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3 years ago

Can You help me understand how to calculate net force please

Physics

1 answer:

VladimirAG [237]3 years ago

8 0

To start off.. add any and all the vectors acting on the same axis (x and y), making sure to pay attention to the director of the vectors. you are often able to ignore many of the forces that cancel each other out. make sure to focus on the forces that are actualy involved in how the object will move. Then you calculate the net force acting on the object in the situation
Since nothing is happening along the y-axis, you can ignore the Fg (300N)and FN (20N) forces.

FNET = Fa + Ff

= 80N + -70N

FNET = 10N

you have to make the friction a negative force
because it is pointing in the direction opposite to the applied force.

When you want to calculate the acceleration of an object, always use the net force acting on it.

you aren’t subtracting the two forces.
Instead, it’s up to you to remember that if one of the forces is pointing in one direction (like to the right), and the other force is pointing in the other direction (like to the left), you will need to make one positive and the other negative.

i don’t really know how to explain this super well. i hope this helps!!

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When the system pressures are lower than the pressure in the refrigerant tank, such as when the system has just been evacuated o

steposvetlana [31]

Answer: True

Explanation:

It should be noted that when refrigerant is added in vapor form to the operating refrigeration system, then the addition of the refrigerant should be to the low-pressure side of the refrigeration system.

Furthermore, when system pressures are lower than the pressure in the refrigerant tank as stated in the question, then the refrigerant can be added to both the high and low pressure sides of the system.

Therefore, the correct option is true.

5 0

3 years ago

A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.

gavmur [86]

Answer:

(a) $E = -1.02 \times 10^5~N/C$

(b) $E = -9.7 \times 10^4~N/C$

Explanation:

(a) The electric field for a point charge is given by the following formula:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}$

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

$\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}$

Hence, dE is now:

$dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}$

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

$\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}$

Therefore, vertical component of dE now becomes

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})$

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

$E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C$

(b) We will have a similar approach, but a simpler integral.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi$

$E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C$

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0

3 years ago

An apple is thrown across the cafeteria with a force of 10 N and at an acceleration of 6 m/s2. What is the mass of the apple? ..

Komok [63]

Answer:

Explanation:

F = m * a

The apple's acceleration is not influenced by the acceleration due to gravity for this question. In real life it most certainly is influenced by gravity.

F = m * a

F = 10 Newtons.

a = 6 m/s^2

m = 10/6 = 1.66 kg. Mighty large apple

7 0

4 years ago

A rock is dropped in a pond, causing circular ripples. The radius increased with a rate 1 foot per second. When the radius is 4

Brrunno [24]

Answer:

true

Explanation:

Area = πr²

Differentiating in respect to r

we get,

dA/dr = 2πr

using chain rule,

$\frac{dA}{dt} = \frac{dA}{dr} .\frac{dr}{dt} = 2\pi \frac{dr}{dt}$

since r = 4feet

so ,

2π(4) = 8π

5 0

3 years ago

Starting with the definition 1.00 in. = 2.54 cm, find the number of kilometers in 8.00 mi .

velikii [3]

Use Factor-Label Method:

8miles 63360 inches
---------- X --------------------- X
1 1 mile

2.54cm 1 meter
X ------------ X ---------------- X
1 inch 100 cm

1 km
----------------- = 12.87 km
1000meters

8 miles = 12.87 km

8 0

4 years ago