.<em>

</em>

4 0

4 years ago

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn doo

Tresset [83]

Answer:

The work done is -209.42 J.

Explanation:

F(x) = (- 20 - 3 x ) N

x = 0 to x = 6.9 m

Here, the force is variable in nature, so the work done by the variable force is given by

$W =\int F dx\\\\W =\int_{0}^{6.9}(-20- 3x dx )\\\\W= \left [ - 20 x - 1.5 x^2 \right ]_{0}^{6.9}\\\\W = - 20 (6.9 - 0) - 1.5(6.9\times 6.9 - 0)\\\\W =- 138 - 71.42\\\\W = - 209.42 J$

7 0

3 years ago

How long does it take to do 300 Joules of work with 10 watts of power? 30 sec 20 sec .03 sec .30 sec

Oksana_A [137]

Answer:

Explanation:

Power=work /time

10watts=300J/time

Time =30 sec

4 0

3 years ago