Suppose that at room temperature, a certain aluminum bar is 1.0000 m long. The bar gets longer when its temperature is raised. The length l of the bar obeys the following relation: l=1.0000+2.4×10−5T, where T is the number of degrees Celsius above room temperature. What is the change of the bar's length if the temperature is raised to 14.1 ∘C above room temperature?

Answers

Answer 1

Answer:

Answer:

0.00034 m

Explanation:

Since the length of the aluminium bar, L is given by , L = 1.0000 + 2.4 × 10⁻⁵T and T = 14.1°C, we substitute the value of T into L. So, we have L = 1.0000 + 2.4 × 10⁻⁵ × 14.1°C = 1.0000 + 0.0003384 = 1.0003384 m. The change in length is thus 1.0003384 - 1.0000 = 0.0003384 m ≅ 0.00034 m

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The force of repulsion between two like-charged particles will increase ifA) the distance between the charged particles is increased.

B) the distance between the charged particles is decreased.

C) the magnitude of the charge on both the particles is decreased.

D) the magnitude of the charge on one of the particles is decreased.

Answers

The force of repulsion between two like-charged particles will increase if the distance between the charged particles is decreased.

What is coulomb's law?

According to Coulomb's law: The magnitude of each of the electric forces with which two point-at-rest charges interact is directly proportional to the product of the magnitude of both charges

And inversely proportional to the square of the distance that separates them and has the direction of the line that joins them.

The formula will be given as:

$F=(kq_1q_2)/(r^2)$

We can see that the force is inversely proportional to the distance of the two charges.

To know more about coulomb's law follow

brainly.com/question/66110

Answer:

B). the distance between the charged particles is decreased.

Explanation:

1. A rifle with a longer barrel can fire bullets with a larger velocity than a rifle with a shorter barrel.

Answers

i dont believe so because thw longer the barrel the longer the bullet is inthe rifle shortening the force from the friction with the barrel so a rifle with a shorter barrel would be betterto use than one with a lomger barrel even though the longer barreled rifle may be more presice

If vx = 0.40 meters/second and vy = 3.00 meters/second, what is the value of θ?

Answers

θ is the angle whose tangent is 3/0.4 = arctan (7.5) = 82.4°

(That's the angle above the horizontal. If that doesn't match
the θ in your diagram, too bad ! For some strange reason,
I wasn't able to see your diagram.)

What is the acceleration due to gravity at a location where a 15 kilogram mass weighs 45 newtons?

Answers

   Weight = (mass) x (acceleration of gravity) .

   45 N = (15 kg) x (acceleration of gravity) .

Divide each side by (15 kg):

   Acceleration of gravity = (45 N) / (15 kg)

   = 3 m/s² .

The acceleration due to gravity at the place where the weight of the body is $45\,{\text{N}}$ is $\boxed{3\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}$ .

Further Explanation:

Given:

The mass of the body is $15\,{\text{kg}}$ .

The weight of the body is $45\,{\text{N}}$ .

Concept:

The weight of an object or a body is the measure of the force of the gravity or the Earth’s gravitational force acting on the body. It is the amount of force with which the body is being pulled towards the center of the Earth.

The weight of the body is expressed as the product of the mass and the acceleration due to gravity on the surface.

The expression for the Weight of any body is written as:

$\boxed{W=mg}$

Here, $W$ is the weight of the body, $m$ is the mass and $g$ is the acceleration due to gravity.

Substitute the values of mass and the weight in above expression.

$\begin{aligned}45&=15*g\ng&=\frac{{45}}{{15}}\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\n&=3\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\n\end{aligned}$

Thus, the acceleration due to gravity at the place where the weight of the body is $45\,{\text{N}}$ is $\boxed{3\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

Learn More:

1. A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 N brainly.com/question/7031524

2. A horizontal rope is tied to a 50 kg box brainly.com/question/2416204

3. Why is it important to define a frame of reference brainly.com/question/526888

Answer Details:

Grade: High School

Subject: Physics

Chapter: Newton’s law of Motion

Keywords:

Weight, mass, acceleration due to gravity, gravity at the location, weighs 45 newtons, 15 kilogram, w=mg, g=3m/s^2, weight of the object.

If The dimensions of a brick that weighs 25N are 0.19m x 0.07m x 0.095m. What pressure does the brick exert on the ground if it is resting on its largest face?

Answers

1 newton per square meter = 1 pascal of pressure.

The largest face of the brick is (0.19 x 0.095) = .01805 square meter

25N/0.01805 = 1,385 newtons per square meter = 1,385 pascal = 1.385 kPa

Audition begins when sound waves strike the ________,causing it to vibrate. A) eardrum B) anvil C) stirrup D) cochlea

Answers

Answer: A) eardrum

Explanation: Audition refers to the sense of hearing. It is the process by which sound waves are perceived and interpreted by our brain. When sound waves enter the ear, they cause the eardrum to vibrate. These vibrations then travel further into the inner ear, leading to a chain of events that ultimately allows us to interpret and understand the sounds that we hear. Therefore, A is correct.

That's it!

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