Khái niệm lực điện,đặc điểm về tương tác điện của lực điện

A box of books is in the floor. The following picture shows a push and a pull acting in the box.

nataly862011 [7]

Net force = 50 + 50 = 100 M
Since two forces act towards the same direction , total force should be their sum amd towards the same direction as the two forces

8 0

3 years ago

9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o

Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

$m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.$

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

$F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).$

Now the force of friction is $F_s=\mu*N,$ where $N$ is the normal force and from the diagram it is $F_y=mg*cos(\theta).$

Thus $F_s=\mu*N=\mu*mg*cos(\theta).$

Therefore,

$F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).$

Substituting the value for $F_H,m,\mu, and \:\theta$ we get:

$F_{tot}= -38.63N.$

Now acceleration is simply

$a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.$

The negative sign indicates that the acceleration is directed up the incline.

Part B:

$d=\frac{1}{2} at^2$

Which can be rearranged to solve for t:

$t=\sqrt{\frac{2d}{a} }$

Substitute the value of $d=0.50m$ and $a=7.7m/s$ and we get:

$t=0.36s.$

which is our answer.

Notice that in using the formula to calculate time we used the positive value of $a$ , because for this formula absolute value is needed.

5 0

4 years ago

The inner planets are relatively small bodies composed mostly of rock and metals. Why did they develop this way instead of becom

sleet_krkn [62]

The temperature of the early solar system explains why the inner planets are rocky and the outer ones are gaseous. As the gases coalesced to form a protosun, the temperature in the solar system rose. In the inner solar system temperatures were as high as 2000 K, while in the outer solar system it was as cool as 50 K. In the inner solar system, only substances with very high melting points would have remained solid. All the rest would have vaoprized. So the inner solar system objects are made of iron, silicon, magnesium, sulfer, aluminum, calcium and nickel. Many of these were present in compounds with oxygen. There were relatively few elements of any other kind in a solid state to form the inner planets. The inner planets are much smaller than the outer planets and because of this have relatively low gravity and were not able to attract large amounts of gas to their atmospheres. In the outer regions of the solar system where it was cooler, other elements like water and methane did not vaporize and were able to form the giant planets. These planets were more massive than the inner planets and were able to attract large amounts of hydrogen and helium, which is why they are composed mainly of hydrogen and helium, the most abundant elements in the solar system, and in the universe

https://lco.global/spacebook/planets-and-how-they-formed/

hope it helps

5 0

3 years ago

What is the acceleration of a proton moving with a speed of 9.5 m/s at right angles to a magnetic field of 1.8 T

Elenna [48]

Explanation:

F = qvBsin(angle)

F = 1.6 x 10^-19 x 9.5 x 1.8 x Sin (90)

F = 273.6 x 10^-20N

a = F/M

M of a proton = 1.67 x 10^-27kg

a = 273.6 x 10^-20/1.67 x 10^-27

a = 1638323353.29m/s

4 0

3 years ago

Un proyectil de masa m (kg) se dispara con una velocidad v (m/s) contra un bloque de masa 4m inicialmente en reposo. Tras la col

Maru [420]

Answer:

x = $\frac{v^2}{10 \mu g}$

Explanation:

Let's start the exercise with the definition of a system formed by the projectile and the block, in this case the forces during the collision are internal and the moment is conserved,

initial instant. Just before the crash

p₀ = m v + 4m 0

final instant. Right after the crash, before the block began to move

$p_{f}$ = (m + 4m) v_{f}

how the moment is preserved

p₀ = p_{f}

m v = 5m v_{f}

v_{f} = v/5

knowing the speed of the system (projectile + block) we can use the relationship between work and energy

W = ΔK

starting point. Just when the projectile + block system starts to move

Em₀ = K = ½ m v_{f}²

final point. When the system is stopped

Em_{f} = 0

The work of the friction force is

W = - fr x

the negative sign is because the friction force opposes the motion, let's use Newton's second law to find the friction force

Y Axis

N- W = 0

N = W = mg

The expression for the friction force is

fr = μ N

substituting

fr = μ mg

W = - μ mg x

using the energy duty ratio

- μ mg x = 0 - ½ m $v_{f}^2$

x = $\frac{v_{f}^{2} }{2 \mu g}$

we substitute speed

x = $\frac{v^2}{10 \mu g}$

7 0

3 years ago