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2 years ago

Can a particle move in a direction of increasing electric potential, yet have its electric potential energy decrease? Explain

Physics

1 answer:

yan [13]2 years ago

3 0

Answer:

Explanation:

Yes , it is possible .

When a negative charge moves towards a positive charge , it is moving in the direction of increasing electrical potential . In the whole process , its electrical potential energy decreases and its kinetic energy increases .

Actually the potential energy of a negative charge near a positive charge is negative and it is inversely proportional to distance .

V = - Qq / R , When R decreases , the negative value of potential increases . That means potential energy decreases .

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A garden hose attached with a nozzle is used to fill a 10‐ gal bucket. The inner diameter of the hose is 2 cm, and it reduces to

nadya68 [22]

Answer:

Explanation:

Given

Volume of bucket $V=10\ gallon$

Time taken to fill the bucket $t=50\ s$

so volume flow rate is $\dot{V}=\frac{10}{50}=0.2\ gal/s$

1 gal is equivalent to $0.133\ ft^3$

$\dot{V}=0.0267\ ft^3/s$

mass flow rate $\dot{m}=\rho \times \dot{V}$

$\dot{m}=62.4\times 0.0267$

$\dot{m}=1.668\ lbs$

(b)Average velocity through nozzle exit

$\dot{V}=Av_{avg}$

$v_{avg}=\dfrac{0.0267}{\frac{\pi}{4}\times (0.0262)^2}$

$v_{avg}=49.51\ ft/s$

8 0

3 years ago

A specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s r

katrin2010 [14]

Answer:

There's a decrease in width of 2.18 × 10^(-6) m

Explanation:

We are given;

Shear Modulus;E = 207 GPa = 207 × 10^(9) N/m²

Force;F = 60000 N.

Poisson’s ratio; υ =0.30

We are told width is 20 mm and thickness 40 mm.

Thus;

Area = 20 × 10^(-3) × 40 × 10^(-3)

Area = 8 × 10^(-4) m²

Now formula for shear modulus is;

E = σ/ε_z

Where σ is stress given by the formula Force(F)/Area(A)

While ε_z is longitudinal strain.

Thus;

E = (F/A)/ε_z

ε_z = (F/A)/E

ε_z = (60,000/(8 × 10^(-4)))/(207 × 10^(9))

ε_z = 3.62 × 10^(-4)

Now, formula for lateral strain is;

ε_x = - υ × ε_z

ε_x = -0.3 × 3.62 × 10^(-4)

ε_x = -1.09 × 10^(-4)

Now, change in width is given by;

Δw = w_o × ε_x

Where w_o is initial width = 20 × 10^(-3) m

So; Δw = 20 × 10^(-3) × -1.09 × 10^(-4)

Δw = -2.18 × 10^(-6) m

Negative means the width decreased.

So there's a decrease in width of 2.18 × 10^(-6) m

6 0

3 years ago

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

vitfil [10]

The acceleration of the sled will be 1.30 m/s². Force is defined as the product of mass and acceleration.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Given data;

m(mass of sled)=8 kg

Θ is the inclination of force= 50°

Force of friction,f=2.4 N.

The applied force at the given angle is resolved into the two-component as;

$\rm F_h=F cos \theta \\\\ F_h= 20 cos 50 ^0 \\\\ F_h= 12.85 \ N$