All categories

3 years ago

Energy is the ability to do work. True or false

1 answer:

4 0

Work can be defined as the energy transferred from a body to its sorroundings, the energy spent to move a body, or the energy you need to alter a charged particle, so no energy, no work; thus, the statement is true.

You might be interested in

A 50 kg crate is pulled across the ice by a rope that is angled at 30 degrees above the horizontal. If the tension in the rope i

denpristay [2]

Let:
Vx = the pulling component of force
Vy = the lifting component of force

Vy:
Sin(n°) = Vy/hypotenuse
hypotenuse * Sin(n°) = Vy
100N*sin(30°) = Vy
50N = Vy

Vx:
Cos(n°) = Vx/hypotenuse
Hypotenuse * cos(n°) = Vx
100N*cos(30°) =Vx
about 86.6N = Vx

4 0

3 years ago

A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a constan

Fantom [35]

Answer:

(a). The emf induced in the loop is 0.005467 V.

(b). The direction of the current is clock wise direction.

Explanation:

Given that,

Circumference = 165.0 cm

Constant rate = 12.0 cm/s

Magnetic field = 0.500 T

Time = 9.0 s

We need to calculate the emf induced

Using formula of emf

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA\cos\theta)}{dt}$

$\epsilon=B\dfrac{dA}{dt}$ ...(I)

Area of the coil is

$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$

Put the value of area in equation (I)

$\epsilon=B\times2\pi r\dfrac{dr}{dt}$ ...(II)

The circumference of the coil is

$C=2\pi r$

$r = \dfrac{C}{2\pi}$

Put the value into the formula

$r=\dfrac{165.0}{2\pi}$

$r=26.26\ cm$

The rate of change of radius is

$\dfrac{dr}{dt}=\dfrac{12}{2\pi}$

$\dfrac{dr}{dt}=1.90\ cm/s$

The radius of the coil in 9 sec

$r=26.26-1.90\times9$

$r=9.16\ cm$

Put the value in the equation (I)

$\epsilon=-0.500\times2\pi\times9.16\times10^{-2}\times1.90\times10^{-2}$

$\epsilon=-0.005467\ V$

The magnitude of the emf

$|\epsilon|=0.005467\ V$

(b). Right hand rule :

According to right hand rule,

The thumb shows the direction of force, the index finger shows the direction of magnetic field and the middle finger shows the direction of current.

So, The direction of the current is clock wise direction.

Hence, (a). The emf induced in the loop is 0.005467 V.

(b). The direction of the current is clock wise direction.

8 0

3 years ago

A 2 kg object with 100 J potential energy (with respect to the ground) is released from rest from the top of a building, the spe

ioda

velocity of object just before hitting the ground = 10 m/s

10 m/sExplanation:

here's the solution : -

=》mechanical energy is always constant, when the object is at the top of the building , the object has potential energy but no kinetic energy,

so, at the top of the building the mechanical energy = potential energy = 100 J

but at when it is dropped from the building, the whole potential energy get converted into kinetic energy, just before reaching the bottom .

so, mechanical energy in the object just before hitting the ground = kinetic energy

( potential energy = 0 )

So, kinetic energy = 100 J

now, we know

$kinetic \: energy = \frac{1}{2} \times m {v}^{2}$

=》

$100 = \frac{1}{2} \times 2 \times {v}^{2}$

( since, mass = 2 kg )

=》

$100 = {v}^{2}$

=》

$v = \sqrt{100}$

=》

$v = 10$

so, velocity of object just before hitting the ground = 10 m/s

4 0

3 years ago

A garden hose has a radius of 0.0120 m, and water initially comes out at a speed of 2.88m/s. Dasha puts her thumb over the end ,

creativ13 [48]

Answer:

v = 12.4 [m/s]

Explanation:

With the speed and Area information, we can determine the volumetric flow.

$V=v*A\\A=\pi *r^{2}$

where:

r = radius = 0.0120 [m]

v = 2.88 [m/s]

$A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\$

Therefore the flow is:

$V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]$

Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.

$v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]$

3 0

3 years ago

A bicyclist is traveling at a speed of 5 m/s when it suddenly accelerates, at a constant rate of 0.6 m/s^2, for a time of 10s .

givi [52]

Answer:

11m/s

Explanation:

$v = v0 + a \times t$

where...

v is current velocity

v0 is starting velocity

a is constant acceleration

and t is time accelerating

plugging in given values gives us

v = 5+.6*10 = 5+6 = 11

therefore the velocity at the end of 10 seconds is 11m/s

4 0

3 years ago