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2 years ago

An object elongates from a length of 45 cm to a length of 55 cm. The percent strain is

Physics

1 answer:

suter [353]2 years ago

5 0

Answer:

22%

Explanation:

55 - 45 = 10

10/45 simplifies to 2/9

2/9 = 0.22222... so 22.22% (rounded 22%)

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A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec

miss Akunina [59]

The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
$W=F d cos \theta$
where the force is F=qE, d=0.556 and $\theta=55.2^{\circ}$ . Using the value of q and E given by the problem, we find
$W=qEdcos\theta = 6.39\cdot10^{-5}J$

3 0

3 years ago

PLEASE ANSWER ASAP!!!!!!!

Viefleur [7K]

Answer:

B. Mechanical energy= 50J+30J=80J

4 0

3 years ago

Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m

Sever21 [200]

Answer:

U₂ = 20 J

KE₂ = 40 J

v= 12.64 m/s

Explanation:

Given that

H= 12 m

m = 0.5 kg

h= 4 m

The potential energy at position 1

U₁ = m g H

U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

The kinetic energy at position 2

KE= 1/2 m V²

From energy conservation

U₁+KE₁=U₂+KE₂

By putting the values

60 - 20 = KE₂

KE₂ = 40 J

lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

160 = v²

v= 12.64 m/s

3 0

2 years ago

Read 2 more answers

C-14 is an isotope of the element carbon. How does it differ from the carbon atom seen here? A) C-14 has two more protons. B) C-

mylen [45]

Answer:

C-14 has two more neutrons.

Explanation:

4 0

3 years ago

Read 2 more answers

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are

igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

$\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2$

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

$I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2$

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

$T = I\alpha = 0.303*0.6454 = 0.196 Nm$

So the force acting on the other end to generate this torque mush be:

$F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N$

4 0

3 years ago