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Artyom0805 [142]
3 years ago
15

Laws of motion are important for the study of objects that are not in motion and objects in motion. Is it true or false?

Physics
2 answers:
Triss [41]3 years ago
8 0
True because objects at rest have a law too
e-lub [12.9K]3 years ago
3 0
Gotta be true because physics
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What type of current is produced by a battery?
padilas [110]
The positive plate is made of lead grid with a lead oxide paste coating and the negative plate is porous lead. Dilute sulphuric acid is the electrolyte. The lead acid battery produced a direct current or DC during a chemical reaction that turns lead dioxide to lead sulphate and sulphuric acid to water.
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3 years ago
A flat loop of wire consisting of a single turn ofcross-sectional area 7.90 cm2is perpendicular to a magnetic field that increas
Westkost [7]

To develop this problem, it is necessary to apply the concepts related to Faraday's law and Magnetic Flow, which is defined as the change that the magnetic field has in a given area. In other words

\Phi = BA Cos\theta

Where

B= Magnetic Field

A = Area

\theta = Angle between magnetic field lines and normal to the area

The differentiation of this value allows us to obtain in turn the induced emf or electromotive force.

In this case we have that the flat loop of wire is perpendicular to the magnetic field, therefore the angle is 0 degrees, since its magnetic field acts parallel to the area:

\theta = 0 then our expression can be written as

\Phi = BA

From the same value of the electromotive force we have to

\epsilon = -\frac{d\Phi}{dt}

Replacing we have

\epsilon = -A\frac{B}{dt}

Replacing with our values we have that

\epsilon = -(7.9*10^{-4}m)\frac{(3.5-0.5)}{0.1}

\epsilon = -0.0237V

Therefore the magnitude of the induced emf in the loop is 0.0237V

On the other hand we have that the current by Ohm's Law can be defined as

I = \frac{\epsilon}{R}

For the given value of the resistance and the previously found potential we have to

I = \frac{0.237}{1.3}

I= 0.0182A

6 0
3 years ago
Robin fired a bullet of mass 100 gm from a gun of mass 5 kg. The bullet leaves the gun with a speed of 400 m/s. After penetratin
Marina CMI [18]

(a) the recoil or backward velocity of the gun is 8 m/s.

(b) the bullet cannot completely penetrate the plank.

<u>The given parameters include;</u>

mass of the bullet, m₁ = 100 g = 0.1 kg

mass of the gun, m₂ = 5 kg

initial velocity of the bullet, u₁ = 400 m/s

thickness of the plank, x = 10 cm

<u>To calculate the following:</u>

(a) the backward velocity of the gun

let the backward velocity of the gun = u₂

Apply the principle of conservation of linear momentum.

m₁u₁  + m₂u₂ = 0

m₂u₂ = -m₁u₁

u_2 = -\frac{m_1u_1}{m_2} \\\\u_2 = - \frac{0.1 \times 400}{5} \\\\u_2 = -8 \ m/s

Thus, the recoil or backward velocity of the gun is 8 m/s.

(b) Can the bullet penetrate the plank of the wood completely ?

  • the bullet traveled 4 cm and lost ¹/₃ of u₁
  • the remaining distance to completely penetrate the plank = 6 cm
  • the final velocity of the bullet at 4 cm, v = 400 - ¹/₃ x 400 m/s  = 266.67 m/s

the acceleration of the bullet is calculated as;

v² = u₁² + 2as

2as = v² - u²

a = \frac{v^2 -u_1^2}{2s} \\\\a = \frac{(266.67)^2 -(400)^2}{2\times 0.04} = -1.111 \times 10^6 \ m/s^2

Finally, determine the distance traveled by the bullet when it comes to a complete stop, that is the final velocity = 0

v_f^2 = v^2 + 2ad\\\\2ad = v_f^2 - v^2\\\\d = \frac{v_f^2 - v^2}{2a} \\\\d = \frac{(0) - (266.67)^2}{2(-1.111\times 10^6)} \\\\d = 0.032 \ m

d = 3.2 cm

The total distance traveled by the bullet inside the plank = 4 cm + 3.2 cm = 7.2 cm

Therefore, the bullet cannot completely penetrate the plank.

<u>To learn more about linear momentum visit: </u>brainly.com/question/15869303

6 0
3 years ago
Can someone help me figure out which one it is
butalik [34]

Answer:

<h2>X=dt</h2>

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hope it helps...

7 0
3 years ago
A baseball is thrown towards home plate. It takes 3.4 seconds for the ball to reach the catcher's glove from the pitcher's glove
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Answer:

57

Explanation:

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