1) A man leans against the wall and...

How do I find net force when something has arrows all four directions with different N values for each?

Nimfa-mama [501]

In the vertical direction, take up to be positive and down to be negative. Then the net vertical force would be

5120 N - 4050 N = 1070 N

(it's positive, so the net vertical force is pointing upward)

In the horizontal direction, take right to be positive and left to be negative. Then the net horizontal force would be

950 N - 1520 N = -570 N

(negative means the net horizontal force points to the left)

So the net force on the balloon is the vector

F = (1070 N) i + (-570 N) j

(where i and j are the unit vectors in the horizontal and vertical directions, respectively)

The magnitude of the net force on the balloon is the magnitude of this vector:

F = √((1070 N)² + (-570 N)²)

F ≈ 1212 N

8 0

3 years ago

In proton beam therapy, a beam of high-energy protons is used to kill cancerous cells in a tumor. In one system, the beam, which

Thepotemich [5.8K]

The number of protons that strike the tumor each second is $1.75 \times 10^8$ protons.

Given the following data:

Energy of proton = $2.8 \times 10^{-11}$ Joules.

Current = 84 nA to A = $84 \times 10^{-9}$ Ampere.

Energy of total dose = $3.4 \times 10^{-3}$ Joules.

Scientific data:

Charge of proton = $1.602 \times 10^{-19}$ C.

To determine the number of protons that strike the tumor each second:

<h3>How to determine the number of protons.</h3>

Mathematically, the quantity of charge per unit time is given by this formula:

$Q =It=Ne\\\\N = \frac{Q}{e}$

Where:

N is the number of protons.

t is the time measured in seconds.

I is the current.

e is the charge of protons.

Substituting the parameters into the formula, we have;

$N=\frac{2.8 \times 10^{-11} }{1.602 \times 10^{-19}} \\\\N = 1.75 \times 10^{8}$ protons.

Read more on charges here: brainly.com/question/14372859

8 0

2 years ago

If Susan exerted a force of 5 newtons and the output force of the machine is 15

Anon25 [30]

Answer:

Mechanical advantage = 3

Explanation:

Mechanical advantage, MA = Output force/Input force

Output force = 15 N. Input force = 5 N

MA = Output force/Input force = 15 N/ 5 N = 3

4 0

4 years ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

4 years ago

What is the magnitude of a point charge in coulombs whose electric field 58 cm away has the magnitude 2.8 N/C?

Yuliya22 [10]

Answer:

q = 1.05 × 10^-10 C

Explanation:

E = 2.8 N/C

r = 58cm = 0.58m

k = 9×10^9

q = ?

E = f/q

E = kq/r²

q = E*r²/k

q = (2.8×0.58²)÷9×10^9

q = 1.05 × 10^-10 C

8 0

3 years ago