If your bedroom is cold, you might feel warmer with several thin blankets than with one thick one

Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction betw

jarptica [38.1K]

Answer:

a) W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d) the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

θ = 90 cos 90 = 0

W=0

In conclusion the work is zero

b) The friction force opposes the displacement whereby the angle is

θ = 180 cos 190 = -1

W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

m g sin θ L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀= + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

ΔU = -U₀

the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

W_Total = W_weight - W_roce

as the body moves down

W_Total> 0

Therefore the total work is positive

3 0

2 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude

Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

$a=\frac{1.2^2}{0.400}=3.6 m/s^2$

7 0

3 years ago

Why does a magnet attract a steel nail

SVEN [57.7K]

Answer:

The nail will stick to the bar magnet because it will become magnetized, and it's metal. The presence of the nearby north pole rearranges the magnetic domains inside the steel so that their south poles all point toward the north pole of the permanent magnet. As a result, the other end of the nail becomes a north pole.Magnets attract iron due to the influence of their magnetic field upon the iron. ... When exposed to the magnetic field, the atoms begin to align their electrons with the flow of the magnetic field, which makes the iron magnetized as well. This, in turn, creates an attraction between the two magnetized objects.

7 0

2 years ago

An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if

maxonik [38]

The electron is accelerated through a potential difference of $\Delta V=780 V$ , so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
$\frac{1}{2}mv^2 = e \Delta V$
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
$\Delta V$ is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
$v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s$

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
$evB=m \frac{v^2}{r}$
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
$B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T$

3 0

3 years ago

A kid on a trampoline has 1,000 J of potential energy when they are at the top of a jump. How much kinetic energy will the kid h

emmasim [6.3K]

C: 500 J
Hope this helped!

7 0

2 years ago