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1 year ago

Calculate the net force acting on the box in the following. refer to the screenshot below.

Physics

1 answer:

Katarina [22]1 year ago

3 0

The net force acting on the box is 6N

The force are unbalanced

<h3>Explain the concept?</h3>

The net force on an object is the sum of all the individual forces acting on it.

If two opposing forces act on an object, the net force is the difference between the two forces.

When two forces act in the same direction on an object, the net force is the sum of the two forces.

Here the two force 3N and 3N acting on same direction, so

net force = 3N + 3N

= 6N

Forces that are balanced are those that are opposite in direction and equal in size, but here the equal forces acts in same direction so it is unbalanced

To learn more about net force refer to :

brainly.com/question/18038995

#SPJ13

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A series circuit has a current of 2 Amperes and an equivalent resistance of 4 Ohms. What is the power of this circuit?

Mekhanik [1.2K]

Power = Current² * Resistance

Power = 2² * 4

= 4 * 4

Power = 16 Watts

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3 years ago

An electron is projected with horizontal speed 105m / s in a downwardly directed 304N / C electric field. Find the vertical posi

Vinil7 [7]

Answer:

-107 m

Explanation:

Sum of forces in the y direction:

∑F = ma

-qE = ma

a = -qE/m

a = -(1.60×10⁻¹⁹ C) (304 N/C) / (9.11×10⁻³¹ kg)

a = -53.4×10¹² m/s²

Given in the y direction:

v₀ = 0 m/s

a = -53.4×10¹² m/s²

t = 2×10⁻⁶ s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (2×10⁻⁶ s) + ½ (-53.4×10¹² m/s²) (2×10⁻⁶ s)²

Δy = -107 m

4 0

3 years ago

A ball is thrown upward at 25 m/s from the ground. what is the balls velocity after 4 seconds?

mixas84 [53]

Answer:

-14.2m/s

Explanation:

Given parameters:

Initial velocity of the ball = 25m/s

Time = 4s

Unknown:

Final velocity of the ball = ?

Solution:

To solve this problem, we use the expression below;

v = u - gt

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

t is the time

v = 25 - (9.8 x 4) = -14.2m/s

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3 years ago

An 8 g bullet leaves the muzzle of a rifle with

Elena-2011 [213]

Answer: 1872 N

Explanation:

This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:

$V^{2}=V_{o}^{2} + 2ad$ (1)

$F=ma$ (2)

Where:

$V=611.9 m/s$ is the bullet's final speed (when it leaves the muzzle)

$V_{o}=0$ is the bullet's initial speed (at rest)

$a$ is the bullet's acceleration

$d=0.8 m$ is the distance traveled by the bullet before leaving the muzzle

$F$ is the force

$m=8 g \frac{1 kg}{1000 g}=0.008 kg$ is the mass of the bullet

Knowing this, let's begin by isolating $a$ from (1):

$a=\frac{V^{2}}{2d}$ (3)

$a=\frac{(611.9 m/s)^{2}}{2(0.8 m)}$ (4)

$a=234013.5063 m/s^{2} \approx 2.34(10)^{5} m/s^{2}$ (5)

Substituting (5) in (2):

$F=(0.008 kg)(2.34(10)^{5} m/s^{2})$ (6)

Finally:

$F=1872 N$

4 0

3 years ago

During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 1.4

Ksju [112]

Answer:

The shell hit at a horizontal distance 149.41 km

Explanation:

The shell had an initial speed of 1.4 km/s

Initial speed = 1400 m/s

Inclination = 65.8° to the horizontal

First let us find the time of travel

Consider the vertical motion,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 1400 sin 65.8 = 1276.97 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = 0 m

Substituting

s = ut + 0.5 at²

0 = 1276.97 x t + 0.5 x (-9.81) xt²

t = 260.34 s

Now let us find the horizontal displacement,

Consider the horizontal motion,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 1400 cos 65.8 = 573.89 m/s

Acceleration, a = 0 m/s²

Time, t = 260.34 s

Substituting

s = ut + 0.5 at²

s = 573.89 x 260.34 + 0.5 x 0 x 260.34²

s = 149407.11 m = 149.41 km

The shell hit at a horizontal distance 149.41 km

5 0

3 years ago