Rohith has been jogging to the bus stop for 2.0 minutes at 3.5 m/s when he looks at

Estimate the force required to bind the two protons in the He nucleus together. (Hint: Model the protons as point charges. Assum

scoray [572]

Answer:

F = 2.30 10⁴ N

Explanation:

The force required to link two gates must be equal to or greater than the electrostatic force of repulsion, because the protons have equal charges.

F = k q₁ q₂ / r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C²

In this case the proton charge is 1.6 10⁻¹⁹ C and the distance between them is approximately the diameter of the core r = 10⁻¹⁵ m

Let's calculate

F = 8.99 10⁹ (1.6 10⁻¹⁹)² / (10⁻¹⁵)²

F = 2.30 10⁴ N

The bond strength must be equal to or greater than this value

3 0

3 years ago

A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output

Arte-miy333 [17]

Answer:

Output power of the charger is 9.85 watts.

Explanation:

It is given that,

Output current of the USB charger, I = 1.97 A

The standard USB output voltage, V = 5 V

We need to find the output power of the charger. It can be determined using the following formula as :

P = V × I

$P=5\ V\times 1.97\ A$

P = 9.85 watts

The output power of the charger is 9.85 watts. Hence, this is the required solution.

3 0

3 years ago

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

3 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

3 years ago

George walks to a friend's house. He walks 850 meters north before he realizes he walked too far.

anyanavicka [17]

Answer:

Explanation:

In this problem we are expected to calculate the velocity of Georges movements.

Given data

Total distance covered by George= 850+250= 1100 meters

Time taken by George to cover the total distance= 25 seconds

We know that velocity is, v= distance/ time

Therefore substituting our data into the expression for velocity we have

v= 1100/ 25= 44 m/s

Hence the velocity in m/s is 44

7 0

3 years ago