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3 years ago

8

You turn on the faucet for your garden hose and let go immediately the end search flying around and spraying water everywhere wh

ich of newtons laws of motion does this motion represent

1 answer:

max2010maxim [7]3 years ago

7 0

It’s Newton’s third law

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A ball is projected into the air with 100 j of kinetic energy which is transformed to gravitational potential energy at the top

raketka [301]

<span>when it returns to its original level after encountering air resistance, its kinetic energy is decreased.
In fact, part of the energy has been dissipated due to the air resistance.

The mechanical energy of the ball as it starts the motion is:
</span> $E=K = 100 J$
<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>

3 0

3 years ago

To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv

Lesechka [4]

Answer:

E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

λ = Q / L

If we derive from the length we have

λ = dq/dx ⇒ dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

dE = k dq / x²2

dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

E = k $\int\limits^{d+L}_d {\lambda/x^{2}} \, dx$

We take out the constant magnitudes and perform the integral

E = k λ (-1/x) ${(-1/x)}^{d+L} _{d}$

Evaluating

E = k λ [ 1/d - 1/ (d+L)]

Using λ = Q/L

E = k Q/L [ 1/d - 1/ (d+L)]

let's use a bit of arithmetic to simplify the expression

[ 1/d - 1/ (d+L)] = L /[d(d+L)]

The final result is

E = k Q / [d(d+L)]

3 0

3 years ago

What quantity is the rate of change of velocity? Displacement Acceleration Final velocity

MariettaO [177]

Answer:

Acceleration

Explanation:

The quantity of the rate of change of velocity is termed the acceleration of the body.

Acceleration is the rate of change of velocity with time;

A = $\frac{v - u}{t}$

A is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

7 0

3 years ago

A 1451 kg car is traveling at 48.0 km/h. How much kinetic energy does it possess? K.E. =

jarptica [38.1K]

   Kinetic energy = (1/2) (mass) (speed)²

BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second. So we'll have to
make that conversion.

KE = (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²

   = (725.5) · (48 · 1000 · 1 / 3,600)² (kg) · (km·m·hr / hr·km·sec)²

   = (725.5) · ( 40/3 )² · ( kg·m² / sec²)

   = 128,978 joules (rounded)

8 0

3 years ago

Which of the following items involve a wedge? Check all that apply A. A scissors blade B. A corkscrew C. An axe D. A wheelchair

Oksanka [162]

Answer:

Axe and Scissor blade

Explanation:

6 0

3 years ago

Read 2 more answers